The sum of the base- 10 logarithms of the divisors of 10^n is 792. What is n?

2008 AMC 12B Problem 23

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23.

The sum of the base-

10

logarithms of the divisors of

10^n

792.

What is

n?

11

12

13

14

15

The sum of the logs of the divisors equals the log of the product of all divisors

10^n

has

(n + 1)^2

divisors, and the product of the divisors of

N

N^{\frac{d(N)}{2}};

this gives

\tfrac12 n(n + 1)^2 = 792

The sum of the base-

10

logs of the divisors is the log of their product. A number

N

with

d(N)

divisors has divisor product

N^{\frac{d(N)}{2}}.

Here

N = 10^n

has

d(N) = (n + 1)^2

divisors, so the product is

(10^n)^{\frac{(n+1)^2}{2}}

and its log is

\frac{n(n + 1)^2}{2} = 792.

Thus

n(n + 1)^2 = 1584

= 11 \cdot 144

= 11 \cdot 12^2,

giving

n = 11.

Thus, the correct answer is A.