1989 AIME Problems/Problem 4
From AoPSWiki
Problem
If
are consecutive positive integers such that
is a perfect square and
is a perfect cube, what is the smallest possible value of
?
Solution
Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know
and
. Thus,
must be in the form of
based upon the first part and in the form of
based upon the second part, with
and
denoting an integers.
is minimized if it’s prime factorization contains only
, and since there is a cubed term in
,
must be a factor of
.
, which works as the solution.
See also
| 1989 AIME (Problems • Resources) | ||
| Preceded by Problem 3 | Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||





