Question

If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, what is the smallest possible value of $c?$

Answer

Answer: 675

Concepts:perfect power perfect square prime factorization

Difficulty rating: 2190

Small Hint:

Express both sums in terms of the middle integer $c$

Big Hint:

Compare the prime exponents in $3c$ and $5c$ modulo $2$ and modulo $3$

Solution:

The two sums are $3c$ and $5c.$ If $c=\prod p^{v_p},$ then $3c$ being a square and $5c$ being a cube impose conditions on every exponent. For $p=3,$ the smallest exponent in $c$ that is odd and divisible by $3$ is $3.$ For $p=5,$ the smallest exponent that is even and congruent to $2\pmod3$ is $2.$ Every other prime exponent can be $0.$ Thus the least possible value is $c=3^3\cdot5^2=675.$

1989 AIME Problem 4

Answer: 675

Difficulty rating: 2190

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years