Show that of any $m$ consecutive integers, exactly one is divisible by $m$. I am finding it difficult to prove that there is only one number among $m$ consecutive integers that is divisible by $m$.


Let the numbers be $b_r=a+r, 0\le r\le m-1$


We can apply Pigeonhole Principle to prove the existence by contradiction.

Let none of them is divisible by $m,$ so they can leave $m-1$ distinct remainder$(r)$s namely, $1\le r\le m-1$

But, as there are $m$ numbers, so at least tow of them leave the same remainders.

Let $b_u,b_v$ leave the same remainders where $1\le u<v\le m-1$

Then $m$ divides $b_v-b_u=v-u$

But, $0<v-u<m-1$ can not be divisible by $m$


If $m$ divides both $b_s,b_t,0\le s\le t\le m-1$

$m$ must divide $b_t-b_s=t-s$ which lies $\in(0,m-1]$ which is impossible

So, there can be at most one $r$ divisible by $m$

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  • $\begingroup$ Did you prove existence? Or just uniqueness? $\endgroup$ – Zubin Mukerjee Nov 27 '14 at 8:21
  • $\begingroup$ @ZubinMukerjee, Updated. Thanks for your observation $\endgroup$ – lab bhattacharjee Nov 27 '14 at 8:25
  • $\begingroup$ I don't think it's actually the pigeonhole principle, since there are $m$ things in $m$ boxes (as opposed to $m+1$ things in $m$ boxes) and we're trying to show that each box has one thing (instead of at least one box having two things). $\endgroup$ – Zubin Mukerjee Nov 27 '14 at 8:25
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    $\begingroup$ @ZubinMukerjee, Let none of them is divisible by $m$. Then there can be $m-1$ distinct remainders $1\cdots m-1$ by $m$ $b_r$s $\endgroup$ – lab bhattacharjee Nov 27 '14 at 8:26
  • $\begingroup$ Okay, you should add this to the answer, +1 $\endgroup$ – Zubin Mukerjee Nov 27 '14 at 8:27


Any integer can be written one of $m$ numbers: $km,km+1,\cdots, km+(m-1)$

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