# Difference between revisions of "1989 AIME Problems/Problem 4"

## Problem

If $a 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 $c$?

## Solution

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers. $c$ is minimized if it’s prime factorization contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = \boxed{675}$, which works as the solution.

## Solution 2

Let $b$, $c$, $d$, and $e$ equal $a+1$, $a+2$, $a+3$, and $a+4$, respectively. Call the square and cube $k^2$ and $m^3$, where both k and m are integers. Then:

$5a + 10 = m^3$

Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 each time.

$m = 5$ gives no solution for k

$m = 10$ gives no solution for k

$m = 15$ gives a solution for k.

$10 + 5a = 15^3$

$2 + a = 675$

$c = \boxed{675}$

-jackshi2006

## See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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