1989 AIME Problems/Problem 4
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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 • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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