2008 AMC 12B Problems/Problem 23

Problem 23

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

Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Using the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ , it suffices to count the total number of 2's and 5's running through all possible $(a,b)$ . For every factor $2^a \times 5^b$ , there will be another $2^b \times 5^a$ , so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since $\log(2)+\log(5) = \log(10) = 1$ , the final sum will be the total number of 2's occurring in all factors of $10^n$ .

There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $0+1+2+3...+n = \frac{n(n+1)}{2}$ total 2's. The total number of 2's is therefore $\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}$ . Plugging in our answer choices into this formula yields 11 (answer choice $\mathrm{(A)}$ ) as the correct answer.