1986 AIME Problem 8

Attempt Problem 8 of the 1986 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AIME solutions, or check the answer key.

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8.

Let SS be the sum of the base 1010 logarithms of all the proper divisors of 1000000.1000000. What is the integer nearest to S?S?

Answer: 141
Concepts:factor countinglogarithmprime factorization
Difficulty rating: 1950
Small Hint:

Factor 10000001000000 and count all of its positive divisors

Big Hint:

Pair every divisor dd with its complementary divisor 1000000d\frac{1000000}{d}

Solution:

Let N=1000000=2656.N=1000000=2^6 5^6. It has (6+1)(6+1)=49(6+1)(6+1)=49 positive divisors. The product of all of them is N492,N^{\frac{49}{2}}, so the sum of their base-1010 logarithms is 492log10N=4926=147. \frac{49}{2}\log_{10}N=\frac{49}{2}\cdot6=147. The proper divisors include 11 but exclude NN itself. Subtracting log10N=6\log_{10}N=6 gives S=141,S=141, already an integer.

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