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 $49-1=48$ ). 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 $141$ times. The answer is thus $S = \log 2^{141}5^{141} = \log 10^{141} = 141$ .

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 $log n+log(10^6/n)=log n+log10^6-logn=6$ . Thus, the sum of the logs of the divisors is half the number of divisors of $10^6 \cdot6 -6$ (since they are asking only for proper divisors), and the answer is $(49/2)\cdot6-6=141$ .

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

Contents

Problem

Solution

Solution 2

See also