# Difference between revisions of "1986 AIME Problems/Problem 8"

## Problem

Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$. What is the integer nearest to $S$?

## Solution

### Solution 1

The prime factorization of $1000000 = 2^65^6$, so there are $(6 + 1)(6 + 1) = 49$ divisors, of which $48$ are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.

Writing out the first few terms, we see that the answer is equal to $$\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).$$ Each power of $2$ appears $7$ times; and the same goes for $5$. So the overall power of $2$ and $5$ is $7(1+2+3+4+5+6) = 7 \cdot 21 = 147$. However, since the question asks for proper divisors, we exclude $2^65^6$, so each power is actually $141$ times. The answer is thus $S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}$.

The

### Solution 2

Since the prime factorization of $10^6$ is $2^6 \cdot 5^6$, the number of factors in $10^6$ is $7 \cdot 7=49$. You can pair them up into groups of two so each group multiplies to $10^6$. Note that $\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6$. Thus, the sum of the logs of the divisors is half the number of divisors of $10^6 \cdot 6 -6$ (since they are asking only for proper divisors), and the answer is $(49/2)\cdot 6-6=141$.