0
$\begingroup$

Let $p,q,r,s,t$ be consecutive positive integers such that $q+r+s$ is a perfect square and $p+q+r+s+t$ is a perfect cube. Find the smallest possible value of $r$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Because $p,q,r,s,t$ are consecutive positive integers, we know immediately that $q=r-1,p=r-2,s=r+1,t=r+2$. This means that $q+r+s=r-1+r+r+1=3r$ and that $p+q+r+s+t=5r$.

If $3r$ is a perfect square, then this means that the prime factorization of $r$ must have $3^{2n+1}$ for some $n$; similarly, for $5r$ to be a perfect cube, we must have that the prime factorization of $r$ must have $5^{3m+2}$ for some $m$.

Clearly having other factors will lead to $r$ being larger, so we can say that $r$ is of the form $r=3^{2n+1}5^{3m+2}$. But $3r$ being a perfect square means that $3m+2$ must be even, so $m$ is even; the least such $m$ is 0. Similarly, $5r$ being a perfect cube means that $2n+1$ must be a (positive) multiple of 3; the least such $n$ to make this possible is $n=1$.

It follows that $r=3^35^2$.

$\endgroup$
2
  • $\begingroup$ Thank you..its really helpful $\endgroup$ Commented Jun 25, 2014 at 16:16
  • $\begingroup$ No problem; don't forget to accept if this is what you needed. $\endgroup$
    – Hayden
    Commented Jun 25, 2014 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .