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