# The sum of $101$ consecutive positive integers is $p^3$, where $p$ is a prime. Find the smallest of the 101 integers.

The sum of $101$ consecutive positive integers is $p^3$ where $p$ is a prime no. What is the smallest of the 101 integers?

My Approach: $$p^3 = (n)+(n+1)+......+(n+100) = 101n + {(100)(101)/2} = 101(n + 50).$$ Now I can't find the characteristics of $p$. Can anybody help?

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Well, $101$ is prime. So the smallest possible value for $n+50$ is...? –  TonyK Jan 26 '14 at 10:40
@TonyK "So the only value for ..." –  Henry Jan 26 '14 at 10:41
(n+50) = (101)^2 . Is it right? Then, is n = 10151 ? –  Anish Bhattacharya Jan 26 '14 at 10:43
Yes, that's right. –  TonyK Jan 26 '14 at 22:01