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 207 Answer

Question

Without taking it apart, how many permutations does a Rubik's Cube have?

8! × 12! × 3⁸ × 2¹² = 519,024,039,293,878,000,000

Problem 207 Answer

Question

Without taking it apart, how many permutations does a Rubik's Cube have?

The six center faces of the cube are fixed. By turning the faces all you can do is rearrange the corners and edges. If you took the cube apart, then there would be 8!=40,320 ways to arrange the eight corners, without respect to the orientation of each piece. Likewise, there are 12!=479,001,600 ways to arrange the 12 edges without regard to orientation.

There are 3 ways each corner can be oriented, for a total of 3⁸=6,561 corner orientations. Likewise there are two ways each edge piece can be oriented, for a total of 2¹²=4,096 edge orientations.

So if we could take the cube apart, and rearrange the edge and corner groups, then there would be 8! × 12! × 3⁸ × 2¹² = 519,024,039,293,878,000,000 possible permutations. However, not all of these permutations can be arrived at from the starting position by rotating the faces.

First, it is impossible to rotate just one corner and leave everything else the same. No combination of turns will achieve that. Basically, every action has to have a reaction. If you wish to rotate one corner, it would disturb the other pieces somehow. Likewise, it is impossible to flop just one edge piece. For these reasons, we have to divide the number of permutations by 3 × 2 = 6.

Second, it is impossible to switch two edge pieces without disturbing the rest of the cube. This is the hardest part of this answer to explain. All you can do with a Rubik's Cube is rotate one face at a time. Each movement rotates four edge pieces and four corner pieces for a total of eight pieces moved. A sequence of rotations can be represented by a number of piece movements divisible by 8. Often a sequence of moves will result in two movements canceling each other out. However, there will always be an even number of pieces moved with any sequence of rotations. To swap two edge pieces would be one movement, an odd number, which can not be achieved with the sum of any set of even numbers. Mathematicians would call this a parity problem. So we have to divide by another 2 because two edge pieces cannot be swapped without other pieces being disturbed.

So there are 3 × 2 × 2 = 12 possible groups of Rubik's Cube permutations. If you disassembled a Rubik's Cube and put it back together randomly, there is a 1 in 12 chance that it would be solvable. So the total number of permutations in a Rubik's Cube is 8! × 12! × 3⁸ × 2¹² / 12 = 43,252,003,274,489,900,000. If you had seven billion monkeys, about the human world population, playing randomly with the Rubik's cube, at a rate of one rotation per second, a cube will pass through the solved position on average once every 196 years.

Problem 208 Answer

Question

A cylinder with an open top, has a surface area of S. What is the maximum volume it can hold?

The answer is [S²×(3×pi/S)^1/2 - pi×(S/(3×pi))^1/2]/(6*pi)

height=(S-pi×r²)/(2×pi×r)

radius=(S/(3×pi))^1/2

Problem 208 Solution

Question

A cylinder with an open top, has a surface area of S. What is the maximum volume it can hold?

Problem 209 Answer

The answer is 8 × ln(2^0.5 + 1) =~ 7.051.

Problem 209 Solution

Please see my solution to problem 209 (PDF -- 384K).

Problem 210 Answer

Question

Given an equal amount of paper, which paper cup would hold more water, a cylinder or cone shape? Assume optimal dimensions in both cases.

The answer is the cone holds 7.4569932% more water than the cylinder.

Problem 210 Solution

Question

Given an equal amount of paper, which paper cup would hold more water, a cylinder or cone shape? Assume optimal dimensions in both cases.

The answer is the cone holds 7.4569932% more water than the cylinder.

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 212 Answer

Answer

If a logitian see both other hats are the same color then he should guess the opposite color.

The probability of winning is 75%.

Problem 213 Answer

Answer

Player X Strategy:

• 0.7 to 1.0: Bet
• 0.1 to 0.7: Check
• 0.0 to 0.1: Bet (bluff)

Player Y Strategy after a bet:

• 0.4 to 1.0: Call
• 0.0 to 0.4: Fold

The expected value of X is $0.10.

Problem 213 Solution

Solution

My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.

First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. In this case, X should obviously raise above a certain point. Call it x2.

Second, X should also bluff below a certain point, call it x1.

With anything above x1 and below x2, X should check.

Y is the last to act, so there is no opportunity to bluff. He should simply fold below a certain point, and raise above it. Call that point y1.

The crux of their solution is to find indifference points where each player perceives the same expected value between two options. Let’s look at this problem as an example.

Next, we must decide which number will be greater, x1 or y1. It seems obvious to me that X will bluff only with a pretty low probability. Meanwhile, Y fold after a raise fairly often. So, based on common sense, I assume that x1

Now we're ready to get our hands dirty with some math. Let's start by finding an equation for x1. We do this by examining the possible outcomes for all possible values of Y, and how much X will win whether he bluffs or checks with exactly x1 points.

To be more specific,

• If Y has 1 to y1, then X will win -2 (or lose 2) if he bluffs and win -1 if he checks. The gain by bluffing is -1. The probability Y has a hand in this range is 1-x2. So, the expected increase in expected value by X bluffing with x1 is (1-y1)*-1 = y1-1.
• If Y has y1 to x1, then X will win 1 if he bluffs and win -1 if he checks. The gain by bluffing is 2. The probability Y has a hand in this range is y1-x1. So, the expected increase in expected value by X bluffing with x1 is (y1-x1)*2 = 2y1-2x1.
• If Y has x1 to 0, then X will win 1 if he bluffs and win 1 if he checks. In other words, X will win either way. So, the gain by bluffing is 0, and the expected value is also 0.

Here is a table the lays out the expected win by bluffing and checking if X has exactly X1, according to the value of Y1.

x1

The sum of the expected value column is -2x1+3y1-1. As stated above, we wish to set x1 to a value where the player is indifferent between bluffing and checking. So, we should set this expected value equation of the gain by bluff equal to zero. In other words, -2x1+3y1-1 = 0.

x2

Next, do the same thing, solving for x2. I will go right to the expected value table, which shows the expected value of raising, as opposed to checking, according to the value of Y.

Setting the sum of the expected value column equal to 0 we get 2x2-y1-1=0.

y1

Next, we do a table for y1, showing the expected gain by calling, as opposed to checking, for every possible value of X.

Setting the sum of the expected value column equal to 0 we get 3x1+x2-1=0.

Now we have three equations and three unknowns. Again they are:

-2x1+3y1-1=0
2x2-y1-1=0
3x1+x2-1=0

With a little matrix algebra we can solve for x1, x2, and y1 as follows:

x1=0.1
x2=0.7
y1=0.4

In other words, X should raise with less than 0.1 or more than 0.7, otherwise check. Assuming X raises, Y should fold with 0.4 or less, otherwise call.

The next question is what is the expected value for X if both players follow this strategy. To answer that question, let's consider how much X wins according to every significant grouping of X and Y.

Next, here is the probability of each combination of X and Y.

Finally, here is the expected value of the how much X will win for each combination, which is equal to the product of the win and the probability.

The sum of each cell in this table is 0.1. So, assuming the ante is $1 per player, X can expect to win 10 ¢ per hand.

Problem 214 Answer

Answer

The starting bankrolls were 12, 21, and 39.

Problem 214 Solution

Solution

Let's call the players P1, P2, and P3, and the starting bankrolls as follow:

• P1: x
• P2: y
• P3: z

Assume that P1 loses the first round. After he pays the winners the bankrolls we be:

• P1: x-y-z
• P2: 2y
• P3: 2z

Assume that P2 loses the second round. After he pays the winners the bankrolls we be:

• P1: 2x-2y-2z
• P2: -x+3y-z
• P3: 4z

Assume that P3 loses the third round. After he pays the winners the bankrolls we be:

• P1: 4x-4y-4z
• P2: -2x+6y-2z
• P3: -x-y+7z

All three of these sums must equal 24. So, we have three equations and three unknowns. That is enough to do a some simple matrix algebra to get x=39, Y=21, Z=12.

My thanks to WongBo for this problem.

Problem 215 Answer

My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.

First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. Let's define some variables to this strategy as follows.

• x1 = If x has below x1 then he will bluff.
• x2 = If x checks, and then y bets, then x will fold with under x2, otherwise call.
• x3 = If x has above x3 then he will bet.
• y1 = If x checks, then y will bluff with under y1.
• y2 = If x bets, then y will fold with under y2, otherwise call.
• y3 = If x checks, then y will bet with over y3.

Here are the values for these indifference points.

Assuming the ante is $1 per player, x can expect to lose 5.55556 ¢ per hand.

Problem 215 Solution

Solution

My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.

First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. Let's define some variables to this strategy as follows.

• x1 = If x has below x1 then he will bluff.
• x2 = If x checks, and then y bells, then x will fold with under x2, otherwise call.
• x3 = If x has above x3 then he will bet.
• y1 = If x checks, then y will bluff with under y1.
• y2 = If x bets, then y will fold with under y2, otherwise call.
• y3 = If x checks, then y will bet with over y3.

x1

Player x should be indifferent to checking and bluffing at x1. Let's take a look at how much x would win or lose both ways according to the value of y. The "difference" column shows the benefit to bluffing if x has exactly x1, and y has a value in the specified range. The "expected value" column shows the expected amount y can expect to win by bluffing, which is the product of the "difference" column and the probability that y has a value in the specified range.

The sum of the expected value column is 3y2-1. As stated above, we wish to set x1 to a value where the player is indifferent between bluffing and checking. So, we should set this expected value equation of the gain by bluff equal to zero. In other words, 3y2-1 = 0.

x2

Next, do the same thing, solving for x2. I will go right to the expected value table, which shows the expected value of calling, as opposed to foldling, according to the value of Y.

Setting the sum of the expected value column equal to 0 we get 3y1+y3-1=0.

x3

Next, do the same thing, solving for x3. I will go right to the expected value table, which shows the expected value of raising, as opposed to checking, according to the value of Y.

Setting the sum of the expected value column equal to 0 we get -y1-y2+y3=0.

y1

Next, we do a table for y1, showing the expected gain by bluffing, as opposed to checking, for every possible value of X.

Setting the sum of the expected value column equal to 0 we get 3x2-x3-2y1=0.

y2

Next, we do a table for y2, showing the expected gain by calling, as opposed to folding, for every possible value of X.

Setting the sum of the expected value column equal to 0 we get 3x1+x3-1=0.

y3

Next, we do a table for y3, showing the expected gain by raising, as opposed to checking, for every possible value of X.

Setting the sum of the expected value column equal to 0 we get -x2-x3+2y3=0.

Solving for x1, x2, x3, y1, y2, and y3.

Now we have six equations and six unknowns. Again they are:

3y2-1 = 0
3y1+y3-1 = 0
-y1-y2+y3 = 0
3x2-x3-2y1 = 0
3x1+x3-1 = 0
-x2-x3+2y3 = 0

Using matrix algebra, we should solve for the following matrix.

0	0	0	0	3	0	1
0	0	0	3	0	1	1
0	0	0	-1	-1	1	0
0	3	-1	-2	0	0	0
3	0	1	0	0	0	1
0	-1	-1	0	0	2	0

Forgive me if I don't go through a matrix algebra lesson, and jump right to the solution, which is:

x1 = 0.111111111
x2 = 0.333333333
x3 = 0.666666667
y1 = 0.166666667
y2 = 0.333333333
y3 = 0.5

In other words, X should bluff with 0 to 0.111111, check with 0.111111 to 0.666667, and bet with 0.666667 to 1.0.

Assuming x bets or bluffs, then y should fold with 0 to 0.333333 and call with 0.333333 to 1.0.

If x checks, then y should bluff with 0.0 to 0.166667, check with 0.166667 to 0.500000, and bet with 0.500000 to 1.0.

Assuming x checks, and y bets, then x should fold with 0 to 0.333333 and call with 0.333333 to 1.0 (same strategy as y when forced to fold or call).

The next question is what is the expected value for X if both players follow this strategy. To answer that question, let's consider how much X wins according to every significant grouping of X and Y.

Next, here is the probability of each combination of X and Y.

Finally, here is the expected value of the how much X will win for each combination, which is equal to the product of the win and the probability.

The sum of each cell in this table is -0.055556. So, assuming the ante is $1 per player, X can expect to lose 5.55556 ¢ per hand.

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 216 Answer

The answer is -1.

Problem 216 Solution

Solution

Solution to problem 217 (PDF -- 112K)

Problem 218 Answer

The answer is e, or approximately 2.7182818.

Problem 218 Solution

Solution to problem 218 (PDF -- 67K)

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 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:

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 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 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 224 Answer

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

Problem 224 Solution

Problem 225 Answer

• Player 1 is indifferent between standing and switching at (1/6)×(81+3×921^1/2)^1/3+(81-3×21^1/2)^1/3 = 0.567364*.
• If player 1 hits, player 2 is indifferent between standing and switching at 0.5.
• If player 1 stands, player 2 is indifferent between standing and switching at 0.660951**.
• Assuming optimal strategy by bother players, the probability player 1 wins is 0.494333***.

Problem 225 Answer

There is some Nash equilibrium to be found between the two players. Let:

• Player 1 = player to act first.
• Player 2 = player to act second.
• x = Point at which player 1 is indifferent to standing and switching.
• y = Point at which player 2 is indifferent to standing and switching, assuming player 1 stood.

It should be obvious that y>x. If player 1 stood, player 2 knows that player 1 has a strong card, so he needs to be more aggressive to beat it. This is where positional advantage comes in. If this isn't obvious to you, then you probably should be attempting an easier problem.

First, let's calculate the probability of winning by standing.

1. It is easy to see that the probability a given number x beats a random number is x.
2. If player 2 has y or less, then the probability of player 1 winning is x.
3. If player 2 has y or more, then the probability of player 1 winning is 0, since y>x.
4. Taking the dot prodoct of the possible values for player 2 and the probability of player 1 winning by standing, the probability of winning by standing is xy.

Second, let's calculate the probability of winning by switching.

1. It is easy to see that if player 1 switches, then player 2 will be indifferent to standing and switching at 0.5.
2. If player 2 has 0.5 or less, then the probability of player 1 winning is 0.5, since both will have switched to a random card.
3. If player 2 has 0.5 or more, then the probability of player 1 winning is 0.25. This can be found using geometry or integral calculus by looking at proption of the area player 1 wins over the area bounded by values of 0 to 1 for player 1 and 0.5 to 1 for player 2. Player 1 wins over 1/4 of that area.
4. Taking the dot prodoct of the possible values for player 2 and the probability of player 1 winning by switching, the probability of winning by switching is 3/8.

Setting the expected value of standing and switching equal, we get xy = 3/8.

Next, solve for y using the same method. Forgive me if I don't through the details this time. Again, using simple geometry or integral calculus, you'll find the probability of player 2 winning by standing at y is (y-x)/(1-x). By the same method, the probability of player 2 winning by switching at y is [(1-x)²/2]/(1-x).

Setting the two equations equal and doing some algebra we get x² - 2y + 1 = 0.

We now have two equations and two unknowns. With some simple algebra to solve for x we get to x³ + x -0.75 = 0. Using one of many cubic equaton solvers on the Internet, we find 0.567364. So, that is the value x is indifferent between standing and switching.

Putting that into xy=3/8, we solve for y as 0.660951.

Finally, find the probability of all four combinations of x and y standing and swithcing and the probability x wins for each one and take the dot product. The following table gives more details. The bottom right cell shows the probability player 1 wins is 0.494333.

Final answers:

• Player 1 is indifferent between standing and switching at 0.567364.
• If player 1 hits, player 2 is indifferent between standing and switching at 0.5.
• If player 1 stands, player 2 is indifferent between standing and switching at 0.660951.
• Assuming optimal strategy by bother players, the probability player 1 wins is 0.494333.

Problem 226 Answer

• Player 1 is indifferent between standing and switching at 0.456260889.
• Player 2 is indifferent between standing and switching at 0.554423955.
• Assuming optimal strategy by bother players, the expected loss of player is -0.13162138 dollars.

My thanks to Joe Shipman for his help with this problem.

Problem 226 Solution

There is some Nash equilibrium to be found between the two players. Let:

• Player 1 = player to act first.
• Player 2 = player to act second.
• x = Point at which player 1 is indifferent to standing and switching.
• y = Point at which player 2 is indifferent to standing and switching.

To begin, we must accept that y>x. Player 1 is going to be conservative about switching because he has to double his bet if he does. So, with a number around 0.5 he is better off to cut his losses and stick with it rather than double his bet and go against player 2 with a random card. Meanwhile, player 2 can be more aggressive because he can switch for free. With a number around 0.5, player 2 is apt to switch because he knows player 1 will have a number with an average over 0.5, because player 1 had the opportunity to switch out of the a bad number.

First, let's find x. We do so by finding the value of x where the expected value of standing and switching are equal.

At x, the expected value of standing is y*(2x-1)-(1-x). The expected value of switching at x is -2*y*(1-y).

Setting the two values equal to each other we get :

(1) 2*y^2 - 2xy - 2y + 1 = 0.

Second, let's find y. We do so by finding the value of y where the expected value of standing and switching are equal.

At y, the expected value of standing is x*(2*(2y -1)) + y - x - (1-y). The expected value of switching at x is -2x*(1 - x).

Setting the two values equal to each other we get:

(2) x^2 - 2y - 4xy +2x + 1 = 0.

We now have two equations and two unknowns. We can rewrite equation (1) as:

(3) x = y - 1 + 1/(2y).

Substituting equation (3) into (2) gives us:

(4) (2y^2 - 2y + 1)^2 - 2y*(4y^2) - 8y^2*(2y^2 - 2y + 1) + 4y*(2y^2 - 2y + 1) + 4y^2 = 0

After a lot of algebra, that reduces to:

(5) 12^4 - 8y^3 - 4y^2 - 1 = 0

You can use a four-order calculator to solve for y as 0.554423955.

You can plug that into equation (3) to get x = 0.456260889.

Finally, find the probability of all four combinations of x and y standing and swithcing and the probability x wins for each one and take the dot product. The following table gives more details. The bottom right cell shows player 1 can expect to lose 0.131621 dollars.

Final answers:

• Player 1 is indifferent between standing and switching at 0.456260889.
• Player 2 is indifferent between standing and switching at 0.554423955.
• Assuming optimal strategy by bother players, the expected loss of player is -0.13162138 dollars.

My thanks to Joe Shipman for his help with this problem.

Problem 227 Answer

The answer is 1700 meters.

Problem 227 Solution

Solution #1

Let's call x the distance between the two meetings points.

We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.

At the first meeting, the two boats combined would have traveled the width of the river.

At the second meeting, the two boats combined would have traveled three times the width of the river.

Since both boats travel at a consistent speed, both boats would have traveled three times as far at the second meeting as the first.

Let's solve for x looking at the fast boat.

3 × (time traveled until first meeting) = (time traveled until second meeting).

3 × (400+x) = 1800 + 2x.

x = 600.

So the total width of the river is 400+600+700 = 1700 meters.

Solution #2

Let's call:

• x = the distance between the two meetings points.
• f = speed of the fast ferry.
• s = speed of the slow ferry.
• t₁ = time of first time ferries cross in the river.
• t₂ = time of second time ferries cross in the river.

We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.

We all know that distance = rate × time. Let's express this equality at the first meeting for both ferries.

• Fast ferry: x + 400 = f * t₁
• Slow ferry: 700 = s * t₁

To reorder the above two equations:

• Fast ferry: t₁ = (x+400)/f
• Slow ferry: t₁ = 700/s

Those two expressions are equal:

t₁ = (x+400)/f = t₁ = 700/s

Let's express the distance=rate*time equality at the second meeting for both ferries.

• Fast ferry: 2x + 1800 = f * t₂
• Slow ferry: x + 1500 = s * t₂

To reorder the above two equations:

• Fast ferry: t₂ = (2x+1800)/f
• Slow ferry: t₂ = (x+1500)/s

Those two expressions are equal:

t₂ = (2x+1800)/f = t₁ = (x+1500)/s

It would seem we have two equations and three unknowns. However, we don't have to actually solve for f and s. We can express the the ratio f/s two ways, as follows:

f/s = (x+400)/700 = (1800+2x)/(1500+x).

Cross multiply to get:

x² + 500x - 660000.

The quadratic formula easily gives us x=600. So the whole river is 700+600+400 = 1700 meters wide.

My thanks to Marvin Gardener, which can be found as problem 11 (with slightly different distances) in his book My Best Mathematical and Logic Puzzles.

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 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 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 237 Answer

Here are the probabilities each ant arrives first:

• Ant A: 20/35 = 57.14%
• Ant B: 9/35 = 25.71%
• Ant C: 6/35 = 17.14%

Problem 237 Solution

Please see my solution to problem 237 (PDF).

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 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 242 Answer

The answer is 1/6.

Problem 242 Solution

Please see my problem 242 solution (PDF -- 97K)

Problem 243 Answer

The answer is 1/1 + 1/2 + 1/3 + ... + 1/100 =~ 5.187377518.

Problem 243 Solution

Consider everybody choosing one at a time.

Let f(n) equal the average number of closed loops with n people.

When the first person picks, there is a 1/100 chance he picks himself and 99/100 he doesn't. In other words, there is a 1/100 chance he closes a loop and 99/100 he doesn't. Either way, it reduces the problem to one of 99 people. In other words, f(100) = (1/100) + f(99).

When the second person goes, he will have a 1/99 chance of closing a loop and 98/99 he doesn't. Whatever happens with the first person, there can be only one name left that started an unclosed loop. Thus, f(99) = 1/99 + f(98). Or, f(100) = (1/100) + (1/99) + f(98).

This process keeps repeating until there is just one person left, who will close the last loop, since the person who started it must be the only name left.

Thus the answer is 1/100 + 1/99 + 1/98 + ... + 1/1 =~ 5.187377518.

An estimate for any sufficiently large n is ln(n).

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 245 Answer

The answer is 10,946.

Problem 245 Solution

Let a jump of one step be a "small jump" and a jump of two steps be a "big jump."

• There is only one way to get to the first step, with a single small jump.
• There are two ways to get to the second step, either two small jumps or one big jump.
• There are two ways to arrive at the third step, either with a big jump from step 1 or a small jump from step 2. There is one way to arrive at step 1 and two ways to arrive at step 2. Thus, the number of ways to get to step 3 is 1 + 2 = 3.
• There are two ways to arrive at the step 4, either with a big jump from step 2 or a small jump from step 3. There are 2 ways to arrive at step 2 and 3 ways to arrive at step 3. Thus, the number of ways to get to step 4 is 2 + 3 = 5.
• There are two ways to arrive at the step 5, either with a big jump from step 3 or a small jump from step 4. There are 3 ways to arrive at step 3 and 5 ways to arrive at step 4. Thus, the number of ways to get to step 4 is 3 + 5 = 8.
• Keep following this logic and you will find that there are 10,946 ways to arrive at the 20th step.

Here is the full Fibbonachi sequence, up to 20:

1. 1
2. 2
3. 3
4. 5
5. 8
6. 13
7. 21
8. 34
9. 55
10. 89
11. 144
12. 233
13. 377
14. 610
15. 987
16. 1597
17. 2584
18. 4181
19. 6765
20. 10946

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 247 Answer

The answer is 10 - 12.5×cos^-1(4/5) =~ 1.95624.

Problem 247 Solution

Please see my problem 247 solution (PDF).

Problem 248 Answer

The answer is approximately 0.302299894.

Problem 248 Solution

1. Let's start by finding the height of any of the sides. Consider the right triangle formed by that height, the side of a triangle, and half of another side. Let the height be x. So we have:
   1^2 = 0.5^2 + x^2
   x^2 = 0.75
   x = sqrt(0.75) =~ 0.866025404
2. Next, let's find the distance from the center of that triangular side to the middle of any edge. Call that distance x. Consider the right triangle with legs x and 0.5. The hypotenuse equals the height less x. Let's call the height h. So, we have a right triangle with sides of 0.5, x, and hypotenuse h-x. Using the Pythagorean Theorem, we can solve for x as (h^2-0.25)/(2*h). We know h from the last step, so the distance from the center of that triangular side to the middle of any edge =~ 0.288675135.
3. Next, let's find the distance from the center of the triangle to a corner. Consider the right triangle formed by the corner of the side of the equilateral triangular side of the tetrahedron, center of that triangle, and the middle of any side of that equilateral triangular. We know the two legs of that right triangle are 0.5 and 0.288675135, so using the Pythagorean Theorem we can find the distance from the corner of the triangle to the center is approximately 0.577350269.
4. Next, let's find the height of the tetrahedron. Imagine the tetrahedron laying flat. Consider the right triangle formed by a corner of the tetrahedron on the base, center of the base, and top of the tetrahedron. We know the hyppotenuse of that triangle is 1 and one leg is the distance from the corner of the triangle to the center, which we solved for in the last step. Using the Pythagorean Theorem yet again, we can find the height of the tetrahedron is 0.816496581.
5. Next, let's find the distance from the center of any side of the tetrahedron to the center of the tetrahedron. Here is a trick to easily finding that. Continue to think of the tetrahedron with a side flat on the ground. The height from the ground of three of the corners is 0. The height of the fourth corner is the height of the tetrahedron, which we just solved for.
6. The height of the center will be the average of the height of each corner. In other words the height is (0+0+0+0.816496581)/4 = 0.816496581/4 = 0.204124145. This is the radius of the sphere inside the tetrahedron.
7. Next, let's find the area of one of the sides of the tetrahedron. The area is (1/2)*side*height = (1/2)*1*0.816496581 = 0.433012702.
8. Next, let's find the area of the tetrahedron. The formula is (1/3)*base*height = (1/3)*0.433012702*0.816496581 = 0.117851130.
9. Next, let's find the volume of the sphere. The formula for that is (4/3)*pi*r^3 = (4/3)*pi*0.204124145^3 = 0.035626384.
10. Finally, our answre is the area of the sphere divided by the area of the tetrahedron = 0.035626384/0.117851130 = 0.302299894.

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 12 Answer

Problem 12 Solution

The total money paid is $9 * 3 = $27.

The total money received is $25 for the room and $2 to the bellboy = $27.

There simply is no extra dollar. The $2 to the belloy is included in the $27 the three men paid. By adding it to the $27 it is being counted twice.

Problem 13 Answer

Problem 13 Solution

U=w²*c.

3*w + 2*c = 100.

c=(100-3w)/2.

U=w²*(100-3w)/2.

dU/dw=2w*(100-3w)/2 + w²*-3/2=0.

Solving for w yields w=200/9.

With wine costing $3 a glass this will cost 3*200/9=$66.67, leaving $33.33 for cigars, for which he can buy 16.67.

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 16 Answer

Problem 16 Solution

To illustrate why the 1/2 answer is wrong let propose a similar problem. Suppose a soup company canned 2,000,000 cans of tomato soup. Suppose further that they put the wrong label on half of the cans. The company then sent 999,999 correct cans and 1 incorrect can to one store and 999,999 incorrect cans and 1 correct can to another store. To determine which store has the correct cans you go to one store at random and open one can. You see it does indeed have tomato soup inside. What is the probability you went to the store with 999,999 correct cans? It is not reasonable to assume the probability is 50/50 that you are in either store. The odds that you would pick the one correct can in the store with 999,999 incorrect cans are extremely remote. It is true that before opening a can the odds were 50/50 but once you do open a can that information gives you a clue that can not be ignorred. In fact the probability that you are in the store with the correctly labled cans is 99.9999%.

Now let me introduce Bayes' theorem on conditional probability to explain why the answer is 2/3:

Let the events A₁, ...,A_k for 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) / [ $\Sigma$ (for j=1 to k) Pr(A_j)*Pr(B|A_j) ] .

For the problem in question we can reword this to: Pr(two headed coin|head chosen) = Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) / [Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) + Pr(choosing one headed coin) * Pr(choosing heads given that one headed coin was chosen)] =

1/2 * 1 / [ (1/2 * 1) + (1/2 * 1/2) ] =

1/2 / (1/2 + 1/4) = (1/2)/(3/4) = (1/2)*(4/3) = 2/3.

If this all went over your head think if it another way. There were 3 heads to begin with. 2 of them have another head on the other side. You could have chosen any head with equal probability and since 2 out of 3 have a head on the reverse side the answer is 2/3. Of course this answer would probably not merit very much in partial credit in probability class.

Some other people have questioned my exact wording of this problem, mainly objecting over the tense of the verb I use for when the coin is chosen. What appears now was taken almost exactly from a similar problem in Probability and Statistics (second edition) by Morris H. Degroot on page 63, problem number 5. I changed cards to coins and eliminated what would be the coin with two tails. The book is a commonly used college text on the subject and should be above reproach. Here is how they stated a similar problem:

A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability taht the other side of the card is also green?

Essentially the same problem, but worded using sunken boats, secret compartments, and bars of gold, is asked on page 603 of Statistics for Business and Economics by Edwin Mansfield.

A reader sent in the following explanation that he thought might help:

You have probably received similar e-mails to the one I am now sending, but
just in case, ...

Perhaps the best way to demonstrate that 2/3 is the correct answer to this
problem is to approach it in a simple intuitive manner, such as:

Denote the double headed coin A and the regular coin B.  Further, denote
the sides of the double headed coin A1 (a head) and A2 (another head), and
the sides of the regular coin B1 (a head) and B2 (a tail).

We then have the following possible outcomes:

coin selected           side shown              other side
         A                 A1 (H)                   A2 (H)
         A                 A2 (H)                   A1 (H)
         B                 B1 (H)                   B2 (T)
         B                 B2 (T)                   B1 (H)

We exclude the last possibility from consideration since it is not a head.
Three possibilities remain, all equally likely, two of which have a head on
the other side.  The required probability is thus 2/3.

Gary.

Here is my favotite e-mail I received on this problem. I x'd out some private information.

Dear Michael,

I am a biology teacher at Xxxxxxx High School in Xxxxxxx, MA. I recently was searching for an extra credit question to put on a test for tenth graders. I used the three coins in a bag problem (#16). Unfortunately, none of my students got the correct answer. They all answered 1/2. I have done my best to explain the problem and some of the students now understand why the correct answer is 2/3. The debate has spread throughout the whole school. Most teachers and students beleive the answer to be 1/2. So I started an organization called Team 2/3 - The Probability Masters. Students take an oath to become official members of the team. The mission of the team is to recruit as many new members as possible. Team members get a membership card that has your original question on the back which helps them recruit new team members. Many people think I have gone off the deep end, but I find it amazing that so many people including most teachers and the principal still do not see the truth. If fact, the day after I started the Team 2/3, the principal got on the morning announcements and stated the the answer is 1/2 and the group that thinks otherwise should rethink the problem.

One student created a computer program that simulates 10,000 coin picks in 1 second. The data clearly show the 2/3 answer. But that same student insists that the probability is 2/3 only if many trials are conducted. He doesn't understand that the probability is the same in the first and last trial. People are stuck on the idea that you must have either the H-H or H-T coin and say it is 1 out of two. I have tried to simplify your solution without the T-T coin to help people understand.

If two coins: H-H, H-T

Conditional Prob = 1/2 divided by 3/4 = 2/3

Team 2/3 will recruit for two more days and then reveal the truth. We will try to get the principal to announce that Team 2/3 was correct all along. All team members will hold their heads high if that happens. I only have one math teacher on my side at this time. One of my student team members had a hat made for me that says "Team 2/3 Captain". I wore it on Friday and will wear it this week. We have almost everyone in the school discussing your coin problem. Many of the students are giving it a great deal of thought.

We are very confident that the answer is 2/3, however most people still disagree.

It would be most helpful if you would place a call to the school and talk to Mr. Xxxxxxx Xxxxxxxxx, the principal. This would be greatly appreciated by the team, your team, Team 2/3.

Xxxxxxx High School 1-(xxx)-xxx-xxxx

Sincerely,

Troy X. Xxxxxxx
Biology Teacher P.S. Please feel free to call me at home: 1-(xxx)-xxx-xxxx

I did call the principal and tried to explain why the answer was 2/3. He seemed pretty entertained that somebody would call all the way from Baltimore about this. After going over the problem he seemed to partially understand but added that "he didn't want to be confused with facts." Later the Biology teacher wrote me back, here is his second e-mail:

I really appreciate you talking to the principal. He did talk to me afterward and he seemed very unsure of the correct answer. Team 2/3 has 42 oath-taking official members including an English teacher and the guidance secretary. We allowed people to join unofficially by signing the unofficial membership list. That way they still get a membership card without taking the oath; it's our "chicken list". The Latin teacher, the US history techer, and the alternative education teacher are on that list. Our membership card states-"TEAM 2/3 the Probability Masters" and there is a picture of a man holding two coins. On the back of the card is your original problem. Another team formed (not as organized as our team) called Team 1/2. They also have membership cards. Just this past Thursday, a good week after your call, I reminded the principal that we needed closure to the debate. I just assumed he believe the answer was 2/3. And I told him it would be nice to recognize the courage of team 2/3 and congratulate them. He got on the announcements and stated "After consulting with several mathematical experts" (and other people heard about the controversy and contacted him with explanations for 2/3) "the answer to the coin probability problem is 'probably' 2/3, so congratulations to Team 2/3, you are probably correct. While we would have liked a stronger statement(without the probablies), we have accepted the principal's announcement as victory and feel that we have been successful in our mission.

Thanks for your help,

Troy

Problem 17 Answer

Problem 17 Solution

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

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 33 Answer

Problem 33 Solution

Mark Perkins suggested an alternative solution:

p = 1/6 + (5/6)*(1-p)
6p = 1 + 5 - 5p
11p = 6
p = 6/11

Problem 34 Answer

Problem 34 Solution

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

Problem 38 Answer

Problem 38 Answer

Problem 38 Answer

Problem 39 Solution

All multiples of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceeding the 00 that will make the entire number divisible by 9.

If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was was likely very easy. The trick is that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also. For example the number 17685 is divisible by 9 because 1+7+6+8+5=27, and 27 is divisible by 9.

To prove this let's consider any five digit number, abcde. This number can be expressed as follows.

a*10000 + b*1000 + c*100 + d*10 + e =

a*(9999+1) + b*(999+1) + c*(99+1) + d*(9+1) + e*1 =

a*9999 + b*999 + c*99 + d*9 + a + b + c + d + e =

9*(a*1111 + b*111 + c*11 + d*1) + a + b + c + d + e

Thus 9*(a*1111 + b*111 + c*11 + d*1) is a multiple of 9. So if a+b+c+d+e is also a multiple of 9 then the entire number must be a multiple of 9. Note also that the remainder of abcde/9 is the same as the remainder of (a+b+c+d+e)/9.

The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.

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 52 Answer

Problem 52 Solution

1. Boy then boy
2. Boy then girl
3. Girl then boy
4. Girl then girl

Note: This is similar to the question in problem 16 which approaches the solution another way.

Problem 54 Answer

Problem 54 Solution

Let h be the number of hours worked per week.

U = h*(80-h).

U^' = 80-2h.

Set the derivative equal to 0:

80-2h=0, thus h=40.

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 63 Answer

Problem 71 Answer

Problem 76 Answer

Problem 76 Solution

In this case A is drawing a black marble and B is having already drawn a black marble.

Pr(A and B) = (1/2) * [(3/4)² + (1/4)²] = 5/16.

Pr(B) = 1/2.

Pr(A|B) = Pr(A and B)/Pr(B) = (5/16)/(1/2) = 10/16 = 5/8.

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 83 Answer

Problem 83 Solution

The table below shows the total suffering given the four possibilities:

                      Your partner
                 -----------------------
                 confess     not confess
  
Y  confess:     10*2+10=30    0+20=20
o
u  not confess: 2*20+0=40     2*3+3=9

10p+20 = 31p+9
11 = 21p
p=11/21 =~ 52.4%.

Thus if you think your partner's probability of confessing is greater than 11/21 you should confess, otherwise you should keep your mouth shut.

Problem 88 Answer

First, surgeon 1 operates with glove A inside glove B.

Second, take glove A out of glove B.

Third, surgeon 2 operates using glove B.

Fourth, turn glove A inside out.

Fifth, surgeon 3 operates with glove A inside glove B.

Note: Another way to ask this is how do you have safe sex with three men, one woman, and two condoms?

Problem 92 Solution

Problem 92 Solution

cos(x)=AD/3, sin(x)=BD/3, cos(y)=CD/7, sin(y)=BD/7.

cos²(x)+sin²(x) = AD²/9 + BD²/9 = 1.

cos²(y)+sin²(y) = CD²/49 + BD²/49 = 1.

BD² = 9 - AD² = 49 - CD².

Remember that AD + CD = 9, or CD = 9 - AD.

9 - AD² = 49 - (9-AD)².

9 = 49 - 81 + 18AD.

AD = 41/18.

(41/18)² + BD² = 9.

BD = (9-(41/18)²)^1/2 = 1.95236

Others have pointed out an alternate solution using Heron's Formula, which states the area of a triange is sqr(s*(s-a)*(s-b)*(s-c)), where s=(a+b+c)/2, and a, b, and c are the lengths of the three sides.

In this case a, b, and c are 3, 7, and 9.

s=(3+7+9)/2=9.5

area=sqr(9.5*(9.5-3)*(9.5-7)*(9.5-9)) = sqr(1235/16) = 8.785641695

The area must also equal (1/2)*9*h, where h is the height.

So sqr(1235/16) = (1/2)*9*h

h=sqr(1235/16)*(2/9) = 1.95236

Problem 94 Answer

Problem 94 Solution

The probability of 0 potholes is e^-6*6⁰/0! = e^-6.

The probability of 1 pothole is e^-6*6¹/1! = 6*e^-6.

The probability of 2 potholes is e^-6*6²/2! = 18*e^-6.

The probability of 3 potholes is e^-6*6³/3! = 36*e^-6.

Take the sum the answer is 61*e^-6 =~ 0.151204 .

This problem was taken from the May 1982 Actuary Exam 110, problem 20.

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 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 100 Answer

Given any right triangle you can create a figure like the one above using four of them. This is because the sum of the three angles always is 180 degrees. The area of the outside square is c². We can also find the area by summing the area of the individual parts: 4*(a*b/2)+(a-b)² = 2ab + a² -2ab + b² = a² + b². Thus a² + b² = c².

I didn't assign a diffulty level because I saw Carl Sagan explain this proof on television so I never had a chance to solve it myself.

Problem 109 Answer

Problem 109 Solution

Pr(AB) = Probability that bag with 99 reds is drawn and 1 red is drawn from that bag = 1/2 * 99/100 = 99/200.

Pr(B) = Probability that a red is drawn = ((1/2)*(99/100) + (1/2)*(1/100)) = 100/200 = 1/2.

Pr(AB)/Pr(B) = (99/200)/(100/200) = 99/100.

Problem 115 Answer

Problem 115 Solution

If the number of glasses is a power of 2 (32, 64, 128, 256, etc.) then the correct procedure on the first step would be to test exactly half the glasses. Then test half of the glasses in the positive half next. Keep repeating this procedure. Each step will narrow down the field of poisoned glasses by a factor of 2.

If the number number of glasses is not a power of 2 then there are lots of ways you can test them and still achieve the same maximum number of tests. One way would be to find the closest power of 2 less than n, set this many glasses aside, and then test the others. The only n for which this remainder is 1, between 100 and 200, is 129. In the case of 129 the first test would likely be negative. Then proceed with the procedure for testing 128. Between the first test and the log₂128 the maximum number of tests is 8.

Credit for this problem belongs to Stephen Barr (Mathematical brain benders; ISBN 0-486-24260- 9 published by mamillan NY 1969), previously published in 'Scientific American.'

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 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 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.

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.

I would like to thank Recreations in the Theory of Numbers by Albert H. Beiler for this problem (#20 on page 296).

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 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 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 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.

Problem 156 Answer

The answer is 55/21 minutes.

Problem 156 Solution

Here is my solution to problem 156.

Problem 163 Answer

Thanks to the Baltimore Actuaries Club and Alan Goldberg for this problem.

Problem 163 Solution

We know that it takes 2.5 hours to empty the tank with 10 valves open. Over the 2.5 hours the sum of the water in the tank initially and the water going into the tank will equal the amount of water leaving the tank. Let's set that up as an equation.

w + 2.5x = 2.5*10y

We also know that it takes 5.5 hours to empty the tank with 6 valves open. Let's set that up as an equation.

w + 5.5x = 5.5*6y

I don't like decimals so let's multiply both equations by 2:

2w+5x=50y
2w+11x=66y

To solve the problem we need to know the relationship between w and x, so lets solve for y in the first equation and subsitute in the the second:

y=(2w+5x)/50.

Substituting this in the second equation...

2w+11x=66*(2w+5x)/50

100w+550x=132w+330x

220x=32w

w=220x/32

w=6.875x

So 6.875 times the input rate equals the capacity of the tank. Thus it would take 6.875 hours to fill the tank.

Andrew N wrote it to add his two cents on this problem, arguing in real life the rate at which the water leaves will depend on the water pressure in the tank. Here are his comments.

Hi Michael,

First off, I'd like to say that you've done a fine job with your math problems website. The problems are interesting, and the solutions are quite thorough and easy to understand for the most part. This is no easy feat. I also love your gambling website, the Wizard of Odds. You taught me how to play blackjack! Thanks for that.

Now, problem 163 about emptying the water tanks has a physical wrinkle which in missing from the problem statement: if water is constantly entering the tank, there is no way the tank can be emptied. This is because the flow rate through the outlets is related to the height of the water in the tank. I can prove this assertion as follows.

Setting up Bernoulli's equation for one outlet, where point 1 is the top of the tank and point 2 is on the free jet at the outlet, we have:

P1/gamma + v1^2/(2g) + h1 = P2/gamma + v2^2/(2g) + h2

where:

P1, P2 are pressures

v1, v2 are flow velocities

h1, h2 are heights

gamma is the specific gravity of water = density * gravity

g is the gravitational constant

We set h2 equal to 0 as a reference height. Now, assuming both water surfaces are exposed to the atmosphere, and the height of the tank is not several miles high so atmospheric pressure is constant, we can cancel P1/gamma and P2/gamma. Also, assuming that the water at the top of the tank is not moving quickly compared to the jet, we can assume v1 = 0. After simplifying, we are left with:

h1 = v2^2/(2g)

Solving for v2 we get:

v2 = (2g*h1)^0.5

By the continuity equation, the flow rate out from the outlet is equal to the flow velocity v2 times the outlet area A, so:

Qout = A*v2 = A*(2g*h1)^0.5

Assuming a cylindrical tank, the volume of water in the tank V is the height of the water h1 times the cross-sectional area R, so:

Qout = A*[2g*(V/R)]^0.5 = k*(V^0.5) for simplicity's sake.

Now that we have an expression for the flow rate out of the outlet as a function of the volume in the tank, we can set up a differential equation to solve for the volume in the tank as a function of time.

change in volume = flow in - flow out

dV/dt = Qin - n*Qout (<-- n outlets, all assumed to flow at the same rate)

dV/dt = Qin - n*k*(V^0.5)

This is a seperable differential equation. Seperating both parts we get:

dV/(Qin - n*k*(V^0.5)) = dt

Integrating both sides, we get:

(1/nk)^2 * [2*(V0^0.5 - V^0.5) + 2*Qin*LN[(Qin - nk*V0^0.5)/(Qin - nk*V^0.5)]] = t

The interesting thing here is that if Qin = nk*V^0.5, or V = (Qin/nk)^2, then the LN part of this equation will be undefined. Values of V less than (Qin/nk)^2 give negative values for t, which are clearly not possible. The physical explanation for this is that as the volume in the tank is reduced, there is not enough driving force (hydrostatic pressure in this case) to evacuate the fluid in the tank at a faster rate than it's coming in. In effect, it is not possible to empty the tank so long as water in flowing in. I used a cylindrical tank as the basis for this proof for simplicity, but this principle applies to any shaped tank.

Just my 2 cents.

Thanks,

Andrew N

Michael Shackleford, ASA - June 22, 2001

Problem 164 Answer

Problem 164 Solution

Then cross multiply...

(1-2x)(1+x) = 2x(2-x)

1-x-2x² = 4x - 2x²

1-x = 4x

5x = 1

x = 1/5

So the first time they meet is 2/5 of the way from Vegas and the second time is 1/5 of the way to Reno.

I find the easiest approach at this point is to just follow the two paths as the birds continue to fly and see where they meet. When v has traveled 2/5 of the distance between Vegas and Reno to a point 3x the distance from Reno r will have traveled 3/5 of this distance to 2x the distance from Reno. When v has traveled another 2/5 he will be exactly in Vegas. Conventiently this is also where r will be.

Thanks to Alan Goldberg and Contingencies magazine for this problem.

Problem 166 Answer

Problem 166 Solution

1. If Miss Maryland and Miss Nevada are in the same column then Miss Maryland would be taller, because she is the tallest in that column.

2. If Miss Maryland and Miss Nevada are in the same row then Miss Maryland would be taller, because Miss Nevada is the shortest in that row.

3. If Miss Maryland and Miss Nevada share neither a row nor column then there must be some other contestant who is both in Miss Maryland's column and Miss Nevada's row. Call this woman Miss Texas. To illustrate her position consider the following diagram.

   xxxxx
   xmxxx
   xxxxx
   xxxxx
   xxxxx
   xxxxx
   xtxxn
   xxxxx
   xxxxx
   xxxxx

   In this digram the m stands for Miss Maryland, the n for Miss Nevada, and the t for Miss Texas.

   Miss Maryland must be taller than miss Texas, because miss Maryland is the tallest in her column. Miss Texas must be taller than Miss Nevada, because Miss Nevada is the shortest in her row. Thus if M>T>N then M>N, or Miss Maryland is taller than Miss Nevada.

   Michael Shackleford, ASA - July 19, 2001

Problem 167 Answer

Problem 167 Solution

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 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 173 Answer

Or, if you can't see the pictures: the king of spades, queen of spades, and queen of hearts.

Credit for this problem goes to Martin Gardner who published it in the November 1965 Scientific American.

Problem 173 Solution

1. To the right of the King there's a Queen or two.
2. To the left of a Queen there's a Queen or two.
3. To the left of a Heart there's a Spade or two.
4. To the right of a Spade there's a Spade or two.

From clues 1 and 2 we can infer that a king is on the left and a queen on the right and the middle card is either a queen or 2.

First consider the possibility that a 2 is in the middle. If there are two consecutive spades (from clue 4) then then three cards must be the king of spades, 2 of spades, queen of hearts. If to the right of a spade there is a 2 (again from rule 4) then the three cards are the kings of spades, a two of unknown suit, and the queen of hearts. So if we assume a 2 in the middle it is not certain what the suit of the 2 is.

Next consider the possibility that a queen is in the middle. Clue 4 tells us there must be two consecutive spades, since there is no 2. Both queens can not be spades since they are from the same deck, thus the left and center cards must be spades. Clue 3 tells us the right queen is a heart. So the three cards would be the king of spades, queen of spades, and queen of hearts.

So a 2 in the middle does not lead to a definitive answer, but a queen in the middle does. The question implies there is just one possible answer. Therefore there is a queen in the middle, leading to the final answer of king of spades, queen of spades, and queen of hearts.

Credit for this problem goes to Martin Gardner who published it in the November 1965 Scientific American.

Problem 180 Solution

1. Pick two adjacent holes and turn both switches to on.
2. Pick two opposite holes and turn both switches to on. Assuming I didn't win then I must have a DUUU configuration.
3. Pick two opposite holes. If one switch is down turn up and win. Otherwise turn one down for a DDUU configuration.
4. Pick two adjacent holes. If the switches are in the same position then switch both (and win). If different switch them for a DUDU configuration.
5. Pick ossosite holes and switch both switches (and win).

Thanks to Stacy Friedman for this puzzle.

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 186 Answer

Problem 186 Solution

The key to this problem is that the host is predestined to open a door with a goat. He knows which door has the car so regardless of which door the player picks he always can reveal a goat behind another door.

Let's assume that the prize is behind door 1. Following are what would happen if the player had a strategy of not switching.

• Player picks door 1 --> player wins
• Player picks door 2 --> player loses
• Player picks door 3 --> player loses

Following are what would happen if the player had a strategy of switching.

• Player picks door 1 --> Host reveals goat behind door 2 or 3 --> player switches to other door --> player loses
• Player picks door 2 --> Host reveals goat behind door 3 --> player switches to door 1 --> player wins
• Player picks door 3 --> Host reveals goat behind door 2 --> player switches to door 1 --> player wins

So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.

The plain simple English reason the probability of winning increases to 66.67% by switching, lays in the fact that the host always reveals a goat. If the host didn't know which door had the car, then the probability of having a win would go up to 50% after revealing one of the goats. Such is the case on the show "Deal or no Deal." On that show, the host does not know where the million dollar case is. So, as cases are eliminated that do not have the million dollars, the probability increases that every remaining case has it, equally. When there are only two cases left and two prizes, each case has a 50% chance of each prize.

Links

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 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 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 5 Answer

Problem 5 Solution

Each wise man initially assumes that he isn't painted and can plainly see the other two are.

However, if only two were painted it would not take long for either of the painted wise men to realize they were painted. For either of the painted ones would see the other painted one laughing and know he is laughing at them, since the other guy is not painted. In other words if you saw a wise man laughing and the other guy wasn't he painted he must be laughing at you, thus you must be painted.

After a sufficient period of time if nobody stops laughing then it can no longer be assumed that you are not painted, thus you must be painted as well. When you realize this you stop laughing.

Problem 7 Answer

Problem 7 Solution

The height of a face is 3^.5/2.

The area of any face is 3^.5/4.

The distance from any corner to the center of a joining face is 1/3^.5.

The height of the tetrahedron is (2/3)^.5

The area is 1/3*base*height =

(1/3) * 3^.5/4 * (2/3)^.5 = 2^1/2/12.

Solution 2

Provided by
Mircea Petrache, of "Liceo scientifico Varano" in Camerino, Italy

The above cube has sides of length 1/sqr(2). The area formed by the diagonals along four of the faces as shown in the diagram is a tetrahedrom. Left over are four pyramids. The base of each pyramid is half of the side of the cube and the height is the height of the cube. Thus the area of each pyramid is: 1/2*(1/sqr(2))² * 1/sqr(2) * 1/3 = (12*sqr(2))^-1

The total area of the cube is (1/sqr(2))³ = (2*sqr(2))^-1

The area of the tetrahedrom is thus (2*sqr(2))^-1 - 4*(12*sqr(2))^-1 =

3/(6*sqr(2)) - 2/(6*sqr(2)) = 1/(6*sqr(2)) = sqr(2)/12

Problem 8 Answer

Problem 8 Solution

Problem 18 Answer

Note: To convert into gallons divide cubic inches by 231 .

Problem 18 Solution

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r²-x²)^1/2.

Next measure the area of the triangle with the level of gas segment as a base and the center of the circle as the opposite angle, which is the the sum of two right triangles, or x×(r²-x²)^1/2.

Next, determine the degree of the angle opposite of the base of gas line segment in the triange above. With a little trigonometry it is found that this angle is 2×cos^-1(x/r).

Next, find the area of the slice of the circle made between the two equal sides of the triangle mentioned above, which is the product of the area of the circule and the ratio of the angle of the wedge to 2×pi: pi×r² × (2×cos^-1(x/r)/2×pi) = r² × (cos^-1(x/r)) .

From this wedge subtract the area of the triangle, leaving r² × (cos^-1(x/r)) - x×(r²-x²)^1/2.

Next, multiply by the length of the tank: l × ((r² × (cos^-1(x/r)) - x×(r²-x²)^1/2).

If g is the depth of the gas then g=r-x. The solution in terms of g is l × ((r² × (cos^-1((r-g)/r)) - (r-g)×(2rg-g²)^1/2).

Thanks for Craig Katz for this solution.

First, ignore the length of the tank until the end of the problem.

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r²-x²)^1/2.

Next, take the integral from 0 to the depth of the gas, measured in distance from the center of the side. For example if the level of gas is y (where y is less than x) then the answer is the integral from r-y to r of 2×(r²-x²)^1/2.

To solve this integral it is helpful to know that the integral of (r²-x²)^1/2 is 1/2×[x×(r²-x²)^1/2 + r²×sin^-1(x/r).

I'm sure if you can get this far you can do the rest yourself.

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 20 Answer

Thanks to The Grey Labyrinth for this one.

Problem 20 Solution

You can also try this problem witha a 15 minute egg and hourglasses of 7 and 11 minutes.

Thanks to The Grey Labyrinth for this one.

Problem 22 Answer

The answer is $2.

Problem 22 Solution

Let the total number of sheep be 10x+y, where y<10. The total money raised is (10x+y)² = 100x² + 20xy + y². Regardless of the values of x and y 100x² + 20xy will be divisible by 20. Because the number of $10 bills is odd y² mod 20 must be greater than 10 and less than 20. The only values of y where this is true is 4 and 6, where y² is either 16 or 36. Either way there will be 6 $1 bills left over. Before the check the first brother will have $4 more than the second brother. A $2 check will balance things out.

Thanks to Bill Feldman for suggesting this problem.

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 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 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 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 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 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 35 Answer

Problem 35 Solution

The density function of the shorter lived life bulbs is f(x)=1/100 * e^-x/100.

The density function of the shorter lived life bulbs is f(x)=1/200 * e^-x/200. The survival is 1 minus the integral of the density function. S(x)=e^-x/100, and S(x)=e^-x/200.

The survival function for the entire group is (e^-x/100)⁵ * (e^-x/200)⁵.

The density function for the group is the derivitive of 1 minus the survival function = e^-3x/40.

The mean for the group is the integral from 0 to infinity of x times the density function for the group = 40/3 hours.

Problem 36 Answer

Problem 36 Solution

The density function of the shorter lived life bulbs is f(x)=1/100 * e^-x/100.

The density function of the longer lived life bulbs is f(x)=1/200 * e^-x/200.

The probability of any given 100 hour bulb burning out first is the integral from 0 to infinity of (1/100)*e^-x/100*(e^-x/100>⁴*(e^-x/200)⁵ =

1/100 * integral e^-15x/200 =

(1/100)*(200/15)=2/15.

The probability that ANY of the 100 hour bulbs burn out first is five times this answer, or 10/15=2/3.

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 40 Answer

Problem 40 Solution

t=(4-x)/6 + (1+x²)^1/2/2.

Set the derivative equal to 0:

dt/dx= -1/6 + 1/2 * 1/2 * 2x * (1+x²)^-1/2 = 0.

The solution is x=1/8^1/2.

y=4-x.

Problem 42 Answer

Problem 42 Solution

Problem 43 Answer

Problem 43 Solution

Let a be the angle between the line from the yellow house to the north gate and the line from the yellow house to the blue house.

sin(a) = r/(r+3)

tan(a) = 9/(2r+3)

cos(a) = sqr(1-sin²(a)) = sqr(1-sin²(r/(r+3))) = sqr(6r+9)/(r+3)

tan(a) = sin(a)/cos(a) = (r/(r+3)) * ((r+3)/sqr(3*(2r+3))) = r/sqr(3*(2r+3))

Equating the two expressions for tan(a):

9/(2r+3) = r/sqr(3*(2r+3))

9 * sqr(3) * sqr(2r+3) = r*(2r+3)

9 * sqr(3) = r * sqr(2r+3)

243 = r²*(2r+3)

2r³ + 3r² - 243 = 0

A visit to The MathServ Calculus Toolkit will factor this to (2r-9)*(r² + 6r + 27) = 0.

The real solution is r=4.5

Here are two more solutions presented by Heng Cheng Suang:

I would like to suggest another soln to problem 43.
- just by using pythagoras thm (and/or similar triangles)
  the solution may be solved too.

Let the triangles be labelled as

        A
        |\
        | \E          It is noted that angle AED is 90 deg,
        | /\          and EC is 9 units (properties of tangents)
        |/  \
       D|    \
        |     \    
        --------
        B       C      


USING ONLY PYTHAGORAS THM
-------------------------------------------------------------------------
        
Based on triangle AED (By pyt. thm.),                           STEP 1

             AE = sqrt[(3+r)^2 - r^2)] = sqrt(9+6r)

Based on triangle ABC (By pyt. thm.),                           STEP 2

             (3+2r)^2 + 9^2 = [sqrt(9+6r) + 9]^2
Solving                                                         STEP 3
             4r^4 + 12r^3 + 9r^2 - 486r - 729 = 0
                                            r = 4.5


                                   OR 

-------------------------------------------------------------------------
USING SIM TRIANGLES PROPERTIES
-------------------------------------------------------------------------
STEP 1 same

STEP 2 is replaced by
             9 / (3 + 2r)   = r / sqrt (9 + 6r)              
            
STEP 3 same

I would like to thank Elliot Hunter for suggesting this problem which originally appeared the the November 1937 issue of Popular Science Monthly. The solution is by Dr. Anthony of the math forum.

Problem 45 Answer

Problem 45 Solution

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 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 50 Answer

Problem 50 Solution

For a series to last exactly n games at least one team must have won exactly 3 out of the last (n-1). Thus:

The probability of the series lasting 4 games is 1/8.

The probability of the series lasting 5 games is 1/4.

The probability of the series lasting 6 games is 5/16.

The probability of the series lasting 7 games is 5/16.

The expected number of games played in team A's home field is: 2*(1/8) + 2*(1/4) + 3*(5/16) + 4*(5/16) = 188/64.

The expected number of games played in team B's home field is: 2*(1/8) + 3*(1/4) + 3*(5/16) + 3*(5/16) = 184/64.

Problem 51 Answer

The answer is 1/6.

Problem 51 Solution

Please see my solution to problem 51 (PDF).

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 65 Answer

If the y/x is an integer then the answer is y/x, otherwise it is y/z.

Problem 80 Answer

Problem 82 Answer

Problem 82 Solution

First review the probability of throwing any given number on any given throw:

• 2: 1/36
• 3: 2/36
• 4: 3/36
• 5: 4/36
• 6: 5/36
• 7: 6/36
• 8: 5/36
• 9: 4/36
• 10: 3/36
• 11: 2/36
• 12: 1/36

From the above the probalility of rolling 7 is 6/36 = 1/6 and the probability of rolling an 11 is 2/36 = 1/18. These are the numbers that win on the first throw.

Next lets assume the point thrown on the first roll is a 4, what is the probability of throwing it again before a 7?

Let pr(x) stand for the probability of event x happening on any given roll. The answer is:

pr(4) +
pr(anything other than 4 and 7) * pr(4) +
pr(anything other than 4 and 7)² * pr(4) +
pr(anything other than 4 and 7)³ * pr(4) +
pr(anything other than 4 and 7)⁴ * pr(4) +
+ ...

Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.

pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)² * 1/12) + ((3/4)³ * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)ⁱ
= 1/12 * (1/(1-3/4))
= 1/12 * 4 = 1/3.

The probability of rolling a 10 is the same as the probability of rolling a 4 so pr(rolling a 10 before a 7) also equals 1/3.

Next assume the point thrown is a 5.

Pr(5) = 4/36 = 1/9, pr(anything other than 5 and 7) = 1-4/36-6/36 = 26/36 = 13/18.

pr(rolling a 5 before a 7)
= 1/9 + (13/18 * 1/9) + ((13/18)² * 1/9) + ((13/18)³ * 1/9) + ...
= 1/9 * sum for i = 0 to infinity of (13/18)ⁱ
= 1/9 * (1/(1-13/18))
= 1/9 * 18/5 = 2/5.

The probability of rolling a 9 is the same as the probability of rolling a 5 so pr(rolling a 9 before a 7) also equals 2/5.

Next assume the point thrown is a 6.

Pr(6) = 5/36, pr(anything other than 6 and 7) = 1-5/36-6/36 = 25/36.

pr(rolling a 6 before a 7)
= 5/36 + (25/36 * 5/36) + ((25/36)² * 5/36) + ((25/36)³ * 5/36) + ...
= 5/36 * sum for i = 0 to infinity of (25/36)ⁱ
= 5/36 * (1/(1-25/36))
= 5/36 * 36/11 = 5/11.

The probability of rolling an 8 is the same as the probability of rolling a 6 so pr(rolling a 8 before a 7) also equals 5/11.

Now we're ready to add all this together...

The probability of winning the pass line bet is:
pr(7) + pr(11) + pr(4)*pr(4 before a 7) + pr(5)*pr(5 before a 7) + pr(6)*pr(6 before a 7) + pr(8)*pr(8 before a 7) + pr(9)*pr(9 before a 7) + pr(10)*pr(10 before a 7)
= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) + (5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)
= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36
= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980
= 976/1980 = 244/495.

The expected return on this wager is the product of the probability of winning and the ratio of what you keep (including your original wager) to the original bet if you do win (in this case 2). Thus the expected return is 244/495 * 2 = 488/495 =~ 98.59%.

The house advantage is what the casino gets to keep, on average, which is 1 minus the expected return, which equals 1-(488/495) = 7/495 =~ 1.41%

In the appendix to this problem I show the house advantage on the don't pass bet.

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 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