Problem 241 Answer

Place ten cards in one pile with the other 42 cards in the other. Then, flip over the ten-card pile.

Problem 241 Solution

Place ten cards in one pile with the other 42 cards in the other. Let's say the 10-card pile has x face up cards and 10-x face down cards. The 42-card pile will have 10-x face up cards.

Then flip over the 10-card pile. It will now have 10-x face up cards, the same as the 42-card pile.

Problem 113 Answer

Problem 113 Solution

Q = The sum for i=1 to 10 of ( (N_i - np_i)²/np_i )

= The sum for i=1 to 10 of ( (N_i - 1000)²/1000 )

= (676+121+16+64+324+64+6241+64+16+1024)/1000 = 8.61

According the the chi-squared table with 9 degrees of freedom a Q statistic of greater than 16.92 shold fail the test. 8.61 is well under 16.92 so it should pass the test. In other words the digits do appear to be random.

At a level of significance such that 10,000 random digits passed 60% of the time the 10,000 digits of e would still pass, but at a 50% level they would not.

Problem 190 Answer

First, move any 20 pennies to the other table, keeping the same side on each penny up.

Second, flip all 20 pennies on the other table.

Explanation

Initially table 1 will have 80 heads and 20 tails, and table 2 will have 0 heads and 0 tails.

Let t be the number tails moved, and h be the number of heads.

After the move table 1 will have 80-h heads and 20-t tails, and table 2 will have h heads and t tails.

After the flip table 1 will have 80-h heads and 20-t tails, and table 2 will have t heads and h tails.

So, at this point table 1 has 20-t tails and table 2 has h tails. We moved 20 coins: h heads, and t tails, so h+t=20, or h=20-t. 20-t and h must be the same amount, thus both tables have the same number of tails, although we don't know what that number is.

Thanks to Dan Florentin for this problem.

Problem 26 Answer

Problem 26 Solution

The probability of winning is sum for i=x+1 to 100 of the probability that the highest dowry is in the ith position multiplied by the probability that you choose it if it is in that position.

This is based on the condition probability formula: Pr(A)=Sum for i=1 to n of Pr( A | B_i ) * Pr( B_i ), where the sum for i=1 to n of Pr(B_i)=1.

For all i the probability that the highest dowry is in that position is 1/100.

The probability that you will choose the dowry if it is in the ith position is equal to the probability that the highest of the first i-1 dowries belongs to one of the first x ladies. This equals x/(i-1). If the highest dowry of the first i-1 were not in the first x ladies you would choose it before getting to the largest dowry, thus losing the game.

For example if you let 30 ladies pass the probability that the highest dowry is in the 75th postion and that you will choose it equals 1/100 * 30/74.

Let A denote the overall probability of winning. The proability of winning given x is the sum for i=x+1 to 100 of x/(i-1) * 1/n. This equals:

Pr(1/n * [ 1 + x/(x+1) + x/(x+2) + ... + x/99 ] ).

Some value of x must be optimal to maximize this formula. This value of x must the first such that Pr(A|x+1) - Pr(A|x) < 0.

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(x+2) + 1/(x+3) + 1/(x+4) + ... + 1/99 - x/(x+1) ].

let y=x+1.

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - (y-1)/y ]. =

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 + 1/y ]. =

Pr(A|x+1) - Pr(A|x) = 1/n * [ 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 - 1 ].

If Pr(A|x+1) - Pr(A|x) = 0, then 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 = 1.

Theorem: The integral from r to n of 1/x = log(n) - log(r) = log(n/r).

The equation above the theorm is an approximation of the integral in the theorem. By applying the theorem log(100/y) = 1.

Taking e to the power of both sides: 100/y = e.

y=100/e.

Since e=~2.7182818 y=~36.79, since x=y-1 x=~35.79.

Since the integral is an underestimate of the series look at some specific values of Pr(A) for values of x close to 35.79:

 x   A
--- ---
35  37.0709%
36  37.1015%
37  37.1043%
38  37.0801%

In general the answer is going to be n/e, where n is the total number of ladies. Because the answer must be an exact integer and error in applying the theorem mentioned above you should check the few integers just above the optimal value. As n increases the error decreases. The probability of winning turns out to approach 1/e =~ 36.79% as n approaches infinity (I'll leave this proof up to you).

Here are some optimal values of x for various values of n and the probability of winning given x:

  n   x   Pr(winning)
---- ---  -----------
   3   2   50.00%
   4   2   43.83%
   5   3   43.33%
   6   3   42.78%
   7   3   41.43%
   8   4   40.98%
   9   4   40.60%
  10   4   39.87%
  15   6   38.94%
  20   8   38.42%
  30  12   37.87%
  50  19   37.42%
 100  38   37.10%
1000 369   36.82%

Here another version of my solution to this problem

Let x be the number of ladies you pass before choosing the next lady with a dowry greater than the first x ladies. Let max_x be the maximum dowry in the first x ladies.

The probability you will select the highest dowry is:

Sum for i=1 to 100-x of: ( Probability that i of the remaining ladies will have a dowry greater than max_x ) * (1/i).

This probability for any i is:

(100-x)*(99-x)*(98-x)*...*(101-x-i) * x
---------------------------------------
100*99*98*...*(100-i)

Pr(n₁) * 1 +
Pr(n₂) * 1/2 +
Pr(n₃) * 1/3 +
Pr(n₄) * 1/4 +
Pr(n₅) * 1/5 +
.
.
.
Pr(n₇₀) * 1/70.

To derive Pr(n_i) consider the probability that the i highest dowrys fall in the last 100-x ladies, multiplied the the probability that the (i+1)th highest dowry falls in the first x ladies. For example:

Pr(n₃) = 70/100 * 69/99 * 68/98 * 30/97.

70/100 = Probability that the highest dowry is not in first 30.
69/99 = Probability that the second highest dowry is not in first 30 (99 positions left, 69 not if first 30).
68/98 = Probability that the third highest dowry is not in first 30.
30/97 = Probability that fourth highest dowry is in first 30.

Personally I had to use a spreadsheet to find that the maximum probability was for x=37, which was 37.1%.

If you change the number of total ladies from 100 to any other number the strategy seems to still be to let 37% of them go by first.

Problem 84 Answer

First weigh any 4 pearls against any other 4.

If one side weighs heavier than the other then:

Label the pearls on the heavy side H, the pearls on the light side L, and the 4 other pearls G. Next, weigh two H pearls and two L pearls against 2 G pearls, 1 H pearl, and 1 L pearl (HHLL vs GGHL).

If the HHLL side goes down then either one of the two H pearls on the left side from step 2 is heavy or the L pearl on the right side of step 2 is light. Finally weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up the L pearl is light, if they stay the same the other H pearl is heavy.

If the HHLL side (from step 2) goes up then either one of the two L pearls on the left side from step 2 is light or the H pearl on the right side of step 2 is heavy. Finally weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up the L pearl is light, if they stay the same the other L pearl is light.

If the two sides from step 2 stay the same you have left one H pearl and one L pearl. Weigh ( HL vs GG ). If the left side goes down the H pearl is heavy, if it goes up the L pearl is light.

If the two sides from step one stay the same then label the 8 pearls used in step 1 as G and the 4 others pearls as B. For the second weighing weigh 3 B pearls against 3 G pearls ( BBB vs GGG).

If the left side goes down one of the B pearls is heavy. Weigh them 2 of them against each other the heavy side will have the heavy pearl, if the sides stay the same the other pearl is heavy.

If the left side goes up one of the B pearls is light. Weigh them 2 of them against each other the light side will have the light pearl, if the sides stay the same the other pearl is light.

If the two sides from the second weighing stay the same then the last B pearl is heavy or light. Use the third weighing against a good pearl to determine if it is heavy or light.

Problem 80 Answer

Problem 118 Answer

Problem 118 Solution

Next put 1716 locks on the safe, one for each way to group 7 pirates. For each lock give 7 keys to a unique group of 7 pirates. This way any given lock will have a keyholder in any group of 7 or more. For any group of 6 there will be exactly one lock in which the other 7 pirates have the key. Obviously any group of less than 6 would also be missing at least one key to at least one lock.

Here are the number of keys required for other numbers of pirates:

Number 
  of      
Pirates   Number of Locks
------- -------------------
3	                  3
5	                 10
7	                 35
9	                126
11	                462
13	              1,716
15	              6,435
17	             24,310
19	             92,378
21	            352,716
23	          1,352,078
25	          5,200,300
27	         20,058,300
29	         77,558,760
31	        300,540,195
33	      1,166,803,110
35	      4,537,567,650
37	     17,672,631,900
39	     68,923,264,410
41	    269,128,937,220
43	  1,052,049,481,860
45	  4,116,715,363,800
47	 16,123,801,841,550
49	 63,205,303,218,876
51	247,959,266,474,050
53	973,469,712,824,060

Problem 204 Answer

Question

On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.

Answer

The answer is P1 + P2 + P3 + P4, where

[ P1 ]                 [12   3   4   5]^153 [1] 
[ P2 ]                 [ 6  27   0   0]     [0]
[ P3 ] = (1/36)^153 ×  [ 8   0  26   0] ×   [0]
[ P4 ]                 [10   0   0  25]     [0]     

P1 = 3.12763 × 10^-11
P2 = 4.63460 × 10^-11
P3 = 4.95558 × 10^-11
P4 = 5.17044 × 10^-11

The sum of these probabilities is 1.78882 × 10^-10 = 1 in 5,590,264,072

Acknowledgement: My thanks to BruceZ for this help with this problem.

Problem 204 Solution

Question

On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.

Answer

Solution

There are four possible states the shooter can be in. Let's define them as follows.

State 1 = Come out roll
State 2 = Point of 4 or 10
State 3 = Point of 5 or 9
State 4 = Point of 6 or 8

The probabilities are recursive. Let p(x,r) represent the probability of being in state x before roll r. Based on simple dice probabilities, p(x,r) can be expressed as follows:

(1) P(1,r) = (12/36)×p(1,r-1) + (3/36)×p(2,r-1) + (4/36)×p(3,r-1) + (5/36)× p(4,r-1)
(2) P(2,r) = (6/36)×p(1,r-1) + (27/36)×p(2,r-1)
(3) P(3,r) = (8/36)×p(1,r-1) + (26/36)×p(3,r-1)
(4) P(4,r) = (10/36)×p(1,r-1) + (25/36)×p(4,r-1)

In more plain simple English, equation (1) is saying that the probability of being in a come out roll before roll r is the sum of the following:

1. Product of the probability of being in a come out roll the previous turn, and the probability of staying in a come out roll (12/36).
2. Product of the probability of rolling for a 4 or 10 the previous roll, and the probability of making the point (3/36), resulting in a come out roll.
3. Product of the probability of rolling for a 5 or 9 the previous roll, and the probability of making the point (4/36), resulting in a come out roll.
4. Product of the probability of rolling for a 6 or 8 the previous roll, and the probability of making the point (5/36), resulting in a come out roll.

Equation (2) is saying the probability of rolling for a 4 or 10 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 4 or 10 (6/36).
2. Product of the probability of rolling for a point of 4 or 10 the previous turn, and the probability of not rolling the desired point or a 7 (27/36).

Equation (3) is saying the probability of rolling for a 5 or 9 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 5 or 9 (8/36).
2. Product of the probability of rolling for a point of 5 or 9 the previous turn, and the probability of not rolling the desired point or a 7 (26/36).

Equation (4) is saying the probability of rolling for a 6 or 8 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 6 or 8 (10/36).
2. Product of the probability of rolling for a point of 6 or 8 the previous turn, and the probability of not rolling the desired point or a 7 (25/36).

Next, let's express the these four equations in the form of a matrix.

[ P(1,r) ]             [12   3   4   5]   [ P(1,r-1) ]
[ P(2,r) ]             [ 6  27   0   0]   [ P(2,r-1) ]
[ P(3,r) ] = (1/36) ×  [ 8   0  26   0] × [ P(3,r-1) ]
[ P(4,r) ]             [10   0   0  25]   [ P(4,r-1) ]

The intial roll is always a come out roll, so the initial state is:

[ 1 ]
[ 0 ]
[ 0 ]
[ 0 ]

The probability of each state before the 154th roll is expressed as follows.  The reason we take the matrix to the 153rd power, and not the 154th, is there are 153 transformations since the initial state.

[ P(1,154) ]                 [12   3   4   5]^153 [1] 
[ P(2,154) ]                 [ 6  27   0   0]     [0]
[ P(3,154) ] = (1/36)^153 ×  [ 8   0  26   0] ×   [0]
[ P(4,154) ]                 [10   0   0  25]     [0]
     

The exact answer is: 

P(1,154) = 3.12763 × 10-11
P(2,154) = 4.63460 × 10-11
P(3,154) = 4.95558 × 10-11
P(4,154) = 5.17044 × 10-11

The sum of these probabilities is 1.78882 × 10-10 = 1 in 5,590,264,072 

Problem 132 Answer

The following are the lengths of the various sides, according to the lables in the image:

a=50
b=35
c=27
d=8
e=19
f=15
g=17
h=11
i=6
j=24
k=29
l=25
m=9
n=2
o=7
p=18
q=16
r=42
s=4
t=37
u=33

Thanks to Terry W. Ryder of San Leandro, CA and New Scientist magazine for the problem and Terry for the graphic.

This problem inspired my friend Dick Tucker to not only solve the problem but create a piece of art with it using acrylic plexiglas.

Problem 132 Solution

Cosinder all the ways you can define the length of an edge of the outer square, where each variable represents the length of the side or the corresponding square in the image:

1. a+b+c
2. a+b+d+e
3. a+f+g+h+e
4. a+f+g+i+j
5. k+l+m+n+g+h+e
6. k+l+m+n+g+i+j
7. k+l+m+o+p+j
8. k+l+q+p+j
9. k+l+q+r
10. k+s+t+r
11. u+t+v
12. a+k+u
13. a+l+s+u
14. a+l+t
15. b+f+l+s+u
16. b+f+l+t
17. b+f+m+q+t
18. b+f+n+o+q+t
19. b+g+o+q+t
20. b+g+q+r
21. b+h+i+p+r
22. b+h+j+r
23. c+d+h+i+p+r
24. c+d+h+j+r
25. c+e+j+r

Then use Cramer's Rule to solve the equations. In Excel the MDETERM(a1:u21) command will give you the determinant of the matrix bounded by a1 and u21, making this process quick and mostly automated. If the determinant used for the denominator is zero then choose a different set of 21 equations.

When solving for a..u ignore the deonominator determinant since it will be the same for all 21 variables. After solving for all 21 looking for common factors among the solutions and divide.

Thanks to Terry W. Ryder of San Leandro, CA and New Scientist magazine for the problem and Terry for the graphic.

Problem 219 Answer

There are various solutions to the problem but most are variants of the one below. I believe my answer will get the prisoners released in the least number of expected days.

1. The prisoner called to the switch room on the first day will be designated the leader. If switch A is in the up position, then he will flip in the down position. If it is already down, he will flip switch 2 (the dummy switch). He will start his running count of visits to 1.
2. Starting with the second day, do the following:
   • If a non-leader is called, and switch A is down position, and he has never fipped switch A, then he will flip switch A to the up position. This is the way to announce that one more person has visited the switch room.
   • Otherwise, if a non-leader is called into the switch room, and any of the conditions are not met, then he should flip switch B.
   • If the leader is called back into the room and switch A is in the up position it means a new prisoner has entered. The leader will flip A down and add one to his running count.
   • If the leader is called back into the room and switch A is in the down position then it will mean no new prisoner has entered the room since the last time he was there. He will flip switch B.
3. When the leader reaches a running count of 23, he can safely declare that everyone has been to the switch room at least once.

This method requires an average of 22*23 + sum for i=1 to 22 of 23/(23-i), which equals 591.8887 days.

Problem 221 Answer

The answer is 7.

Problem 221 Solution

1. Divide the 25 horses into five groups of five.
2. Races 1-5: Race each group and note the top three horses, in order. Call these the heats.
3. Race 6: Race the winner of each race from step 2. The winner of this race will be the fastest horse.
4. Race 7: Race the the following five horses:
   
   1. Second and third place horses from race 6.
   2. The two horses from the group of five second place horses from step 1 that finished second in its heat to the winning two horses in race 6.
   3. The horse from the third place group that finished third in its heat to the winning horse in race 6.
   
   

There is a good explanation at A Collection of Quant Riddles With Answers of why this solution works.

Problem 239 Answer

The first should go either a smidge to the left of the point 1/4 from the left end of the beach and a smidge to the right of the point 3/4 from the right end.

The second vendor will pick the spot from A's two choices that A didn't pick.

The third vendor will go in the middle.

This will give A and B 3/8 of the beach each, and C 1/4.

Problem 239 Solution

It stands to reason that A and B will want C to go in the middle, least he be the closest to an edge and cut them off from that section of the beach. How far can they extend away from an edge and still keep C in the middle?

Let's call x the distance from either edge that would put C indifferent between going right next to A and B but closer to an edge and the middle.

At x, assuming C goes in the middle, A will get x + (0.5 - x)/2 = x/2 + 1/4 of the beach. Same with B. C will bet (1-2x)/2 = 1/2 - x of the beach.

If C located right next to A or B, but closer to an edge, he would get territory of length x. So, we want to equate x and (1/2 - x) to find that indifference point for C.

x = 1/2 - x
2x = 1/2
x = 1/4

To make sure C isn't indifferent, and barely prefers the middle, A and B should go just a smidge closer to the edges than the 1/4 and 3/4 points. This will give A and B a smidge less then 3/8 of the beach each and C a smidge more than 1/4.

Problem 240 Answer

A will set is stand a smidge less than 1/6 of a mile from either end.

B will do the same as A, but at the opposite end.

C will set his stand exactly in the middle of the beach.

D will be indifferent between placing his stand between either (1) A and C and (2) B and C.

Problem 240 Solution

A will want to pick a spot close to either edge, maximizing his space without giving another vendor the incentive to put his stand right next to his on the side close to the edge. It is good to be close to an edge, because you get every customer between you and that edge. Let's just say he picks the left side of the beach.

B will want to do the same, but on the right side.

C will pick a spot in the middle, putting D to an indifference point which side of C to go on.

D will arbitrarily pick a spot between A and C or B and C.

The question is how far from the edge should A and B pick?

We want D to pick on either side of C. Let's ask the question at what point would D be indifferent between going just to the left of A, between A and C, and between B and C.

If A sets his stand at point x, then D would get x miles just to the left of A or (1/2-x)/2 by going between A and C or B and C. Let's set these two equations equal to each other and solve, to find the indifference point for x.

x = (0.5-x)/2
2x = 0.5 - x
3x = 0.5 x = 1/6

However, we don't want D to be indifferent, we want him to pick a spot on either side of C. So, we pick a spot a smidge to the left of 1/6, to get D to go next to C.

Same logic for B -- a smidge to the right of 5/6.

C will want to make D indifferent to going on either side of him, which is obviously 1/2.

D will then arbitrarily picks between 1/3 and 2/3.

A will then get 1/4 if D picks 3/8 and 1/3 if he picks 5/8. The average of those two is 7/24.

B is in the same situation as A, so 7/24 on average to him too.

Let's say D flips a coin and goes in the 1/3 spot. Then C will get everything from the 5/12 to the 2/3 point, which is 1/4 mile.

With D and the 1/3 spot, he will everything from the 1/4 to the 5/12 points, which is 1/6 mile.

Problem 165 Answers

1. At time t=0 light fuse 1 at both ends and fuses 2, 3, 4, and 5 at one end. Fuse 1 will have 32 minutes remaining, fuses 2-5 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 3-5 will have 32 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4,5 will have 16 minutes remaining.
4. At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end. Fuse 4 will have 4 minutes remaining, fuse 5 will have 8 minutes remaining.
5. At time t=60 fuse 4 will have burned out. At this moment start fuse 6. Fuse 5 will have 4 minutes remaining, fuse 6 will have 64 minutes remaining.
6. At time t=64 fuse 5 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining when relit.

Next is another solution submitted by Michael Nolan.

Solution 2

1. At time t=0 light fuse 1 at both ends and fuse 2 at one end. Fuse 1 will have 32 minutes remaining and fuse 2 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light the other end of fuse 2 and fuse 3 at one end. Fuse 2 will have 16 minutes remaining and fuse 3 will have 64 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end and light fuse 4 at one end. Fuse 3 will have 24 minutes remaining and fuse 4 will have 64 minutes remaining.
4. At time t=72 fuse 3 will have burned out. At this moment light fuse 5 from both ends. Fuse 4 will have 40 minutes remaining and fuse 5 will have 32 minutes remaining.
5. At time t=104 fuse 5 will have burned out. At this moment light fuse 4 from the other end and light fuse 6 from one end. Fuse 4 will have 4 minutes remaining and fuse 6 will have 64 minutes remaining.
6. At time t=108 fuse 4 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining.

Next is another solution submitted by Michael Nolan and Andy 'Deuce' Smith.

Solution 3

1. Light #1 at both ends, and #2 and #3 at one end. Unlight #2 and #3 when #1 burns out. This makes #2 and #3 both 32-minute fuses.
2. Light #2 at both ends, and #4 at one end. Unlight #4 when #2 burns out. This makes #4 a 48-minute fuse.
3. Light #4 at both ends, and #3 at one end. Unlight #3 when #4 burns out. This makes #3 an 8-minute fuse.
4. Light #3 at both ends, and #5 at one end. Unlight #5 when #3 burns out. This makes #5 a 60-minute fuse.

Solution 4

1. At time t=0 light fuse 1 at both ends and fuses 2, 3, and 4 at both ends. Fuse 1 will have 32 minutes remaining, fuses 2-4 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 3,4 will have 32 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4 will have 16 minutes remaining.
4. At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end and fuse 5 at one end. Fuse 4 will have 4 minutes remaining and fuse 5 will have 64 minutes remaining.
5. At time t=60 fuse 4 will have burned out. At this moment snuff out fuse 5. Fuse 5 will have 60 minutes remaining when relit.

Credit and thanks for this problem goes to Craig Olson, Alan Goldberg, and the Nov/Dec 1999 Contingencies magazine.

Problem 97 Answer

Problem 97 Solution

Since the radius of the lighthouse is 1 the length of the straight part of the chain is the same as x. Through a little trigonomtry we can get the coordinates of the dog given x as (cos(x)+x*sin(x),sin(x)-x*cos(x)). The distance from (0,0) to the dog, given x, conveniently works out to (1+x²)^1/2, using the length of a line segment formula.

Lets call y the angle fromed by (1,0), (0,0), and the location of the dog. Through some more trigonomty (again left up to you) we can solve for y in terms of x as y=x-tan^-1. Now we are ready to integrate.

The area in terms of polar coordinates is 1/2 * the integral over the range of the angle of the radius squared. We can not simply take the integral from 0 to Π of 1+x² because this is not the radius formed by the angle x, but the radius formed by y. We could take the integral over the range of the angle y but finding the coordinates of the dog in terms of y is very hard (I couldn't do it). However we can take the ingregal as x goes from 0 to Π of 1+x² multiplied by the change in y given a change a change in x. Since y=x-tan^-1(x). dy/dx = 1-1/(1+x²)!

So now the area becomes the integral as x goes from 0 to Π of 1/2 the product of 1+x² and 1-1/(1+x²) which equals 1/2 the integral from 0 to Π of x² which equals Π³/6.

This integral, however, does not cover the area bounded by the triangle (0,0), (-1,0) and (-1,Π). So we must add this area which is Π/2. Yet we must subtract the area inside the lighthouse which the dog may not go in, which is also Π/2. So these two modifications conventiently cancel each other out.

The area of the quarter circle to the left of x=-1 and above y=0 is Π³/4, since the length of the chain is Π and the dog covers a semicircle to the left of the lighthouse. So the total area above y=0 is 5/12 * Π³. Finally double this for the area under y=0 and the total area is 5/6 * Π³

Here is another solution I received from Patrick.

I would like to thank Steven Lutz for sending me this problem.

Problem 32 Answer

The shadow over your yard is (540^2/3-36)^1/2 =~ 5.505698 feet.

The angle formed is approximately 47.46 degrees.

Problem 32 Solution

Let y be the shadow over neighbor's yard.

By similar triangles: x/(36+x²)^1/2=(x+y)/15.

y=15x/(36+x²)^1/2-x.

y^'=(((15*(36+x²)^1/2)-(15x²*(36+x²)^-1/2)) / 36+x²) - 1.

Then set y^'=0 and solve for x.

Finally subsitute x in the similar triangles formula to solve for y.

Problem 150 Answer

Problem 150 Solution

Next consider E. E*4 must end in the digit 2. The only numbers that works for are 3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number that satisifies both conditions.

Next consider B. We already know that 2BCD8*4 is at least 80000 and less than 90000. B can not be more than 2 because then 2BCD8 * 4 would be more than 80000. 2 is already taken so B must be 0 or 1. Lets consider the case that B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies this condition. So B must be 1.

Next consider D. D8*4 must end in the digits 12. The only possiblity is D=7 (78*4=312).

Now solve for C:

21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300=9.

So ABCDE=21978.

I'd like to thank "Marisa" for this problem.

Problem 179 Answer

Problem 179 Solution

Let a be the acceleration.

Let t be the total time traveled.

Basic calculus tells us that the velocity at time t is a*t and the total distance traveled is a*t²/2.

So we know:

(1) 1=a*t²/2 and

(2) 0.75 = a*(t-3)²/2

Solving both sides for a we have :

(3) 2/t² = 3/(2*(t-3)²)

4*(t-3)² = 3*t²

t² - 24t + 36 = 0

t = (24 +/- 432^1/2)/2 (11) t = 6*3^1/2 + 12

Add 3 for the total distance of 12 + 6*3^1/2 =~ 22.39230485

Problem 27 Answer

Problem 27 Solution

Consider the two ladders two lines on a graph. Let the shorter ladder extend from (0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^1/2). The square root of 108 can be found using the pythagorean formula. Then solve for the slope and y intercept to find the equations of the two lines:

The shorter ladder have the equation y=(4/3)*x.

The longer ladder has the equation y=(-108^1/2/6)*x + 108^1/2.

The lines meet where the ladders cross.

Use substitution to solve for y.

Acknowledgement: Thanks to Steve Lutz for suggesting this method of solution.

Scott R. Walshon pointed out that if x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.

To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders. From similar triangles we get:

p/x = a/(a+b)
p/y = b/(a+b)

Add the two equations:

p/x + p/y = a/(a+b) + b/(a+b)
py/xy + px/xy = (a+b)/(a+b)
p(x+y)/xy = 1
p = xy/x+y
1/p = x+y/xy
1/p = x/xy + y/xy
1/p = 1/y + 1/x

Thanks Scott for this observation.

Problem 152 Answer

Visualize the box folded out. The spider has to crawl 22"+1"+1"=24" vertically and 4.5"+9"+4.5"=18" horizontally. The total distance traveled is (24"² + 18"²)^1/2 =
(576" + 324")^1/2 =
900"^1/2 =
30".

Thanks to Anil Rhemtulla for this problem.

>

Problem 4 Answer

If the spider started at a corner diagonally on the same face as the ant the answer would be 9, and if the spider started at an adjacent corner the answer would be 7.

Here are answers for other figures:

Square: 4
Octahedron: 6
Dodecahedron: 35
Icosahedon: 15

Problem 206 Answer

Question

Imagine an infinitely elastic rubber band that is 1 km. long unstretched. It expands at a rate of 1 km. per second.

Imagine an ant at one end of the rubber band. At the moment the rubber band starts expanding the ant crawls towards the other end at a speed, relative to his current position, of 1 cm. per second.

Will the ant ever reach the other end? If so, when?

Answer

Yes. The ant will reach the other end after e^100,000 - 1 seconds.

Problem 206 Solution

Question

Imagine an infinitely elastic rubber band that is 1 km. long unstretched. It expands at a rate of 1 km. per second.

Imagine an ant at one end of the rubber band. At the moment the rubber band starts expanding the ant crawls towards the other end at a speed, relative to his current position, of 1 cm. per second.

Will the ant ever reach the other end? If so, when?

Solution #1

Let f(t)=distance traveled from origin at time t.
So f'(t)=speed of ant at time t.
f'(t)= 1 + 100,000×ratio of progress.
The ratio of progress is the distance the ant has covered divided by the length of the rubber band = f(t)/(100,000*(1+t))
So f'(t) = 1 + 100,000 * f(t)/(100,000*(1+t))
f'(t) = 1 + f(t)/(1+t)

So, for what f(t) is this true? Here is where you pretty much need to have some linear differential equations in your back pocket. What works is f(t)=(1+t) × ln(1+t)

Let g(t)=length of rubber band time t.
g(t)=100,000×(1+t)
The question is at what time t is f(t)=g(t)?

(1+t) × ln(1+t) = 100,000×(1+t)
ln(1+t) = 100,000
1+t = e^100,000
t = e^100,000-1

My thanks to Doc for solution #1.

Solution #2

The ratio of the rubber band the ant covers at time t is 1/[100000×(1+t)]. To help visualize this, think of the ant's position as your relative point of reference, and both ends of the rubber band moving away from the ant. So the ratio of progress is the ant's speed of 1 cm/sec compared to the total length of the rubber band at time t of 100,000×(t+t).

The question is, at what time T will the sum of the progress be 1?

Integral from 0 to T of 1/[100000×(1+t)] dt = 1
10^-5×ln(1+t) from 0 to T = 1
10^-5×(ln(1+T)-ln(0)) = 1
10^-5×ln(T) = 1
ln(1+T)=100,000
1+T=e^100,000
T=e^100,000-1

My thanks to PapaChubby for solution #1.

General Answer

Let: a = ant's speed.
b = rubber band's expansion speed.
c = initial length of rubber band.

Then the ant will get to the end in (c/b)×(e^b/a-1) units of time.

Problem 195 Answer

Problem 195 Solution

In problem 194, we see the solution for the 2-number case is 1/3. The following shows the solution for the 3-number case is 1/4, and for the 4-number case is 1/5. Following the pattern, the solution for the n-number case is 1/(n+1).

If it isn't clear from the scan below, first I find the area of integration, and then multiply the expected average by the inverse of that area. In the 3-number case, x is the largest number, and ranges from 0 to 1. y is the second largest, and ranges from 0 to x. z is the smallest and ranges from 0 to y. Same concept in the 4-number case, but we start with w as the largest number.

Problem 194 Answer

Problem 194 Solution

See the following scan. The reason for multiplying by 2, is the area of integration is 1/2, as shown by the shaded area. x is the larger of the two numbers, which can range from 0 to 1. y is the smaller number, which can range from 0 to x.

Problem 56 Answer

Problem 56 Solution

Let events A₁,...,A_k form a partition of the space S such that Pr(A_j)>0 for j=1,...,k, and let B be any event such that Pr(B)>0. Then for i=1,...,k,

Pr(A_i|B) = Pr(A_i)*Pr(B|A_i) / Sum for j=1 to k of Pr(A_j)*Pr(B|A_j).

The source of this problem is the May 1997 issue of the Actuarial Review. Applying this theorm:

B=boy selected
A₁=boy added
A₂=girl added
g=number of girls before baby is added.

Pr(A₂|B) = (1/2)*(2/(3+g)) / [(1/2)*(2/(3+g)) + (1/2)*(3/(3+g))] = 0.4

Problem 24 Answer

Problem 24 Solution

dc = 100-c dt

dc 1/(100-c) = dt

-ln(100-c) = t + K₁

t = ln(K₂) - ln(100-c) (where K₁ = ln(K₂)

t = ln(K₂/(100-c))

When t=0 c=15 so K₂ must be 85

t = ln(85/(100-c))

Solving for t when c=25:

t = ln(85/(100-25)) = ln(85/75) =~ ln(17/15) = 0.1252

Problem 19 Answer

Problem 19 Solution

After the first turn you will have a configuration like 1,1,2,3. Call this configuration 1.

From configuration 1 if you draw a 1 on the first ball you will end up with the same or similar configuration. The probability of this happening is 1/2.

From configuration 1 if you draw a 3 or 4 first and then a 1 you will have a configuration like 1,1,1,2. The probability of this happening is 1/3. Call this new pattern configuration 2.

From configuration 1 if you draw a 3 or 4 first and then draw the other non 1 you will have a configuration like 1,1,2,2. The probability of this is 1/6. Call this new pattern configuration 3.

From configuration 2 you will end up with the same pattern with probability 1/2, configuration 3 with probability 1/4, and ending the experiment with probability 1/4.

From configuration 3 you will end up with the same thing with probability 1/3, and with configuration 2 with probability 2/3.

From this information you can draw the following equations, where a is the expected number of turns from equation 1, b from equation 2, and c from equation 3:

• a=(a+1)/2 + (b+1)/3 + (c+1)/6
• b=(b+1)/2 + (1)/4 + (c+1)/4
• c=(c+1)/3 + (2/3)*(b+1)

Problem 200 Answer

American East 27.492507%
American Central 27.492507%
American West 31.268731%
National East 25.934066%
National Central 23.443223%
National West 25.934066%

Problem 200 Solution

Question

In Major League Baseball, the National League has three divisions, the East and Central divisions with 5 teams each, and the West division with 4 teams. Four teams in the National League will make the playoffs, the three division leaders, and a wild card team. The wild card team is the team with the best record in the league, not including the three division leaders. Assuming all teams are equally good, what is the probability of a team in each division making the playoffs? Assume any ties are broken randomly.

Solution

The difficult part of this question is determing the probability that the Wild Card team will fall in any given division. Let's consider the American League East first. There are six ways the wild card team could be in the American League East, as follows.

• American League East has both first and second place teams in the entire American League. The probability of this is (5/14)×(4/13) = 10.9890%.

• American League East has both first and third place teams, but not the second, in the entire American League. The probability of this is (5/14)×(4/13)×(9/12) = 8.2418%.

• American League East has both second and third place teams, but not the first, in the entire American League. The probability of this is (5/14)×(4/13)×(9/12) = 8.2418%.

• American League East has both first and fourth place teams, in the entire American League, and the second and third place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.

• American League East has both second and fourth place teams, in the entire American League, and the first and third place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.

• American League East has both third and fourth place teams, in the entire American League, and the first and second place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.

Add these probabilties together and you get 37.4625%. It is obvious the probability should be the same for the American League Central. The probability for the American League West thus has a probability of 100% - 2×37.4625% = 25.074925% of having the wild card team.

The probability any given team will be a division leader is 20% in the East and Central divisions, and 25% in the West division. These are also the probabilities for the wild card spot, given that it is in the same division. So, the probability that a team from the East or Central divisions makes the playoffs is 20% + 37.4625%×20% = 27.492500%. The probability that a team from the West division makes the playoffs is 25% + 25.074925%×25% = 31.268731%.

Extra Credit

The probability the National League East or West division will get the wild card is 29.670330% each, and 40.659341% for the Central division. The probability any given team in the East or West will make the playoffs is 25.934066%, and 23.443223% for Central.

Summary

American East 27.492507%
American Central 27.492507%
American West 31.268731%
National East 25.934066%
National Central 23.443223%
National West 25.934066%

Problem 154 Answer

This problem is from the March 2000 issue of The Actuary, published by the Society of Actuaries.

Problem 154 Solution

p₁ = .4 + .6*p₂
p₂ = .4*p₁ + .6*p₃
p₃ = .4*p₂ + .6*p₄
p₄ = .4*p₃ + .6*p₅
p₅ = .4*p₄ + .6*p₆
.
.
.

Now sum the equations.

Sum for i=1 to infinity of p_i = .4 + .4*(sum for i=1 to infinity of p_i) + .6*(sum for i=2 to infinity of p_i)

Sum for i=1 to infinity of p_i = .4 + (sum for i=1 to infinity of p_i) -.6*p₁)

.6*p₁=.4

p₁=.4/.6 = 2/3

At this point you can use substitution to find that p_i=(2/3)ⁱ.

So p₁₀=(2/3)¹⁰=~0.017342

This problem is from the March 2000 issue of The Actuary, published by the Society of Actuaries.

Problem 155 Answer

This problem is from the March 2000 issue of The Actuary, published by the Society of Actuaries.

Problem 155 Solution

p_i given eventual ruin = (p_i&pr(ruin at $i+1))/pr(ruin at $i)=

.6*(.4/.6)¹¹/(.4/.6)¹⁰ (see solution to problem 154 for probability of ruin) =

.6*(.4/.6) = 0.4

So the probability that any given flip will be heads is 0.4, thus the expected total flips that are heads is also 0.4 .

This problem is from the March 2000 issue of The Actuary, published by the Society of Actuaries.

Problem 202 Answer

Question

Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.

Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?

Answer

Problem 202 Solution

Question

Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.

Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?

Solution

Next, from the first piece of information, "Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning," we can put in equation form:

r = 1 + 0.05×5×r + 0.1×3×b + 0.01×(5r+3b)
r = 1 + 0.25r + .3b + 0.05r + 0.03b
r = 1 + 0.3r + 0.33b

Next, from the second piece of information, "Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning," we can put in equation form:

b = 0.03×5b + 0.02×10r + 0.01×(5b+10r)
b = 0.15b + 0.2r + .05b + 0.1r
b = 0.2b + 0.3r

Using basic algebra, we can solve for r and b , given the two equations, and get:

r = 800/461 =~ 1.7354
b = 300/461 =~ 0.6508

So, one initial red amoeba will spawn into 1.7354 on average. We also know that one blue amoea will spawn into 0.6508 red amoeba on average. However, that doesn't tell us how many blue amoeba a red amoeba will spawn into. To answer that question, let's redefine the variables as follows:

Let r = expected number of blue amoeba observed per initial red amoeba.
Let b = expected number of blue amoeba observed per initial blue amoeba.

Putting the initial conditions into equations:

r = 0.3r + 0.33b
b = 1 + 0.2b + 0.3r

Using algebra to solve the equations we get:

r = 330/461 =~ 0.7158 b = 700/461 =~ 1.5184

So, one red amoeba will spawn into 0.7158 blue amoeba on average.

Simulation

I did a random simulation of 74,016,000,000 initial states of one red amoeba. Following were the total counts observed:

Red amoeba: 128,445,045,020
Blue Amoeba: 52,983,662,357

That is an averoage of 1.735368637 red and 0.715840661 blue.

Problem 149 Answer

Thanks to the Geometry Forum for this problem.

Problem 149 Solution

Thanks to the Geometry Forum for this problem.

Problem 110 Answer

Let the corners of the square be at (0,0),(1,0),(1,1), and (0,1).

The solution is to dig the following ditches:

• (0,0) to (x,x)
• (1,0) to (x,x)
• (0,1) to (x,x)
• (1/2,1/2) to (1,1)

Where x = (3-sqr(3))/6 =~ 0.2113248654

The total length of all ditches is aproximately 2.6389584338

Problem 99 Answer

Problem 99 Solution

From the beginning you have enough grain to make 19 one-way trips to some cache point along the way. The first question is, far out should you transport that grain? The answer is far enough so that you leave a cache somewhere of 9000 pounds of grain. Why 9000? Every cache point should have some amount of grain evenly divisible by 1000. This way the camel can start out every trip with 1000 pounds, and not have a remainder on the last trip. So you have 1000 pounds of grain to let the camel eat, divided by 19 one-way trips. 1000 pounds divided by 19 trips is 1000/19=~52.63 pounds, or 52.63 miles per trip. Each trip you will eat 52.63 miles along the way there, deposit 894.73 pounds of grain, and take back 52.63 pounds of grain to eat on the way home, except the last trip in which you don't need to return. The total deposit works out to (10*894.73)+52.63 = 9000 pounds, as arranged. So you will be able to cache 9000 pounds of grain 52.63 miles.

Next repeat the above but only allowing 17 one-way trips. 1000 divided by 17 is 1000/17 =~ 58.82 . So you will be able to cache 8000 pounds of grain 111.46 miles from city A.

Below is a summary of the entire journey:

1. Cache 9000 pounds over 19 one-way trips, moving 52.63 miles, or 52.63 miles from city A
2. Cache 8000 pounds over 17 one-way trips, moving 58.82 miles, or 111.46 miles from city A
3. Cache 7000 pounds over 15 one-way trips, moving 66.67 miles, or 178.12 miles from city A
4. Cache 6000 pounds over 13 one-way trips, moving 76.92 miles, or 255.04 miles from city A
5. Cache 5000 pounds over 11 one-way trips, moving 90.91 miles, or 345.95 miles from city A
6. Cache 4000 pounds over 9 one-way trips, moving 111.11 miles, or 457.07 miles from city A
7. Cache 3000 pounds over 7 one-way trips, moving 142.86 miles, or 599.92 miles from city A
8. Cache 2000 pounds over 5 one-way trips, moving 200.00 miles, or 799.92 miles from city A
9. Take 1000 pounds directly to city B, comsuming 200.08 pounds on the way, leaving 599.84 pounds and take back 200.08 pounds to eat on the way back.
10. Take the last 1000 pounds directly to city B, comsuming 200.08 pounds on the way, and depositing an additional 799.92 pounds of grain, for a total deposit of 1399.77 pounds.

I would like to thank Scott Morris for this problem.

Problem 64 Answer

Problem 64 Solution

Let a be the distance from the point (0,0) to a point on the figure (known as a cardoid). Next consider the angle t formed from the points (-1,0), (0,0) and a at (0,0).

After working through the trigonomtry (this step is left to you) the length of a in terms of t is 2*(1-cos(t)).

The formula for arc length is the integral of ( a² + (da/dt)² )^1/2.

In this case it is the integral from 0 to 2*pi of:

( 4*(1-cos(t))² + 4*sin²(t) )^1/2. =

8^1/2 * (1 - cos(t))^1/2.

After consulting a table of integrals we find that the integral of (1 - cos(t))^1/2 is -2*2^1/2*cos(t/2).

Taking the integral the answer works out to 16.

I'd like to thank Andrea for correcting an earlier mistake in my answer.

Problem 57 Answer

For n=3 (8 by 8) use four sets of the 4 by 4 configuration. Use three of them in such a way that the three extra squares form one extra triominoe, the fourth one should have the extra square in the corner.:

For each additional n repeat the n-1 by n-1 configuration four times, rotating in such as way as to create one extra triominoe.

Here is a nice graphic from Chris Johnson that explains the solution more clearly.

Thanks to Guy de Kindler for this one.

Problem 58 Answer

Problem 9 Answer

If the number is not divisible by 3 then use one 20 pack. If the remaining number is divisible by 3 then use the above method for the rest.

If the number still isn't divisible by 3 use a second 20 pack. The remainder must be divisible by 3, in which case use the 6 and 9 packs as above.

The largest impossible number would be such that you would have to subtract 20 twice to get a remainder divisible by 3. However, you can't make 3 itself with 6 and 9 packs. So the largest impossible number is 2*20+3=43.

Problem 189 Answer

Orange =(3*3^1/2+2*pi)/6 =~ 1.91322

Green = (4*pi - 3*3^1/2)/6 =~ 1.22836

Thanks to Eugene Wilson for this problem and image.

Problem 189 Answer

Consider the sub-triangle formed by the three points where the circle touches the large triangle. This is also an equilateral traingle because each angle is 60 degrees. Once you realize that is it just basic trigonomtry to get the rest as shown in this diagram.

Orange = 6a+2b = (3*3^1/2+2*pi)/6 =~ 1.91322

Green = 4b = (4*pi - 3*3^1/2)/6 =~ 1.22836

Problem 125 Solution

Björn from Germany suggested another solution. Let both hands be at 12. Over the next 12 hours they will cross 11 times. Thus they every 12/11 hours.

Problem 66 Answer

Problem 66 Solution

Next order the two points such that p₁ <= p₂.

Next put the two points in cartesial coordinates (x,y,z).

Through a little geometry it can be found that:
(x₁,y₁,z₁) = (cos(p₁)*sin(q₁), sin(p₁)*sin(q₁), cos(q₁)) and
(x₂,y₂,z₂) = (cos(p₂)*sin(q₂), sin(p₂)*sin(q₂), cos(q₂)).

The angle between two vectors is defined as cos(t)=(v dot w)/(|v| * |w|).

Let v donate the vector from (0,0,0) to (x₁,y₁,z₁) and
let w donate the vector from (0,0,0) to (x₂,y₂,z₂).

The dot product of v and w is:
cos(p₁)*sin(q₁)*cos(p₂)*sin(q₂) + sin(p₁)*sin(q₁)*sin(p₂)*sin(q₂) + cos(q₁)*cos(q₂).

If we rotate the longitude of both points by p₁ then p₁ becomes 0 and p₂ becomes p₂-p₁, in which case the dot product is:
cos(0)*sin(q₁)*cos(p₂-p₁)*sin(q₂) + sin(0)*sin(q₁)*sin(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂). =
sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂).

Now, both |u| and |v| must be 1 since they lie on the surface of a unit sphere. Thus the angle between the two points must be cos^-1[sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂) / (1 * 1). = sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂)].

This angle is the ratio of the distance along the surface between the two points to the circumference of the earth. Thus the distance is the angle times the circumference of the earth which equals:
cos^-1[(sin(q₁)*cos(p₂-p₁)*sin(q₂) + cos(q₁)*cos(q₂))] * r.

Problem 184 Answer

Problem 184 Solution

We know from the second man that that total must be one less than a number divisible by 9.

After considering all 10 people the final number must be one less than a number divisible by every number from 1 to 10. Certainly 10!=3628800 works but there is smaller solutions. The smallest answer is LCM(1,2,3,4,5,6,7,8,9,10)-1 = 7*2*2*2*3*3*5-1 = 2519.

Note that LCM stands for Least Common Multiple.

Thanks to David Brown for suggesting this problem.

Problem 11 Answer

Problem 11 Solution

Think of this problem instead as a random walk along the number line, starting at zero. If the walker ever returns to zero, then success is achieved. Let x_i be the probability of success from point i on the number line. Let p be the probability that the walker moves to the right. Let's also let p be the lesser of the probabilities of flipping heads and tails.

I hope I won't have to prove that x_-1=1. In other words, the gravity of the greater probability of q will pull the walker to the right, over the long run. So if his first move is to the left, he will eventually make it back to point 0. For the same reason, given a casino game with a house advantage, the house always wins in the long run. If the player wins his first bet, if he plays indefinitely, he will eventually lose it back.

That said, we can set up the following equations:

x₀ = p + q×x₁
x₁ = p + q×x₂
x₂ = p×x₁ + q×x₃
x₃ = p×x₂ + q×x₄
.
.
.

Next, take the sum of both sides.

Let E = x₀ + x₁ + x₂ + ...

E = 2p + pE -p×x₀ + qE - q×x₀
E = 2p + E×(p+q) - x₀×(p+q)
E = 2p + E - x₀ (because p+q=1)
0 = 2p - x₀
x₀ = 2p So the answer is 2p.

Below are the results of computer trials which have verified the answer above. For each probability p 100,000 trials were conducted, in which a trial ended when either the number of heads equalled the number of tails, or the difference was more than 25. It seems reasonable to assume that for p<=.4 if the difference ever equals 25 it is very unlikely that the gap ever be closed. A 'win' means the number of heads was ever equal to the number of tails. The number of losses would be 100,000 minus the number of wins.

  p    wins 
-----  ------
 0.4   80058
 0.3   59799
 0.2   39957
 0.1   19803
 

If you don't buy my solution, I have posted altnernate solutions by Richard Tucker and David Beim.

Thanks to Guy de Kindler for providing the problem and Richard Tucker and David Beim for the alternate solutions.

Problem 44 Answer

Problem 44 Solution

The payoff for getting x heads followed by a tails is $2^x.

The expected return is the sum of the products of the probabilities and the payoffs is the sum for x=0 to infinity of (1/2) * (1/2)^x * 2^x.

(1/2)^x * 2^x = 1. So the answer is an infinite sum of 1/2's, or infinity.

This problem is a well known paradox in mathematics. The final result must be finite yet the expected value is infinite.

Problem 46 Answer

Problem 46 Solution

Next determine the expected number of turns given such barriers using the same method of interlocking expected values as in the ant and spider problem. The answer turns out to be 2x.

Thus, when x is infinity the expected number of turns must be 2*infinity=infinity.

Related question: What is the expected number of turns it would take a random walker on a number line to revist any point? The answer is here.

Problem 61 Answer

Problem 61 Solution

It may not seem reasonable to take the last step on faith but millions of computer trials have verified the answer.

Problem 128 Solution

E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(x))/2

Solving for E(x) and E(y) yields E(x)=6 and E(y)=4. Since the first state is x the answer is 6.

Approach the second problem in the same way, however this time from state y you either end or stay in state y, thus the equations are:

E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(y))/2

In this case E(x)=4 and E(y)=2, the answer being E(x) or 4.

Thanks to Alon Amit for this problem.

Problem 246 Answer

The answer is 0.0880680413.

Problem 246 Solution

1. First a note about terminology, let C(x,y) equal the number of ways to choose y items out of x. The formula is x!/(y! × (x-y)!). For example, the number of way to choose five cards out of 52 is C(52,5) = 52!/(5! × 47!) = 2,598,960.
2. Rather than keep using the word "birthday," let's say each person will get one "day."
3. Next, let's consider the case of five people sharing a common birthday. The logic will work for any number of people sharing a common birthday.
4. To make a specific example, consider the probability that in an office of 200 people there will be at least one day of the year shared by five or more people.
5. To find that answer, first find the probability that in 200 people nobody day will be shared by five or more people. Subtracting that answer from 1 will give us the probability of a common birthday to five or more people.
6. There are 365²⁰⁰ possible sequences.
7. To find the number of sequences where there is no day celebrated by five or more people, consider all the ways to clump the 200 people into days shared by 4 people, shared by 3, shared by 2, and shared by 1.
8. Let a = the number of days shared by 4 people. This number can be 0 to 50.
9. Let b = the number of days shared by 3 people. This number can be 0 to int((200-4a)/3).
10. Let c = the number of days shared by 2 people. This number can be 0 to int((200-4a-3b)/2).
11. Let d = the number of days occupided by just one person. This number must be 200-4a-3b-2c.
12. The number of ways you can assign specific days to each of these days with at least one birthday is C(365,a)×C(365-a,b)×C(365-a-b,c)×combin(365-a-b-c,d). Let's call that number p.
13. The number of ways to order the 200 birthdays is 200!. Multiply that by p.
14. If we stopped here, p would be too big. Why? Let's say two people share May 23. Let's go onto say that if you lined up all 200 people they took spots 81 and 144. It wouldn't make any difference which May 23 person got 81 and who got 144. The sequence of birthdays would still look the same.
15. Likewise, we would divide by 3!=6 for each group of 3 people, and 4!=24 for each group of 4 people.
16. To adjust for the c groups of 2 people, divide p by 2^c.
17. To adjust for the b groups of 3 people, divide p by 6^b.
18. To adjust for the a groups of 4 people, divide p by 24^a.
19. That will give us the correct value of p, the number of sequences of birthdays where no day is shared by 5 or more people.

Thus, the probability of no common birthday is C(365,a)×C(365-a,b)×C(365-a-b,c)×combin(365-a-b-c,d)/(2^c × 6^b × 24^a)/365²⁰⁰, for all possible combinations of a, b, c, and d. This works out to 0.9119319587.

So, the probability of at least one day common to five or more people is 1 - 0.9119319587 = 0.0880680413.

As a practical note, there is no easy way to get at this number without a computer. Here is my code that shows how to calculate the probability of no common birthday, where p is the number of 200:

Problem 37 Answer

Problem 37 Solution

One minus this figure is the probability that there is at least one common birthday. The probability that there is at least one common birthday in 22 people is 47.57%, for 23 people it is 50.73%. Since the percentage must be at least 50% the answer is 23.

Problem 31 Answer

Problem 31 Solution

Below is another approach:

The value of $1 invested for n years at interest rate i, compounded x times per year is (1+(i/x))^nx.

In this case f(x) = the value of $1 invested at 100% interest compounded n times per year = (1+(1/x))^x, where x approaches infinity.

The following table shows values of f(x) for various values of x:

     x              f(x)
-----------     ----------
          1	2.00000000
          4	2.44140625
         12	2.61303529
         52	2.69259695
        365	2.71456748
      1,000	2.71692393
     10,000	2.71814593
    100,000	2.71826824
  1,000,000	2.71828047
 10,000,000	2.71828169
100,000,000	2.71828180

As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.

Problem 49 Answer

Problem 49 Solution

However, if you were willing to not touch your investment and make the payments out of your pocket there is still a problem. You have to consider the time value of money. Lets assume inflation equal to the 6% interest rate, compounded monthly. The time value of the car payments made would be $10,364.77, however the time value of your $13,488.50 after five years would only be $10,000. In other words after five years your initial investment of $10,000 has gone down in value more than the interest savings.

Note: The payments on a loan of $x, at intrest rate i, over n periods is $xi/(1-vⁿ), where v=1/(1+i). In this case x=$10,000, i=.075/12=.00625, and n=60.

Problem 86 Answer

Problem 86 Solution

f'(t)=t*f(t).

Let y=f(t).

dy/dt=ty.

dy=ty dt.

dy/y = t dt.

Integrating both sides:

ln(y) = t²/2 + C.

y = exp(t²/2 + C).

y = exp(t²/2). Since y=1 when t=0

Problem 244 Answer

Here is the general shape of the road network:

The minimum road distance is 200 * (3*sqrt(3)/3 + 1) =~ = 546.4102 kilometers.

Problem 244 Solution

Here is the general shape of the road network:

The total length of the roads will be 4x+z.

We're told the length of the side of the state is 200, However, let's assume a side length of 1, to be more elegant, and multiply by 200 at the end.

If a side of the square is 1, then 2y+z = 1.

Rearranging, z = 1 - 2y.

So, the total distance can be expressed as 4x + 1 - 2y.

From the Pythagorean formula, 0.5^2 + y^2 = x^2.

Rearranging, y^2 = x^2 - 0.25.

y = (x^2 - 0.25)^0.5

To put the total distance as a function of just x:

f(x) = 4x + 1 - 2*(x^2 - 0.25)^0.5.

Next, set the derivative to equal zero, to find the value of x that maximizes f(x).

f'(x) = 4 - 2*0.5*2x*(x^2 - 0.25)^-0.5 = 0.
2*0.5*2x*(x^2 - 0.25)^-0.5 = 4
x*(x^2 - 0.25)^-0.5 = 2
x = 2*(x^2 - 0.25)^0.5
x^2 = 4 * (x^2 -0.25)
3x^2 = 1
x^2 = 1/3
x = sqrt(3)/3

Our total road length, based on a side of 1, is:

4*sqrt(3)/3 + 1 - 2*(1/3 - 1/4)^0.5 =
4*sqrt(3)/3 + 2*(1/12)^0.5 =
4*sqrt(3)/3 + 1 + 2*0.5*sqr(3)/3 =
3*sqrt(3)/3 + 1 =~ 2.7321

So, for a side of length 200 kilometers, the answer is 200 * (3*sqrt(3)/3 + 1) =~ = 546.4102 kilometers.

Problem 224 Answer

ln(200)/ln(2) =~ 7.6439 hours.

Problem 224 Solution

Problem 168 Answer

Problem 168 Solution

By the time the cows have eaten all the grass total grass consumed will equal total initial grass plus total grass regrowth. Next lets define some terms.

x=initial amount of grass in an acre.
y=amount of grass grown in one acre in one day.

Putting the given information in the form of equations we get:

50=2x+20y
210=3x+90y

To put the first equation in more simple English, 5 cows eating for 10 days results in a consumption of 50 units of grass. This is equal to the sum of 2x initial units of grass and 20y units of grass growth (10 days times 2 acres).

Next we must solve for x and y. Rewriting the above equations we get:

150=6x+60y (multiplying by 3)
420=6x+180y (multiplying by 2)

Subtracting the first equation from the second we get 270=120y, so y=9/4. Plugging this into either equation we get x=5/2.

The question to be answered is "How many days will it take 16 cows to eat 7 acres of grass?" Let's let d be the number of days. So setting this up as an equation we get.

16d=7x + 7dy

16d=35/2 + 63d/4

d/4=35/2

d=70.

So it will take 70 days for 16 cows to eat 7 acres of grass.

Problem 107 Answer

Problem 103 Answer

Thanks go to Stephanie Mullen for this one.

Problem 102 Solution

f(x)=2*pi*x/pi = 2x.

The mean distance to the center is the integral from 0 to 1 of f(x)*x:

Integral from 0 to 1 of 2x*x = 2x³/3 from 0 to 1 = 2/3.

Problem 3 Answer

Problem 3 Solution

So the total arc length is the integral from 0 to 2pi of ( (dx/dt)² + (dy/dt)² ) ^1/2.

After a few steps this integral becomes:

r×2^1/2 × (1-cos(t))^1/2.

Using the hint:

r×2^1/2 × 2^1/2 × integral of sin(t/2) dt from 0 to 2×pi

= 2×r × (-2×cos(t/w) from 2×pi to 0)
= 8r

Reference: Example 3, page 550, Calculus and Analytic Geometry, 1982 edition.

Problem 223 Answer

The radius which achieves the maximum volume of the cylinder is (2/3)^0.5 =~ 0.816496581.

Problem 223 Solution

Call the radius of the cylinder 1. The distance from the center of the clinder to the edge is 1, because it is in a unit sphere. By phythagerous, we know the height of the cylinder must be (1-r^2)^0.5.

The volume of the clinder V is equal to pi*r^2*(1-r^2)^0.5.

Take the derivative and set it equal to zero:

The volume of said cylinder is 2*pi*sqrt(3)/9 =~ 1.209199576.

Problem 228 Answer

The answer is 36 × pi.

Problem 228 Solution

Let r = radius of the sphere.

To find the volume of the sphere left after you drive the hole through, consider the sphere centered at (0,0). Then find the volume of the part of the sphere ranging in x value from -3 to +3. Then we'll subtract out the area of the cylinder in the middle that the drill removed.

The area of the sphere from x = -3 to +3 equals:

So, it makes no difference what the radius of the sphere is, there will 36*pi left over after drilling through it.

Problem 17 Answer

Problem 17 Solution

Integral from 0 to 1 of d*2d = 2d³/3 from 0 to 1 = 2/3.

Problem 102 Answer

Problem 102 Solution

f(x)=2*pi*x/pi = 2x.

The mean distance to the center is the integral from 0 to 1 of f(x)*x:

Integral from 0 to 1 of 2x*x = 2x³/3 from 0 to 1 = 2/3.

Problem 174 Answer

Problem 174 Solution

m = men before increase
w = women before increase
m' = men after increase
w' = women after increase
d = m-w
d' = m'-w'

The fact that a random drawing will result in a date 50% of the time tells us that 2*(m/(m+w))*(w/(m+w-1)) = 0.5

(m/(m+w))*(w/(m+w-1)) = .25

m/(m+w) = (m+w-1)/(4w)

4mw = (m+w)*(m+w-1) 4mw = m² + mw -m + mw + w² - w

4mw = m² + w² -2mw - m - w

(m-w)² = m + w, or d² = m+w

By the same reasoning:

(m'-w')² = m' + w', or d'² = m'+w'

We are also given:

m'+w'-m-w=100

d'² - d² = 100

(d'+d)*(d'-d) = 100

We know that d and d' must be integers so the only possible values for d'+d and d'-d are: (100,1), (50,2), (25,4), (20,5), (10,10).

Let's consider the first possibility of (100,1). If this were the solution then d'+d=100 and d'-d=1. Adding the two equations together yields 2d'=101 --> d'=101/2. However since m' and w' are integers then d' is also an integer and can not be 101/2. For the same reason (25,4) and (20,5) are not possible because the sum of the two factor is odd.

Let's consider (10,10). Then d'+d=10, d'-d = 10 --> d=0 --> m-w=0. However the problem stated m>w, so (10,10) doesn't work.

That leaves only (d'+d,d'-d)=(50,2). Adding the two equations gives us d'=m'-w'=26.

Remember, d'² = m'+w', so m'+w'=26² = 676.

Remember also that m'-w'=26

Adding the two equations give us 2m'=702 --> m'=351 --> w'=325.

So after in increase there are 325 women and 351 men. From here it can also be easily found that before the increase there were 300 men and 276 women.

Thanks to Ady Tzidon for suggesting a very similar problem and to Rob Pratt for this solution (which is better than how I initially solved it).

Problem 141 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 141 Solution

• If the first roll is a 12 then A wins.
• If the first roll is a anything other than a 7 and 12 then neither side has improved their odds and the probability of A winning is still p.
• If the first roll is a 7 and the second roll is a 12 then A wins.
• If the first roll is a 7 and the second roll is anything other than a 7 and 12 then neither side has improved their odds and the probability of A winning is still p.

We can now express p as follows:

p = 1/36 + (29/36)*p + (1/6)*(1/36) + (1/6)*(29/36)* p

p = 7/216 + (203/216)*p

13p = 7

p = 7/13

So the probability that A will win is 7/13 =~ 0.538462 .

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 142 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 142 Solution

1/6 + (29/36)*(6/13) = 252/468 = 7/13.

So the probability that A will win if the last roll was a 7 is 6/13, otherwise it is 7/13. Likewise the probability that B will win if the last roll was a 7 is 7/13, and 6/13 otherwise.

By jumping ahead x rolls the probability that the (x-1)^th roll was a 7 is 1/6 and the probability of a non 7 is 5/6. Thus the probability of of A winning the next bet resolved is:

pr((x-1)^th roll = 7)*(6/13) + pr((x-1)^th roll != 7)*(7/13) =

(1/6)*(6/13) + (5/6)*(7/13) = 41/78 =~ 0.525641 .

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 143 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 143 Solution

The probability that the (x-1)^th roll will be a 7 is 1/6 and the probability the x^th roll will be a 7 is 1/6, thus the probability of the x^th roll being a win for B is also (1/6)*(1/6) = 1/36.

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 144 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 145 Answer

A's expected gain over a million bets resolved is 8.33 cents.

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 145 Solution

The probability A wins the first bet is 7/13.

The probability A wins the second bet is (7/13)*(7/13)+(6/13)*(6/13) = 85/169, using logic from earlier problems.

The probability A wins the second bet is (85/169)*(7/13)+(84/169)*(6/13) = 1099/2197.

Continuing this pattern, the probability A wins the n^th bet is (((13ⁿ+1)/2)/13ⁿ)*(7/13)+(((13ⁿ-1)/2)/13ⁿ)*(6/13) =

((13ⁿ+1)/2)/13ⁿ⁺¹.

The expected gain by A on the first bet is (7/13)*(+1) + (6/13)*(-1) = 1/13.

The expected gain by A on the second bet is (85/169)*(+1) + (84/169)*(-1) = 1/169.

The expected gain by A on the n^th bet is 1/13ⁿ.

Taking the sum from 1 to one million the total gain by A is:

$\sum$(for i=1 to 1,000,000) 1/13ⁱ =~ (1/13)/(12/13) = 1/12 =~ 8.33 cents.

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 146 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Problem 34 Answer

Problem 34 Solution

In this specific case this is [(.1)(.9)]/[(.1)(.9)+(.9)(.1)] = 1/2.

Problem 42 Answer

Problem 42 Solution

Problem 222 Answer

The man who contributed five loaves should get seven coins and the man who contributed three loaves should get one coin.

Problem 222 Solution

Think about who contributed what to what the third man ate. Assuming each man ate the same amount, each ate 8/3 loaves. The man who contributed 5 loaves ate 8/3 of them, leaving 7/3 loaves to the third man. The man who contributed 3 loaves ate 8/3 of them, leaving 1/3 loaves to the third man.

So, of the 8/3 loaves eaten by the third man, 7/3 were from the man with five loaves and 1/3 from the man with three loaves. Thus split the coins according to whose bread you ate -- 7 coins to the man with five loaves and one coin to the man with three loaves.

Problem 147 Answer

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.

Problem 147 Solution

Obviously at least one number will be needed. If the first number is less than x then another will will need to be drawn with a lesser total needed to finish.

So E(x) = 1 + integral from 0 to x of E(y) dy.

Take the derivative of both sides...

E'(x) = E(x)

What function is a derivative of itself? e^x of course! So the answer is e¹ = e.

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.

Problem 218 Answer

The answer is e, or approximately 2.7182818.

Problem 218 Solution

Solution to problem 218 (PDF -- 67K)

Problem 216 Answer

The answer is 61.217385.

Problem 216 Solution

Solution

This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.

1. First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.

2. Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.

   The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.

   So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.

3. Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.

   What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:

   pr(A or B) = pr(A) + pr(B) - pr(A and B)

   You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,

   pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3) - pr(4 before 2 and 3) = (3/4)+(3/5)-(3/6) = 0.85.

   The probability of not getting the four along the way to the two and three is 1.0 - 0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15×12 = 43.8.

4. Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.

   What is the probability of getting the five before achieving the two, three, or four? The general rule is:

   pr (A or B or C) = pr(A) + pr(B) + pr(C) - pr(A and B) - pr(A and C) - pr(B and C) + pr(A and B and C)

   So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)-pr(5 before 2 and 3)-pr(5 before 2 and 4)-pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)-(4/7)-(4/8)-(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1 - 83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)×9 = 44.5.

5. Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.

   Here is the general rule for pr(A or B or C or ... or Z)

   pr(A or B or C or ... or Z) =
   pr(A) + pr(B) + ... + pr(Z)
   - pr (A and B) - pr(A and C) - ... - pr(Y and Z) Subtract the probability of every combination of two events
   + pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
   - pr (A and B and C and D) - pr(A and B and C and E) - ... - pr(W and X and Y and Z) Subtract the probability of every combination of four events

   Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.

The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Problem 158 Answer

Problem 175 Answer

Problem 175 Solution

It is intuitive that at all times the ants will form the corners of a square, ever decreasing in size and rotating about the center.

Let C be the center of the square at point (0,0). Let the ants be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let ant 1 bet at (.5,-.5) who is chasing ant 2 at (.5,.5). Consider the line from C to ant 1. Next consider the line from ant 1 to ant 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of ant 1 would be be at a 270 degree angle. The line of sight from ant 1 to ant 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e^-x.

At the moment the ants start walking x equals 0. At this moment the distance from the center to any of the ants is 2^-1/2. So r(x)=2^-1/2*e^-x.

The arc length is given by the formula:
Integral from negative infinity to 0 of ((f(x))²+(f^'(x))²)^1/2 =
Integral from negative infinity to 0 of ((2^-1/2*e^-x)² + (-2^-1/2*e^-x)²)^1/2 =
Integral from negative infinity to 0 of (2^-1*e^-2x + 2^-1*e^-2x)^1/2 =
Integral from negative infinity to 0 of e^-x =
-e^-x from negative infinity to 0 =
=0 - (-e⁰) =
0 -(-1) = 1.

The easy way to solve this problem is to first note that the direction of one ant will always be perpendicular to the path of the ant he is chasing. If ant A is chasing ant B then B has zero velocity in the direction of A's path. In other words B neither approaches nor recedes from A during the walk. So it is like A is walking directly towards an unmoving ant B, in which case the total distance is simply the intial separation of 1.

No we're ready to consider the pentagon case. Whether you solve the square case the easy or the hard way it should be clear that as ant A chases ant B what is important is the velocity of ant B relative to the path of ant A.

Before we go further we have to do some trigonometry. Consider an equilateral pentagon of side 1 with the first vertice at (0,0) and the second at (1,0). Where is the third vertice? Given that the interior angles of a pentagon are 108 degrees the next vertice is at (1+cos(72),sin(72)).

So as ant A inches 1 unit towards ant B, ant B will move cos(72) units directly away from ant A in the same direction and sin(72) in a perpenticular direction. We know form the four corners problem that the perpenticular direction can be ignorred. So for every micron ant A moves towards ant B, ant B will move away from ant A by cos(72) microns.

The total distance covered by A, or any ant, is the sum for i=0 to infinity of cos(72)ⁱ. We know from infinite sums that this equals 1/(1-cos(72)).

If you prefer to express the answer another way cos(72)=1/(1+5^1/2). So 1/(1-cos(72)) = 1/(1-1/(1+5^1/2)) = 1/((5^1/2)/(1+5^1/2)) = (1+5^1/2)/5^1/2 = 1 + 1/5^1/2 =~ 1.4472 .

For more information on this problem please see the November 1957 and July 1965 issues of Scientific American. The July 1965 issue notes that for the general case of an n-gon the total distance is as follows. Let A and B be the initial points of any two consecutive ants, where A is chasing B. Let C be the center of the n-gon. Consider the line segment AB extended to a point X, such that the angle XCA is a right angle. The total distance any ant will travel is the same as the distance from A to X.

Thus, if the interior angle of the n-gon is x then the total distance is 1/(1+cos(x)). So for an n-gon the total distance is 1/[1+cos(180-360/n)].

Problem 230 Answer

Dan should pick x coins with probability pr(x), as follows:

Problem 230 Solution

It stands to reason that Sam's expected win should be the same regardless of whatever he picks.

Let p₁ = Probability Dan picks 1 coin
Let p₂ = Probability Dan picks 2 coins
Let p₃ = Probability Dan picks 3 coins
Let p₄ = Probability Dan picks 4 coins
Let p₅ = Probability Dan picks 5 coins

If Sam picks one coin, then Sam's expected value is p₁×2 - p₂×3 - p₃×4 - p₄×5 - p₅×6
If Sam picks two coins, then Sam's expected value is p₁×-3 + p₂×4 - p₃×5 - p₄×6 - p₅×7
If Sam picks three coins, then Sam's expected value is p₁×-4 - p₂×5 + p₃×6 - p₄×7 - p₅×8
If Sam picks four coins, then Sam's expected value is p₁×-5 - p₂×6 - p₃×7 + p₄×8 - p₅×9
If Sam picks five coins, then Sam's expected value is p₁×-6 - p₂×7 - p₃×8 - p₄×9 + p₅×10

Setting the expected value of Sam picking one to Sam picking two:

p₁×2 - p₂×3 - p₃×4 - p₄×5 - p₅×6 = p₁×-3 + p₂×4 - p₃×5 - p₄×6 - p₅×7

p₁×5 - p₂×7 + p₃×1 + p₄×1 + p₅×1

Setting the expected value of Sam picking one to Sam picking three:

p₁×2 - p₂×3 - p₃×4 - p₄×5 - p₅×6 = p₁×-4 - p₂×5 + p₃×6 - p₄×7 - p₅×8

p₁×6 + p₂×2 - p₃×10 + p₄×2 + p₅×2

Setting the expected value of Sam picking one to Sam picking four:

p₁×2 - p₂×3 - p₃×4 - p₄×5 - p₅×6 = p₁×-5 - p₂×6 - p₃×7 + p₄×8 - p₅×9

p₁×7 + p₂×3 + p₃×3 - p₄×13 + p₅×3

Setting the expected value of Sam picking one to Sam picking five:

p₁×2 - p₂×3 - p₃×4 - p₄×5 - p₅×6 = p₁×-6 - p₂×7 - p₃×8 - p₄×9 + p₅×10

p₁×8 + p₂×4 + p₃×4 + p₄×4 - p₅×16

It goes without saying that p₁ + p₂ + p₃ + p₄ + p₅ = 1

Now we have five equations and five unknowns. Here is our matrix for these equations:

A little matrix algebra gives us:

Problem 108 Answer

Problem 108 Solution

After the first person is finished the remaining pile will have 4, 8, 12, 16, 20, ...

The remaining pile after the first person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 16, 36, 56, 76, 96, ...

After the second person is finished the remaining pile will have 12, 28, 44, 60, 76, ...

The remaining pile after the second person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 76, 156, 236, 316, 396, ...

After the third person is finished the remaining pile will have 60, 124, 188, 252, 316 ...

The remaining pile after the third person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 316, 636, 956, 1276, 1596, ...

After the fourth person is finished the remaining pile will have 252, 508, 764, 1020, 1276 ...

The remaining pile after the fourth person must be a number such that you can subtract one then multiply by 4/5 and get an integer. The smallest possibilities is 1276

After the fifth person is finished the remaining pile will have 1020 ...

So the fifth person will get 1276-1020-1 = 255.

The fifth person will leave behind a pile of 1020.

The fourth person will have left behind a pile of 1020+255+1 = 1276.

The fourth person will get 1276/4 = 319.

The third person will have left behind a pile of 1276+319+1 = 1596.

The third person will get 1596/4 = 399.

The second person will have left behind a pile of 399+1596+1 = 1996.

The second person will get 1996/4 = 499.

The first person will have left behind a pile of 499+1996+1 = 2496.

The first person will get 2496/4 = 624.

The original pile must have had 624+2496+1 = 3121.

Problem 85 Answer

Problem 85 Solution

If only pirate 5 were left he would get everything.

If only pirate 4 and 5 were left then pirate 4 has no hope, regardless of what he suggests pirate 5 will vote to throw him overboard, and overboard he will go. Thus it doesn't make any difference what pirate 4 suggests.

If only pirates 3,4, and 5 were left then pirate 3 could count on pirate 4's support as long as pirate 4 got at least 1 coin. This is because if pirate 3 goes overboard pirate 4 knows he will get nothing. So pirate 3 would suggest (0,1,999) to which he and pirate 4 would vote 'yes.' Note that according to the problem pirate 4 would vote 'no' to (0,0,1000), even though a 'yes' vote would save his own life his pleasure of throwing pirate 3 over would be worth his own life.

If only pirates 2,3,4, and 5 were left then pirate 2 could count on pirate 3 voting 'no' unless pirate 3 got more than 999 coins, which would leave everybody else with 0, which obviously wouldn't pass. So pirate 2 should not count on pirate 3's support so he shouldn't waste any coins on him. If pirate 2 goes overboard he knows the money will be distributed (0,1,999). So if he offers 2 coins to pirate 4 and 1 coin to pirate 5 then he will get their support since they will come out better voting 'yes.' Thus pirate 2's suggestion would be (1,2,0,997) to which pirates 2, 4 and 5 would vote 'yes.'

Thus if all five pirates were left pirate 1 could count on pirate 2 voting 'no' since pirate 2 would get 997 if pirate 1 went overboard. So pirate 1 should not waste any coins on pirate 2 since he will vote 'no' anyway and concentrate on winning 2 votes from the other 3. If pirate 1 goes overboard the money will be distributed (1,2,0,997). So it would cost 2 coins to get pirate 5's support, 3 coins to get priate 4's support, and 1 coin to get pirate 3's support. Since he only needs to buy two votes the cheapest two are pirates 5 and 3. Thus pirate 1 would suggest (2,0,1,0,997) to which pirate 1, 3, and 5 would vote 'yes'.

Thanks to The Ultimate Puzzle Site which has also posted this problem, click here to see their solution if this one was too confusing.

Problem 162 Answer

Thanks to Elizabeth Gomez for this problem.

Here are the four possible answers, the last one appears in the picture below.

1. 1x9, 2x8, 3x6, 4x7, 5x10 = 11x11 square
2. 1x6, 2x10, 3x9, 4x7, 5x8 = 11x11 square
3. 1x2, 3x7, 4x6, 5x10, 8x9 = 13x13 square
4. 1x2, 3x8, 4x5, 7x9, 6x10 = 13x13 square

Go back to Problem 98

Go back to Shack's Math Problems

This solution is hard to explain so if you don't understand my explanation please forgive me.

First consider the minimum and maximum possible areas the rectangles can cover, not neccessarily forming a square. The maximum area would be (10*9)+(8*7)+(6*5)+(4*3)+(2*1)=190. The minimum area would be (10*1)+(9*2)+(8*3)+(7*4)+(6*5)=110. The three possible squares between these two bounds are 121 (11 by 11), 144 (12 by 12), and 169 (13 by 13).

Try to tile a 12 by 12 square. It makes common sense that one rectangle should fit in each corner of the square with the fifth one in the middle. There is no other possibility since the sides of the rectangles are all different lengths. Each side of the square shall consist of the sides of two different rectangles. Thus the sides will consist of these sums: 10+2, 9+3, 8+4, 7+5. The middle rectangle will be the leftover two numbers, 6 by 1.

It also stands to reason that each side of the square will consist of the long side of one rectangle and the short side of another. To form a middle rectangle of side 6 we must subtract the side of one rectangle from the side of another, the possibilities being 10-4=6 and 8-2=6. 9-3=6 is not possible since these two dimensions must form a side of the square. In the 10-4 case we would have the 10+2 side of the square opposite of the 8+4 side. The other two opposite sides being 9+3 and 7+5, all odd numbers. Note that it is impossible to subtract one odd number from another and obtain 1. Thus the 10-4 case is impossible. Under the 8-2 case we still have the 10+2 and the 8+4 on opposite sides of the square. Again it will be impossible to form a 1 remainder with odd numbered sides only the other way. Thus the 12 by 12 square is impossible to tile.

The 13 by 13 square will have sides of lengths 10+3, 9+4, 8+5, and 7+6, with a 2 by 1 rectangle in the middle. Through a process of elimination as in the 12 by 12 case we can't help but stumple upon the solutions.

The 11 by 11 square can have sides of lengths 10+1, 9+2, 8+3, 7+4, and 6+5. Four of these will go along the sides and the fifth shall form the middle rectangle. Again it is a process of trial and error to find the solutions.

This is a picture of just one solution:

Go back to Problem 98

Go back to Shack's Math Problems

Problem 112 Answer

Problem 112 Solution

The standard deviation of the number of reds is sqr(1,000,000 * 0.4 * (1-.4)) =~ 489.9

Let l denote the lower bound of the range, let u denote the upper bound, and let x denote the number of reds drawn:

Pr (l <= x <= u) = .95

Pr (l-400,000 <= x-400,000 <= u-400,000) = .95

Pr ((l-.5-400,000)/489.9 <= Z <= (u+.5-400,000)/489.9) = .95 where Z denotes a random variable distributed according to the standard normal distribution.

Next we want the probability that the number of reds will fall on either side of this range to be .025 per side.

Pr( Z <= (u+.5-400,000)/489.9) = .975

(u+.5-400,000)/489.9 = 1.96

u-399,999.5 = 960.2

u =~ 400,960

Pr( Z <= (l-400,000.5)/489.9) = .025

(l-400,000.5)/489.9 = -1.96

l = 399,040

So the range is 399,040 to 400,960

Problem 229 Answer

The around the world flight can be accomplished with as few as three planes.

Problem 229 Solution

The around the world flight can be accomplished with only three planes.

The following table shows how it can be done. I'm sure there are variations of this that will also work. The time column is relative to the time required to fly around the earth non-stop. Fuel remaining is relative to a full tank. So a value of 1 would be full and 0 would be empty. The location is relative to a 360-degree circle, starting at the 0 point.

Notes:
* It was not required to transfer this much fuel. As long as all planes had at least 1/8 of a tank to get back all would have enough to make it back to the island. I chose 7/24 so that each plane would have the same amount of emergency fuel, in the interests of safety.

I found this problem in the book My Best Mathematical and Logic Puzzles by Martin Gardner, which I highly recommend. It is problem 19.

Problem 55 Answer

Problem 55 Solution

First reword the problem such that the coordinates of the bird's initial position are (100,0) and the nest is at (0,0). In polar coordinates the inital position of the bird is (100,0^o). Let p stand for the distance to the origin in polar coordinates, and let a stand for the angle.

Considering that the bird always flied towards the nest (p direction) and the wind blows north (y direction) the following graph shows the change in the variables with respect to a change of one infinitesimal unit of time in units of time. Note that as an angle approaches zero its arc approaches a straight line.

(1) dp/dt = -5 + 2sin(a).
(2) da/dt = 2cos(a)/p.

(3) dx/dt = -5cos(a).
(4) dy/dt = 2 - 5sin(a).

Divide eq. (3) by eq. (2):

(5) dx/da = (dx/dt)/(da/dt) = -5pcos(a)/2cos(a) = -5p/2 = -(5/2)*(x/cos(a)). (see graph above)

Rearrange:

(6) (2/x)dx = (-5/cos(a))da.

Integrate both sides:

(7) ln(x²) = -ln(sec(a)+tan(a))⁵ + C you might need an integral table for the integral of 1/cos(a).

Exp() both sides:

(8) x²*(sec(a) + tan(a))⁵ = C.

Put in the intial condition of x=100, a=0 to solve for the constant:

(9) x²*(sec(a) + tan(a)) = 10,000.

(sec(a) + tan(a))⁵ = (100/x)².

(sec(a) + tan(a)) = (100/x)^2/5.

Let b=(100/x)^2/5, then:

(10) sec(a) + tan(a) = b.

(11) (1+sin(a))/cos(a) = b (sec(a)=1/cos(a), tan(a)=sin(a)/cos(a)).

(12) (1+(1-cos²(a))^1/2) / cos(a) = b.

1+(1-cos²(a))^1/2 = b*cos(a).

(1-cos²(a))^1/2 = b*cos(a) - 1.

1-cos²(a) = b²*cos²(a) - 2*b*cos(a) + 1.

2*b*cos(a) = (1+b²)*cos²(a).

2*b = (1+b²)*cos(a).

(13) cos(a) = 2*b / (1+b²).

Substitute equation (13) into equation (3):

(14) dx/dt = -5*cos(a) = -10*b / (1+b²).

Rearranging:

dt = -(1+b²)/(10*b) dx.

-10 dt = b+(1/b) dx

(15) -10 dt = (100/x)^2/5 dx + (x/100)^2/5 dx, remember b=(100/x)^2/5

Integrating:

(16) Constant -10t = (5/3)*100^2/5*x^3/5 + (5/7)*(1/100)^2/5*x^7/5.

Put in the initial constants t=0 and x=100:

Const = (5/3)*100^2/5*100^3/5 + (5/7)*(1/100)^2/5*100^7/5.

Const = (5/3)*100 + (5/7)*100 = 5000/21.

So the function of x and t is:

(17) 5000/21 - 10t = (5/3)*100^2/5*x^3/5 + (5/7)*(1/100)^2/5*x^7/5.

The original question was how long does it take the bird to reach the nest, in other words what is t when x=0. By substituting x=0 in the above equation t = Const/10 = 500/21.

Tristan Simbulan sent in the following solution which you may find better:

The initial position of the bird is ( 100, 0 ) and the nest is ( 0, 0 ) I am using the velocity vector component parallel to y and x.

Let a be the angle between x axis and bird's line of sight towards ( 0, 0 ).

(1)The wind velocity minus y component of the bird's velocity = dy/dt = 2 - 5 sin a = 2 - 5 * y/sqrt(x^2 + y^2)

(2) Velocity x component = dx/dt = - 5 cos a = - 5 * x / sqrt( x^2 + y^2 )

(3) (dy/dt)/(dx/dt) = dy/dx = [ 5y -2 * sqrt(x^2 + y^2) ] / 5x 5xdy - 5ydx + 2 * sqrt(x^2 + y^2) dx = 0 Let y = vx, dy = vdx + x dv , the result by substitution is dv/sqrt(1 + v^2) + 2/5 * dx/x = 0.

ln[ v + sqrt(1 + v^2) ] + 2/5 ln x = C since v = y/x, by substitution ln [ y/x + sqrt(1 + y^2/x^2) ] + 2/5*ln x = C At y = 0 and x = 100 , then C = ln 100^(2/5).

The equation of the curve is sqrt(x^2 + y^2) = x^(3/5) *100^(2/5) - y.

Square both sides then y = [ 100^(4/5) * x^(3/5) - x^(7/5) ] / [ 2 * 100^(2/5)].

Solve (2) Velocity x component = dx/dt = - 5 cos a = - 5 * x / sqrt( x^2 + y^2 ) in terms of y.

By algebraic manipulation using fractional exponent and factoring the result is 100^(4/5) * x^(-2/5) dx + x^2/5 dx = - 10^(9/5) dt 5/3 *100^(4/5) * x^(3/5) + 5/7 * x^(7/5) = - 10^(9/5) * t + C At x =100 and t = 0 , C = 100^(7/5) * 50/21.

At x = 0 the result is 0 = - 10^(9/5) * t + C , substitute C = 100^(7/5) * 50/21 Solve: 10^(9/5)*t = 100^(7/5) * 50/21.

Therefore t = [100^(7/5)] / [10^(9/5)] * 50/21

t = [ 10^(14/5) ] / 10^(9/5) * 50/21 = 10 *50/21 = 500/21.

Go back to Problem 55
Go back to Shack's Math Problems

Problem 23 Answer

Answer 2: You point to either path and say, "Are you from this village?" If the person answers 'yes' then you are pointing to the truthful village, if they answer 'no' then you are pointing to the lying village.

Answer 3: Ask "Which path would a person from the other village say leads to the truthful (or untruthful) village?" Regardless of whom you speak to the answer will filter through one lie, so if you ask for the truthful village the person will point to the lying village, and vise versa.

Problem 23 Solution

This is not the only answer that works but is the simplist I have heard so far.

I continue to receive a lot of e-mail on this problem from people who don't understand or disagree with my solution. Before you question me on this answer these four questions:

1. If you point to the truthful village what will a truthful person say in answer to your question?
2. If you point to the truthful village what will a untruthful person say in answer to your question?
3. If you point to the untruthful village what will a truthful person say in answer to your question?
4. If you point to the untruthful village what will a untruthful person say in answer to your question?

Note that in the two cases where the answer is "yes" you are pointing to the truthful village and in the two cases where the answer is "no" you are pointing to the untruthful village. Thus you can tell from the reply to which village you are pointing. You do not have to know to whom you are speaking.

Another possible solution which I have heard is to ask "Which path would a person from the other village say leads to the truthful (or untruthful) village?" Regardless of whom you speak to the answer will filter through one lie, so the person will point to the lying village if you ask for the truthful village, and vise versa.

Problem 111 Answer

The radius of the exterior solution is 6.

Problem 111 Solution

The triangle marked by the centers of the three outer circles conveniently forms a right triangle with sides 3, 4, and 5 (3² + 4² = 5²) as seen by the diagram below.

Next draw segments from the center of the interior circle and perpendicular to the outer triangle sides of length 3 and 4.

There are four right triangles in this picture, two of which are the same. Using the pythagorean formula we know:

(1): x² + y² = (1+r)²
(2): x² + (3-y)² = (r+2)²
(3): y² + (4-x)² = (r+3)²

Combining equations (1) and (2) we get y=(3-r)/3
Combining equations (1) and (3) we get x=(2-r)/2

Substituting these values for x and y back into equation 1 we get:

((2-r)/2)² + ((3-r)/3)² = r² + 2r + 1

This leads to...

23*r² + 132r - 36 = 0

Putting this through the quadratic formula we get r=6/23, -6

So the interior circle has radius 6/23 and the outer circle has radius 6.

Here is my old solution, as well as some comments from another reader.

The solution to this problem is rather hard to explain so I'm only going to go over it briefly.

First arrange two of the circles so that their centers lay on the y axis. Then use your trigonomtry to determine the coordinates of the center of the third circle (2.4,-.2) if the large circle is on top, the middle circle on bottom, and the third circle is to the right. See problem 92 for help in how to find such coordinates.

Second, invert all three circles such that the third circle is mapped onto itself. Note that the top and bottom circles are mapped to a horizontal lines that are tanjent to the third circle.

Note: To invert any figure take all points (t,r) and map them onto the point (t,c/r) where t is the angle in polor coordinates and r is the distance from the center. In this case choose the constant c carefully such that the third circle is mapped onto itself (okay, c=4.8 in this case).

The image of the center circle will be tanjent to both horizontal lines and the third circle. Going to the right of the third circle will lead us to the interior circle.

Consider the line that goes through (0,0) and the center of the circle in the above step. The distance from the center of the points where this line intersect the circle are sqr(19.4)-1 and sqr(19.4)+1. Invert these two points and their distance from the center are 4.8/(sqr(19.4)-1) and 4.8/(sqr(19.4)+1). Both these points lay on a diameter of the interior circle so take their difference and divide by 2 to get the radius. The answer simplifies to 6/23.

To find the exterior solution shift the third circle to the left rather than the right. The radius of the exterior solution is exactly 6!

Note: The general case if the third circle has coordinates (x,y) and radius r then the radii of the solution circles are (x² + y² - r²) / (4xr +/- (x²+y²+3*r²)).

The + will determine the interior solution, the - the exterior solution.

I would like to propose a solution, also using some trigonometry. First of
all, some definitions:

   The centres of the four circles:
     'A' for radius 3.
     'B' for radius 2
     'C' for radius 1
     'Z' for the middle circle.

   Two angles:
     'a' between CA and CZ
     'b' between CB and CZ

   Our unknown:
     'x' is the radius of the middle circle

Then, we make some obvious statements:
   CA = 1 + 3 = 4
   CB = 1 + 2 = 3
   AB = 2 + 3 = 5
   AZ = 3 + x
   BZ = 2 + x
   CZ = 1 + x

Followed by a trivial consequence:   
(1)   a + b = PI/2  (because CA^2 + CB^2 = AB^2)

Now we establish two equations using trygonometry:
(2)   BZ^2 = CZ^2 + CB^2 - 2*CZ*CB*cos(b)
(3)   AZ^2 = CZ^2 + CA^2 - 2*CZ*CA*cos(a)

By replacing the values we know and introducing 'x', (2) and (3) become:
(2)   (2 + x)^2 = (1 + x)^2 + 9 - 2*3*(1 + x)*cos(b)
(3)   (3 + x)^2 = (1 + x)^2 + 16 - 2*4*(1 + x)*cos(a)

By expanding the squares and simplifying (I also omit some '*' to make the
formulae more readable) we obtain:

(2)   -6 + 2x + 6(1 + x)cos(b) = 0
(3)   -8 + 4x + 8(1 + x)cos(a) = 0

By adding '+2 -2' to the left side of (2) and '+4 -4' to the left side of
(3), and then grouping '(1 + x)' terms, we obtain:

(2)   (1 + x)(1 + 3cos(b)) = 4
(3)   (1 + x)(1 + 2cos(a)) = 3

>From (1), we can express cos(a) in terms of cos(b):
(1)   cos(a) = cos(PI/2 - b) = sin(b) = sqrt(1 - (cos(b)^2))

By substituting cos(a) in (3):
(3)   (1 + x)(1 + 2sqrt(1 - (cos(b)^2))) = 3

Now we can calculate cos(b) from (2) and substitute it into (3):
(2)   cos(b) = (4/(1 + x) - 1)/3    ==>
(3)   (1 + x)(1 + 2sqrt(1 - ((4/(1 + x) - 1)/3)^2)) = 3

For better reading, we can define y = 1/(1 + x), with which (3) becomes (we
can multiply both sides by (1 + x) because x > 0; therefore, (1 + x) > 0 too):

   1 + (2/3)sqrt(9 - (4y - 1)^2) = 3y

A bit of simple algebra brings us to:

   4sqrt(2 + y - 2y^2) = 9y - 3

We will verify later that '(2 + y - 2y^2) >= 0'. For the time being, we
square both sides and we get (after some simplification):

   145y^2 - 86y -23 = 0

This resolves into: (43 +- 72)/145.  By taking the positive solution, we
obtain:

   y = 23/29   ==>   x = 29/23 - 1 = 6/23.

To check that '2 + y - 2y^2 >= 0', we calculate:

   2 + y - 2y^2 = 2 + 23/29 - 2(23/29)^2 = 1.5+ > 0

I haven't worked on the "external" circle (yet), and my solution relies on
the fact that the centres of the original circles form a right triangle.
You do show a more general solution, but I confess (please, do not take
offense) that I am a bit skeptical that such a simple equation would apply.
In fact, I did not manage to understand it. Perhaps (time permitting), you
could add an explanatory figure.

You say that the two centres A and B are along the Y-axis. What you don't
say is that the X-axis is the tangent between the two bigger circles. This
implies the two following equations (C must be below the X-axis, because
the biggest circle is above it. Therefore, we can write the formulae in
such a way that y calculates to a negative number):
(1)   BC^2 - (2 + y)^2 = x^2
(2)   AC^2 - (3 - y)^2 = x^2

This system provides x = 2.4 and y = -0.2, which is what you say in your
solution.

So far so good, but then I really get lost...


Ciao,    Giulio

Problem 159 Answer

Here is an e-mail I received from Dave Stigant before adding the last two clues.

I think there's another solution with 2 cycles instead of 1: Bruce loves Mary loves Bruce (By my reading this fulfills the requirement that Bruce loves the woman (Mary) who loves the man (Bruce) who loves Mary) Arthur loves Hazel loves Chad loves Ellen loves David loves Gloria

Incidently, assuming everybody is heterosexual there are no other 2 cycle solutions and no 3 cycle solutions. Arthur cannot love Ellen: either Bruce loves Mary (in which case you must have Hazel loves x loves Gloria) or not (in which case either H loves x loves G or G loves B). Chad and David must be in the same cycle, so the only way to have 3 cycles is A loves E and B loves M and the only way to have 2 cycles is if either A loves E or B loves M.

If you allow for homosexuality, we know that Chad, Arthur and Bruce must be straight. We know that Hazel is straight. We know that at least 2 women are straight (A loves girl, Gx, who loves the man who loves E... and B loves the girl, Gy, who loves the man who loves M. Gx and Gy must be straight but not the same since each is loved by a different man) Also, the number of gay men must be the same as the number of gay women (the number of straight men + the number of gay women = the number of women = number of gay men + the number of straight women = the number of men = 4). So if anybody is gay it must be David and one woman other than Hazel.

This is not an exhaustive list, but here are a few possible solutions:

• Chad loves Mary loves David loves Arthur loves Hazel loves Bruce loves Ellen loves Gloria loves Chad
• Chad loves Mary loves David loves Arthur loves Hazel loves Bruce loves Ellen loves Chad and then Gloria loves herself (narcissism is not explicitly ruled out either)
• Chad loves Ellen loves David loves Arthur loves Hazel loves Chad and Bruce loves Gloria loves Mary loves Bruce.

This problem appears in Games for the Superintelligent by James F. Fixx (page 54).

Problem 29 Answer

The total distance covered is a.

Problem 29 Solution

Hard Solution

Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e^-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2^-1/2. So r(x)=2^-1/2*e^-x.

The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))²+(f^'(x))²)^1/2 =
Integral from 0 to infinity of ((2^-1/2*e^-x)² + (-2^-1/2*e^-x)²)^1/2 =
Integral from 0 to infinity of (2^-1*e^-2x + 2^-1*e^-2x)^1/2 =
Integral from 0 to infinity of e^-x =
-e^-x from 0 to infinity =
=0 - (-e⁰) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e^-2*pi/2^1/2 = 0.03055663 times it's size before the revolution.

Easy Solution

Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics.

For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)). So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309 units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

Thus the general formula is 1/(1+cos(t)), where t is an interior angle of the shape.

Number of sides = 3
Lengh of initial side = 1
Each angle = 180*(1-(2/3)) = 60°
Length of path = 1/(1+cos(60)) = 1/(1+0.5) = 2/3

Number of sides = 4
Lengh of initial side = 1
Each angle = 180*(1-(2/4)) = 90°
Length of path = 1/(1+cos(90)) = 1/(1+0) = 1

Number of sides = 5
Lengh of initial side = 1
Each angle = 180*(1-(2/5)) = 108°
Length of path = 1/(1+cos(108)) = 1/(1-.309) = 1.4472

Number of sides = 6
Lengh of initial side = 1
Each angle = 180*(1-(2/6)) = 120°
Length of path = 1/(1+cos(120)) = 1/(1-.5) = 2

Number of sides = 10
Lengh of initial side = 1
Each angle = 180*(1-(2/10)) = 144°
Length of path = 1/(1+cos(144)) = 1/(1-.809) = 5.2361

Number of sides = 15
Lengh of initial side = 1
Each angle = 180*(1-(2/15)) = 156°
Length of path = 1/(1+cos(156)) = 1/(1-.9135) = 11.5668

Problem 151 Answer

• Alice: applied and insane.
• Bob: pure and sane.
• Charlie: pure and insane.
• Dorothy: pure and insane.

I'd like to thank Mathematics and Informatics Quartly 9:4 Dec 1999, page 162 for this problem.

Problem 151 Solution

1. Pure and sane
2. Pure and insane
3. Applied and sane
4. Applied and insane

The first step is to determine what each kind of mathematician would say about themselves:

1. Pure and sane: "I am pure and sane."
2. Pure and insane: "I am applied and sane."
3. Applied and sane: "I am pure and insane."
4. Applied and insane: "I am applied and insane."

Note that the applied and insane mathematician is lying about a lie so is actually speaking the truth.

Next consider the first statement by Alice, "I am insane." Based on the four groups there are two that will say they are insane:

1. The applied and sane.
2. The applied and insane.

By the same logic we can determine that Bob is sane, Charlie is insane, and Dorothy is pure.

Now we know one of the two characteristics of each mathematician. Next jump to the last statement by Dorothy, "Charlie is sane." We already know that Dorothy is pure so she actually believes that Charlie is sane. However we also know that Charlie is infact insane so Dorothy's belief must be incorrect. Since she is not deliberately telling a lie her belief is incorrect, making her insane. So Dorothy is pure and insane.

Let review what we know thus far:

1. Alice: applied.
2. Bob: sane.
3. Charlie: insane.
4. Dorothy: pure and insane.

Next look at what Bob says about Dorothy, "Dorothy is insane." This is a true statement and since Bob is already sane we can conclude that he is also pure. Now we have:

1. Alice: applied.
2. Bob: sane and pure.
3. Charlie: insane.
4. Dorothy: pure and insane.

Next consider what Charlie says about Bob, "Bob is applied." This is not true since Bob is pure. We already know Charlie is insane thus he must also be pure. If he were applied he would be lying about a lie, thus telling the truth. Now we have:

1. Alice: applied.
2. Bob: sane and pure.
3. Charlie: pure and insane.
4. Dorothy: pure and insane.

Finally consider Alice's second statement, "Charlie is pure." Charlie is pure thus Alice is making a true statement. However Alice is applied, thus she thinks she is telling a lie. So Alice is lying about an incorrect belief, making a true statement. Since Alice's belief is incorrect she must be insane. Thus:

1. Alice: applied and insane.
2. Bob: pure and sane.
3. Charlie: pure and insane.
4. Dorothy: pure and insane.

I'd like to thank Mathematics and Informatics Quartly 9:4 Dec 1999, page 162 for this problem.

Problem 95 Answer

Second, A and B cross together and D returns (11 minutes consumed).

Third, C and D cross together (2 minutes consumed).

Thanks to Antonio Cordoba for this problem. He says that Microsoft applicants are tested with this one.

Problem 95 Solution

Problem 148 Answer

1. Ronald and Jimmy cross in 2 minutes, total time 2 minutes.
2. Ronald crosses back in 2 minutes, total time 4 minutes.
3. Bill and George cross in 10 minutes, total time 14 minutes.
4. Jimmy cross back in 1 minute, total time 15 minutes.
5. Ronald and Jimmy cross in 2 minutes, total time 17 minutes.

Answer 2:

1. Ronald and Jimmy cross in 2 minutes, total time 2 minutes.
2. Jimmy crosses back in 1 minutes, total time 3 minutes.
3. Bill and George cross in 10 minutes, total time 13 minutes.
4. Ronald cross back in 2 minutes, total time 15 minutes.
5. Ronald and Jimmy cross in 2 minutes, total time 17 minutes.

Thanks to Glenn Yoshimura for this problem. He says it has been asked to prospective employees at Microsoft, with five minutes given to solve it.

Problem 14 Answer

  3     5
pint  pint
glass glass
----- -----
  0     5
  3     2
  0     2
  2     0
  2     5
  3     4

  3     5
pint  pint
glass glass
----- -----
  3     0
  0     3
  3     3
  1     5
  1     0
  0     1
  3     1
  0     4

In other words, fill the 3 pint glass and pour it into the 5 pint glass. Then fill it again and from it top off the 5 pint glass, leaving 1 pint in the 3 pint glass. Then empty the 5 pint glass and move the 1 pint to the 5 pint glass. Then fill the 3 pint glass and pour that into the 5 pint glass. A problem like this appeared in the movie Die Hard III.

Problem 14 Solution

  3     5
pint  pint
glass glass
----- -----
  0     5
  3     2
  0     2
  2     0
  2     5
  3     4

At this point there are four pints in the five pint glass.

A problem like this appeared in the movie Die Hard III.

Problem 73 Answer

Problem 73 Solution

Number   Probability
  of         of
Trials     Success
------   -----------
  1           p
  2           pq
  3           pq2
  4           pq3
  .           .
  .           .
  .           .

The mean number of trials (m) is: m = p + 2pq + 3pq² + 4pq³ + ...

mq = pq + 2pq² + 3pq³ + ...

m-qm= p + pq + pq² + pq³ + ...

m(1-q) = 1.

m = 1/(1-q) = 1/p.

Second, the answer to the problem can be express as the sum of the following:

• Number of trials to get a first toy
• Number of trials to get a second toy once you have one toy
• Number of trials to get a third toy once you have two toys
• Number of trials to get the final toy once you have three toys

Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Summing these yields 1 + 4/3 + 2 + 4 = 25/3.

I'd like to thank Michael Brasher for providing this solution. Below is my old solution which is much less elegant, I recommend you ignore it.

The probability that in j boxes there will be three or less kinds of toys is 4*(3/4)^j where j>3.

The probability that in j boxes there will be exactly three kinds of toys is 4*(3/4)^j - 12*(1/2)^j + 12*(1/4)^j. This is rather hard to explain why but basically there are 4 different sets of 3 toys, 6 different sets of 2 toys, and 4 different sets of 1 toy, use this information to not double count combinations of 2 prizes or 1 prize.

Thus the answer is 1/4*sum for j=3 to infinity of j*(4*(3/4)^j - 12*(1/2)^j + 12*(1/4)^j).

Theorem: The sum for j=1 to infinity of i*(1/n)ⁱ = n/(n-1)².

From there is is just simple math to find that the answer is 25/3.

Problem 74 Answer

I got this answer from a Usenet discussion on probability, I can't prove it myself at this time.

Problem 134 Answer

Problem 134 Solution

Although the United States can make either product more efficiently it has a comparative advantage in food, and Mexico has a comparative advantage in clothing.

Without free trade 2 units of food will be worth 1 unit of clothing in the United States. In Mexico 4 units of food will be worth 3 units of clothing. So the exchange rate of food to clothing in the U.S. will be 2:1, or 6:3, in the US and 4:3 in Mexico.

Next lets assume the traders set a compromise exchange rate of 5:3, or 5 units of food for 3 units of clothing. Before free trade if someone in the U.S. had 60 units of food they could exchange that for 30 units of clothing. With free trade they could exchange the 60 units of food for 36 units of clothing. Before free trade if someone in Mexico had 60 units of clothing they could exchange that for 80 units of food. With free trade they could exchange the 60 units of clothing for 100 units of clothing. The trader will be taking in food from the U.S. and clothing from Mexico, allowing for supply and demand for both products and no shortages or surpluses of either.

The net effect will be that the price of clothing will drop in the U.S. from 2 days labor to 5/3 of a day, and the price of food will drop in Mexico from 3 days labor to 2.4 days.

Yes, there will be a temporary loss of clothing jobs in the U.S. and farming jobs in Mexico but with a movement in the labor force into more farming in the U.S. and more clothing in Mexico everybody will be able to enjoy a higher standard of living.

For more information please read the chapter 'International Trade and the Theory of Comparative Advantage' in Economics by Paul J. Samuelson which is seems to invariably be the textbook of choice for freshmen in economics.

Problem 28 Answer

Problem 28 Solution

Let c be the number of chairs produced per hour and t the number of tables produced per hour.

The number of saws limit the combination of chairs and tables to 600=10c+5t.
The number of lathes limit the combination of chairs and tables to 360=5c+5t.
The number of sanding machines limit the combination of chairs and tables to 1080=5c+20t.

Next graph these three lines. It should be expected that the answer will lie on the intersection of two of these lines or to make all chairs or all tables. The intersection of the saw and sanding machine line occurs outside of how many chairs the lathe can make so this combination is not a viable answer. The saw and lathe lines cross at 48 chairs and 24 tables. The lathe and sanding machine lines cross at 24 chairs and 48 tables.

Next determine the revenue at all points of intersection.

So the optimal answer is to make 24 chairs and 48 tables for revenue of $1200 per hour.

To check it will take 24*10 + 48*5 = 480 minutes of saw time. There are 600 minutes available so the saws will be idle 20% of the time.

It will take 24*5 + 48*5 = 360 minutes of lathe time which is exactly what we have.

It will take 24*5 + 48*20 = 1080 minutes of sanding machine time which is exactly what we have.

I would like to thank Brain Storm for this problem.

Problem 75 Answer

Here are some approxmiate probabilities for number of people n:

n       Prob(no matches)
--      ----------
2	50.000000%
3	33.333333%
4	37.500000%
5	36.666667%
6	36.805556%
7	36.785714%
8	36.788194%
9	36.787919%
10	36.787946%
11	36.787944%
12	36.787944%
13	36.787944%
14	36.787944%
15	36.787944%

Problem 75 Solution

There are 12! ways of drawing names.

Next determine the number of these 12! ways that have no matches.

Subtract from 12! the number of ways that one person matches, which would be 12*11! (12 people times 11! ways to arrange the other 11 names).

However this would be double counting those combinations in which two people match, which is (12:2)*10! = 12!/2!. Note that (x:y) means x choose y or x!/(y!*(x-y)!). So add this group back in.

Yet we must take out from those we put in those combinations in which three people match, which is (12:3)*9! = 12!/3!.

So keep repeating this process and the numerator or number of combinations in which there are no matches is 12! - 12! + 12!/2! - 12!/3! + 12!/4! - 12/5! ... + 12!/12!.

Divide this by the total number of combinatios, 12!:

1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! ... +1/12!.

This same method works for any n.

Note that it approaches 1/e.

Problem 114 Answer

Problem 114 Solution

Let S_A² = The sum for i=1 to 8 of (s_i-1.5125)² = .10875, where S_i is the ith sample from location A.

Let S_B² = The sum for i=1 to 6 of (s_i-1.5125)² = .167673, where S_i is the ith sample from location B.

Let U be the statistic for testing the the hypothesis that the mean of location A does not equal the mean of location B.

U = (sqr(8+6-2)*(1.5125-1.6683)) / (sqr(1/8 + 1/6)+sqr(.10875+.167673)) = -1.693354 .

The t value, with 12 degrees of freedom, at the .95 level of significance is 1.782 .

Thus if -1.782 <= U <= 1.782 the test will pass, which it does.

Problem 220 Answer

Here is the answer, including two patterns Will didn't get.

Problem 220 Solution

This solution is nothing very scientific but my attempt at a common sense answer.

1. First, there is the case of just one intersection:
   
   
   
   
2. Second, let's consider all the cases of two intersection points.
   
   Let's start with the case of two nodes on one side. Considering both intersection points, and the two nodes on one side, there are six nodes left for the other side:
   
   
3. Next, move a node from the side with six to the side with two, for a 3-5 split:
   
   
   
   
4. Next, move a node from the side with five to the side with three, for a 4-4 split:
   
   
   
   
5. Third, let's consider all the cases of three intersection points.
   
   Let's start with the case of two nodes off one side. In the middle we can have just one node, since it has two intersection lines braching off of it. Consider the three intersections and three nodes, there are four nodes left for the other end:
   
   
   
   
6. Next, move a node from the side with four to the other end with only two:
   
   
   
   
7. Next, move a node from either side to the middle:
   
   
   
   
8. Next, move a node from the other side (that still has three nodes) to the middle:
   
   
   
   
9. Fourth, let's consider all the cases of four intersection points in a row.
   
   Considering the two ends must have at least two nodes and the middle two must have at least one each, and the four intersection points themselves, we've already spoken for all ten points:
   
   
   
   
10. Fifth, let's consider all the cases of four intersection points where there is one in the middle and three other branching off.
    
    We've already spoken for four intersection points and each must have at least two nodes branching off, so we've already spoken for all 4+6=10 nodes:
    
    
    
    

Here is the full answer:

Binomial Distribution Solution

The table below shows the probability of specific numbers of death using this formula:

The table shows that if 14 graves were dug then the probability of not running out would be 0.91758768 (the smallest number greater or equal to 90%). So they must dig 14 graves to have a 90% of not running out. To have a 95% chance they must dig 15 and to have a 99% chance they must dig 18.

Normal Distribution Solution

Pr(d <= g+.5) = 0.90 (The 0.5 is added because the number of deaths must be an integer)
Pr(d-9.5 <= g-10) = 0.90
Pr((d-9.5)/3.1464 <= (g-10)/3.1464) = 0.90
Pr(Z <= (g-9.5)/3.1464) = 0.90
(g-9.5)/3.1464 = 1.28
g =~ 13.527

The number of graves must be an integer, thus we round up to 14.

Problem 25 Answer

Problem 25 Solution

1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

From the statement that the sum equals the house number it is possible to eliminate all but two possibilities. The sums of the rest are unique and would allow for an immediate answer. For example if the house number were 16 the ages must be 1, 3, and 12. The two remaining possibilities are 2, 2, and 9; or 1, 6, and 6.

After the clue that the oldest has red hair you can eliminate 1, 6, and 6 because the oldest two have the same age thus there is no oldest son. The only remaining posibility is 2, 2, and 9.

Problem 211 Answer

Question

Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place.

What is the speed of the river current?

Answer

Three miles per hour.

Problem 211 Solution

Question

Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place.

What is the speed of the river current?

Easy Solution

If Joe paddles upstream and them downstream the same amount of time in both directions then his effort paddling in both directions an equal amount of time would cancel out. In other words, he would end up in the same place if he floated the whole time.

We know he paddled upstream for ten minutes. So, after paddling downstream for another ten minutes he would end up in the same spot as he would had he floated for 20 minutes. We also know from the problem that in this same 20 minutes the hat floated one mile. So if the hat travels a mile in 20 minutes then it would travel three miles in an hour, for a speed of 3 m.ph..

Hard Solution

Let:

p=paddling speed in still water.
c=current of the water.
d=distance Joe paddled upstream since his hat fell out of the canoe.

Recall that distance = rate * time. Consider the unknown distance Joe went upstream since his hat fell out:

(1) d=(p-c)*(1/6)

This is because his net speed is the paddling speed less the current speed. We know he paddled without his hat upstream for 1/6 of an hour.

We're also given that it took the same time for the hat to float downriver a mile as it took for Joe to travel d miles upstream and 1+d miles downstream. Let's set up an equation, balancing for time.

(2) Time for hat to float one mile downstream = Time for Joe to travel d miles upstream + Time for Joe to travel 1+d times downstream.

Time for hat to float one mile downstream = one mile/rate of current = 1/c.

Time for Joe to travel d miles upstream = 1/6 (in hours), as provided in the question.

Time for Joe to travel 1+d times downstream = distance Joe traveled downstream / rate Joe traveled downstream = [1 + (1/6)*(p-c) ]/ (p+c).

The distance is the one mile the hat floated in the river + the (1/6)*(p-c) from equation (1). Joe’s speed going downstream is p+c, which is the sum of his paddling speed and current speed.

Now we’re ready to solve for equation (2)

1/c = (1/6) + [1 + (1/6)*(p-c) ]/ (p+c)
1/c = (1/6) + (6+p-c)/(6p+6c)
6/c = 1 + (6+p-c)/(p+c) Multiplying both sides by 6.
6/c = (p+c)/(p+c) + (6+p-c)/(p+c) Finding a common denominator
6/c = (6+2p)/(p+c)
6p + 6c = 6c + 2pc
6p = 2pc
3=c
So the current is 3 miles per hour.

Problem 156 Answer

The answer is 55/21 minutes.

Problem 156 Solution

Here is my solution to problem 156.

Problem 123 Answer

Problem 123 Solution

Solution #1

Solution #2

Let x be the height of the pentagon and r be the distance from the center to any corner. The height of the pentagon was solved in problem 47 as (2*5^1/2+5)^1/2/2. The area of the pentagon equals 10 times the area of the triangle shown in the following diagram.

We already know r so let's solve for x in terms of r using the pythagorean formula on the triangle in the diagram above.

(1/2)² + (x-r)² = r²

1/4 + x² -2xr + r² = r²

1/4 + x² -2xr = 0

r = 1/(8x) + x/2

So x-r = x/2 - 1/(8x)

The area of each triangle is 1/2*base*height = x/8 - 1/(32x)

The area of the pentagon is 10 of these triangles = 10x/8 - 10/(32x) = 5x/4 - 5/(16x) = (4x² - 1)/(32x).

We know from problem 47 that x = (2*5^1/2+5)^1/2/2. Substituting this value for x in the above equation results in an area of (25+10*5^1/2)^1/2/4 =~ 1.7205.

Solution #3

The next step is hard to explain without a picture, however I will try. Consider three consecutive corners of the pentagon and label them a, b, and c. Next draw a line from a to c. Call d the point in the middle of segment ac. Now consider the triangle abd. This triangle has angles 36, 54, and 90. We also know the hypotenuse is 1 and longer base is half the distance between any two non-adjacent corners, or (1+sqr(5))/4. From this we can deduce that cos(36)=(1+sqr(5))/4.

Next extend a segment from all five corners of the pentagon and all five midpoints of each side. This will divide the pentagon into 10 equal triangles. Each of these triangles conveniently also has angles of 36, 54, and 90. Next lets define the following:

r=distance from any corner of pentagon to center of pentagon
y=distance from the midpoint of any side to the center of pentagon

r+y is the height of the pentagon, which from problem 47 is sqr(5+2*sqr(5))/2.

The ratio y/r = cos(36) = (1+sqr(5))/4.

Now we have two equations and two unknowns. At this point we grind out the algebra to find that:

r=2*sqr(5+2*sqr(5))/(5+sqr(5))=~0.850651

y=sqr(25+10*sqr(5))/10=~0.688191

The pentagon is composed of 10 right triangles, each of area 1/2*1/2*y. The area of the pentagon is thus (10/4)*y = sqr(25+10*sqr(5))/4 =~ 1.720477.

Thanks to Mark Hoyle for some help with this problem.

Problem 47 Answer

Problem 47 Solution

Let x be the height of the pentagon and a be the length of the extensions of the sides, as shown in the following diagram:

Using the Pythagorean formula we know x² + (a+1/2)² = (a+1)², or

(1) x² = a + 3/4.

Now consider the added two chords in the next diagram:

The isosceles triangle formed by these two chords and one side of the pentagon is congruent to the triangle formed by two extensions of length a and the same side of the pentagon.

Thus the distance between two non-adjacent corners of the pentagon is also a. Consider the chord going from either corner of the base to the top of the pentagon. We can use the Pythagorean formula, with this chord as the hypotenuse to obtain:

x² + (1/2)² = a².

Substituting for x² from above we can solve for a:

a + 3/4 = a² - (1/2)², or

a² - a - 1 = 0, or

a=(1+sqr(5))/2.

Now that we know a we can solve for x:

x² = ((1+sqr(5))/2)² - 1/4, or

x² = (5+2*sqr(5))/4, or

x = sqr(5+2*sqr(5))/2 =~ 1.538842

Old Solution

• sin(5x)=5*sin(x)-20*sin³(x)+16sin⁵(x).

• tan(x)=sin(x)/(1-sin²(x))^1/2.

• The interior angles of an n-gon are each 180*(n-2)/n.

With some simple trig we can express the height as 1/2*[1/sin(36) + 1/tan(36)].

From the formula in the hint we can express tan(36) as sin(36)/sqr(1-sin²(36)).

After a few simple steps we can express the height as (1) 1/2*[(1+sqr(1-sin²(36))/2*sin(36)].

Next lets work on expressing sin(36).

From the formula in the hint we know sin(180) = 16*sin⁵(36) - 20*sin³(36) + 5*sin(36)

= 16*sin⁴(36) - 20*sin²(36) + 5

= 16x⁴ - 20*x² + 5 (where x=sin(36)).

= 16y² - 20*y + 5 (where y=x²).

y = (5-sqr(5))/8.

x = sqr((5-sqr(5))/8) = sin(36).

Now put this expression for sin(36) into equation (1):

1/2 * [ (1+sqr(1-(5-sqr(5))/8)) / sqr((5-sqr(5))/8) ] =

1/2 * [ (1+sqr(3+sqr(5))/8)) / sqr((5-sqr(5))/8) ] =

1/2 * [ sqr(8/(5-sqr(5))) + sqr((3+sqr(5))/(5-sqr(5))) ] =

1/2 * [ (sqr(8) + sqr((3+sqr(5)))/(5-sqr(5))) ] =~ 1.5389

Problem 78 Answer

Thanks to Marilyn Vos Savant for this one.

Problem 78 Solution

George can paint at a rate of 1/30th of a house per hour.

Combine their efforts and they can paint 1/20 + 1/30 = 1/12th of a house in an hour.

So if their rate is 1/12th of a house per hour it will take 12 hours to paint an entire house.

Thanks to Marilyn Vos Savant for this one.

Problem 87 Answer

Problem 87 Solution

The plane will run out of fuel after 10/3 hours.

The distance the plane can cover is the integral from 0 to 10/3 of (6,000,000/(20,000-3,000t)) dt.

= 6,000,000 * -1/3000 * ln(20,000-3,000t) from 10/3 to 0.

= -2,000 * ln(1/2)

= 2000 * ln(2) =~ 1386.3 miles .

Problem 171 Answer

Problem 171 Solution

Let's define the number of direct routes from A to B as x.
Let's define the number of direct routes from B to C as y.
Let's define the number of direct routes from A to C as z.

From the information given we know:
x + yz = 82
y + xz = 62

Adding them we get:
x + xz + y + yz = 144
(x + y) * (z + 1) = 144

and subtract them to get:
x - xz - y + yz = 20
(y - x) * (z - 1) = 20

The factors of 20 are {1, 2, 4, 5, 10, 20}. So (z - 1) is one of {1, 2, 4, 5, 10, 20}. then (z + 1) is one of {3, 4, 6, 7, 12, 22}.

The factors of 144 are: {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. So (z + 1) must be one of these factors.

Common possible (z + 1) values are {1, 2, 3, 4, 6, 12}. Given (x + y) * (z + 1) = 144, possible values of (x + y) are: {144, 64, 48, 36, 24, 12}.
Possible values of (z - 1) are: {-1, 0, 1, 2, 4, 10}.
Concurrently since (y - x) * (z - 1) = 20, then possible values of (y - x) are: {20, 10, 5, 2, -20}.

This leads to five possible solutions. Note that x+y=64 is not listed because then z would be 1 and (y-x)*(z-1) could not equal 20 for any values of x and y.

1. If x+y=144 then z=0, y-x=-20, y=62, x=82
2. If x+y=48 then z=2, y-x=20, y=34, x=14.
3. If x+y=36 then z=3, y-x=10, y=23, x=13.
4. If x+y=24 then z=5, y-x=5, y=14.5, x=9.5.
5. if x+y=12 then z=11, y-x=2, y=7, x=5

The fourth solutions is not possible because there can not be half a path. The problem limits us to z+xy<300, so that eliminates the first three. Thus the fifth solution is the correct one. If x=5, y=7, and z=11 then the final answer is z+xy=11+5*7=46.

Alternative Solution

Let's define the distance from A to B as x.
Let's define the distance from B to C as y.
Let's define the distance from A to C as z.

The information given tells us that:

x+yz=82
y+xz=62
z+xy<300

Combining the first two equations and solving for z we get:

z=(82-x)/y, and z=(62-y)/x

Equating the two we get x/y = (62-y)/(82-x)

Unfortunately trial and error seems the only solution from here. The way I did it was to create a spreadsheet with values of x along one axis and values of y along the other. In the body of the spreadsheet I had the following formula (62-y)/(82-x) - (x/y). Then I simply looked for a zero.

If your ranges for x and y are small then the values x=5 and y=7 should pop out, which lead to the correct solution. However other values that work are (x,y)=(13,23), (14,28), (14,34), (12,42), (5,55).

The problem tell us that z+xy<300. Multiplying x and y easily eliminates all possible values except (5,7) and (13,23). If x=23 and y=23 then z is easily solved as 3. In this case the total paths from A to C would be 3+13*23=302, which is too many.

So x=5 and y=7. Simple algebra tells us that z=11. So 11+5*7 = 46.

I would like to thank Recreations in the Theory of Numbers by Albert H. Beiler for this problem (#42 on page 299).

x*(15/11) + y*(16/11) + z*(17/11) = 11. Multiplying by 11:

15x + 16y + 17z = 121.

From here we have to use trial and error or educated guessing. Personally I divided 121 by 15 to get 8 plus a remainder of 1. It was then obvious that x=7, y=1, and z=0 would solve the equation.

Problem 170 Answer

Problem 170 Solution

i = 2*((i+j)/2 + (j-i)) + (j-i)
= 2*(3j-i)/2 + (j-i)
= 4j - 2i
= (4/3)*j

The second piece of information tells us...

j = (j+10)/2 + (i-j)
= i - j/2 + 5 = (2i+10)/3

Combining the two equations we get...

i = (4/3)*(2i+10)/3
= (4/9)*(2i+10)
= (8/9)*i + 40/9
= 40

So I am 40 (and John is 30).

Problem 169 Answer

Problem 169 Solution

The probability that 0 flies were eaten is combin(5,0)*(1/2)⁵=1/32.

The probability that 1 fly was eaten is combin(5,1)*(1/2)⁵=5/32.

The probability that 2 flies were eaten is combin(5,2)*(1/2)⁵=10/32.

So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 1-1/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .

Problem 63 Answer

Problem 188 Answer

Problem 188 Answer

This could be rephrased as x=1+(1/x).

x-1 = 1/x

x²-x-1 = 0

x=(1+5^1/2)/2

Problem 133 Answer

• 1 person $1
• 1 person $7
• 3 people $49
• 3 people $343
• 3 people $2401
• 3 people $16807
• 1 person $117649
• 1 person $823543.

I would like to thank Recreations in the Theory of Numbers by Albert H. Beiler for this problem (#52 on page 300).

Problem 133 Solution

1,000,000 = a + 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.

Next divide 1,000,000 by 7. If 1,000,000 is divisible by 7 we could divide the money without giving anyone only $1. However 1,000,000/7 = 142857 plus a remainder of 1. By giving one person $1 (a=1) we now have:

999,999 = 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.

Next divide by 7:

142,857 = b + 7c + 49d + 343e + 2401f + 16807g + 117469h.

Dividing 142,857 by 7 we get 20,408 plus a remainder of 1. So let b=1, divide by 7, and we get:

20,408 = c + 7d + 49e + 343f + 2401g + 16807h .

Dividing 20,408 by 7 we get 2,915 plus a remainder of 3. So let c=3, divide by 7, and we get:

2915 = d + 7e + 49f + 343g + 2401h.

Dividing 2915 by 7 we get 416 plus a remainder of 3. So let d=3, divide by 7, and we get:

416 = e + 7f + 49g + 343h.

Dividing 416 by 7 we get 59 plus a remainder of 3. So let e=3, divide by 7, and we get:

59 = f + 7g + 49h.

Dividing 59 by 7 we get 8 plus a remainder of 3. So let f=3, divide by 7, and we get:

8 = g + 7h.

Obviously g=1 and h=1.

I would like to thank Recreations in the Theory of Numbers by Albert H. Beiler for this problem (#52 on page 300).

Problem 153 Answer

This is a Society of Actuaries sample problem for course 100.

Problem 153 Solution

E(a) = 200*.01=2
E(a²) = 200²*.01=400
Var(a)=E(a²) - (E(a))² = 400 - 4 = 396

E(b) = 100*.05=5
E(b²) = 100²*.05=500
Var(b)=E(b²) - (E(b))² = 500 - 25 = 475

The variance of total claims is 500*396+300*475 = 340,500
The standard deviation of total claims is 340500^1/2 =~ 583.52

Expected total claims is 500*2 + 300*5 = 2500

Pr(total claims <= total revenue) = .95
Pr(C<=R) = .95 (where C=total claims, R=total revenue)
Pr(C<=R) = .95
Pr(C-m<=R-m) = .95 (where m is the expected total claims)
Pr(C-2500<=R-2500) = .95
Pr((C-2500)/s<=(R-2500)/s) = .95 (where s is the standard deviation)
Pr((C-2500)/583.52<=(R-2500)/583.25) = .95

We can now use the central limit theorem because (C-2500)/583.52 has a normal distribution with mean of 0 and stanrdard deviation of 1.

(R-2500)/583.25 = 1.645
R = 3459.89

So the insurance company needs 3459.89 in revenue. The expected costs of claims is 2500. So k = 3459.89/2500 = 1.38

This is a Society of Actuaries sample problem for course 100.