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Romanian Mathematical Olympiad

Number theory Difficulty 5.1 AIME, harder Prove it Romania

Let $n$ be an integer, $n \ge 2$ . Find the remainder of the division of the number $n(n+1)(n+2)$ by $n-1$ .

Solution

We notice that $n(n+1)(n+2) = (n-1+1)(n-1+2)(n-1+3) = (n-1)^3 + 6(n-1)^2 + 11(n-1) + 6$ . If $n-1 > 6$ , the residue is $6$ .

If $n = 2, 3, 4, 7$ , the residue is $0$ . If $n = 5$ , the residue is $2$ . If $n = 6$ , the residue is $1$ .

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