1990 AIME Problems/Problem 5
Problem
Let be the smallest positive integer that is a multiple of
and has exactly
positive integral divisors, including
and itself. Find
.
Solution
The prime factorization of . For
to have exactly
integral divisors, we need to have
such that
. Since
, two of the prime factors must be
and
. To minimize
, we can introduce a third prime factor,
. Also to minimize
, we want
, the greatest of all the factors, to be raised to the least power. Therefore,
and
.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |