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 $n$ ?

$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

Solution

Every factor of $10^n$ will be of the form $2^a * 5^b , a\leq n , b\leq n$ . Logarithmically, addition and multiplication are interchangeable (i.e. $\log(a*b) = \log(a)+\log(b)$ ), so we need only count the number of 2's and 5's occurring in total. For every factor $2^a * 5^b$ , there will be another $2^b * 5^a$ , 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 $\frac{n*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}$ . Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2008 AMC 12B Problems/Problem 23

Problem 23

Solution

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