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 $1000000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) = 49$ divisors (note the question asks for proper divisors, so we ignore $1000000$ and get $\displaystyle 49-1=48\displaystyle$ ). 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 $\log 1 + \log 2 + \log 4 + \log 5 + \ldots + \log 1000000 = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots (2^65^6)$ . Each power of $2$ appears $7$ times; and the same goes for $5$ . So they appear $7(1+2+3+4+5+6) = 7 \cdot 21 = 147$ times. However, since the question asks for proper divisors, we exclude $2^65^6$ , so each number appears $\displaystyle 141$ times. 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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