Prove that any $n+1$ integers has a pair of integers differ by a multiple of $n$?

Question

My attempt : Take any $n+1$ integers $1,2,.....,n+1$ , Then there is a pair of integers $1, n+1$ such that :

$$(n+1)-1 = n$$, which is a multiple of n.

Answer 1

Divide each of the $n+1$ numbers by $n$. Since there divisor is $n$ you will get $n$ distinct remainders. But there are $n+1$ numbers hence two numbers will leave the same remainder when divided by $n$. The difference of these two numbers will be divisible by $n$ since their remainders will cancel out.