1986 AIME Problems/Problem 8

Problem

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

Solution

The prime factorization of $100000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) - 1 = 48$ proper divisors (the subtracted 1 to ignore $1000000$ itself). 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 $\log 1 + \log 2 + \log 4 + \log 5 \ldots = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots$ . Each powers of 2 from 0 to 5 in this equation appear $7$ times (excluding 6, which only appears 6 times due to the exclusion of $100000$ ). Therefore, it appears $(0 + 1 + 2 + 3 + 4 + 5) \cdot 7 + 6 \cdot 6 = 15 \times 7 + 36 = 141$ . The same goes for $5$ .

The answer is thus $\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141$ .

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1986 AIME Problems/Problem 8

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