1986 AIME Problems/Problem 8
Problem
Let be the sum of the base
logarithms of all the proper divisors (all divisors of a number excluding itself) of
. What is the integer nearest to
?
Solution
The prime factorization of , so there are
proper divisors (the subtracted 1 to ignore
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 . Each power of 2 from
to
in this equation appear
times (excluding 6, which only appears
times due to the exclusion of
). Therefore, it appears
. The same goes for
.
The answer is thus .
See also
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 | ||
All AIME Problems and Solutions |