1989 AIME Problem 4

Attempt Problem 4 of the 1989 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AIME solutions, or check the answer key.

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4.

If a<b<c<d<ea<b<c<d<e are consecutive positive integers such that b+c+db+c+d is a perfect square and a+b+c+d+ea+b+c+d+e is a perfect cube, what is the smallest possible value of c?c?
Answer: 675
Concepts:perfect powerperfect squareprime factorization
Small hint:
Express both sums in terms of the middle integer cc
Big hint:
Compare the prime exponents in 3c3c and 5c5c modulo 22 and modulo 33
Solution:
The two sums are 3c3c and 5c.5c. If c=∏pvp,c=\prod p^{v_p}, then 3c3c being a square and 5c5c being a cube impose conditions on every exponent. For p=3,p=3, the smallest exponent in cc that is odd and divisible by 33 is 3.3. For p=5,p=5, the smallest exponent that is even and congruent to 2(mod3)2\pmod3 is 2.2. Every other prime exponent can be 0.0. Thus the least possible value is c=33⋅52=675.c=3^3\cdot5^2=675.
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