Competition problem — Number theory

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Number theory Difficulty 6.0 National olympiad Prove it Ukraine

When dividing with remainder some four consecutive positive integers by some three-digit integer it turned out, that the sum of these four remainders is equal to $983$ . Find the remainder under the division of the smallest of these four numbers by $109$ .

Solution

Denote these four consecutive integers by $n$ , $n+1$ , $n+2$ and $n+3$ . Denote the three-digit number that we were dividing by as $b$ . Let $n = bq + r$ . Consider possible values of $r$ .

If $r \le b - 4$ , then these remainders are $r$ , $r+1$ , $r+2$ and $r+3 \Rightarrow r + r+1 + r+2 + r+3 = 983 \Rightarrow 4r = 973$ , contradiction.

If $r = b - 3$ , then these remainders are $r$ , $r+1$ , $r+2$ and $0 \Rightarrow r + r+1 + r+2 + 0 = 983 \Rightarrow 3r = 980$ , contradiction.

If $r = b - 2$ , then these remainders are $r$ , $r+1$ , $0$ and $1 \Rightarrow r + r+1 + 0 + 1 = 983 \Rightarrow 2r = 981$ , contradiction.

If $r = b - 1$ , then these remainders are $r$ , $0$ , $1$ and $2 \Rightarrow$
$r + 0 + 1 + 2 = 983 \Rightarrow r = 980 \Rightarrow b = r + 1 = 981.$
It remains to find the required remainder:
$n = bq + r = 981q + 980 = 109(9q + 8) + 108.$

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