1986 AIME Problem 8

Question

Let

S

be the sum of the base

10

logarithms of all the proper divisors of

1000000.

What is the integer nearest to

S?

Answer

Factor

1000000

and count all of its positive divisors

Pair every divisor

d

with its complementary divisor

\frac{1000000}{d}

Let

N=1000000=2^6 5^6.

It has

(6+1)(6+1)=49

positive divisors. The product of all of them is

N^{\frac{49}{2}},

so the sum of their base-

10

logarithms is

\frac{49}{2}\log_{10}N=\frac{49}{2}\cdot6=147.

The proper divisors include

1

but exclude

N

itself. Subtracting

\log_{10}N=6

gives

S=141,

already an integer.

Problem 8 in other years