Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.