# Problem 133 Solution

Lets call a the number of people he leaves \$1 to, b the number of people he leaves \$7 to, c the number of people he leaves \$49 to, etc. The solution equation is thus:

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).

Michael Shackleford, A.S.A., May 27, 1999