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 . Thus, for
to have exactly
integral divisors, we need to have
. Since we know that
is divisible by
, two of the factors must be
and
. To minimize
, the remaining factor must be
; also to minimize it, we want
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 |