number theory - Prove that one integer among $m$ consecutive integers is divisible by $m$

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 $m$.

Answer

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

Existence:

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

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

But, $0

Uniqueness:

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 $m$