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Problem 1139

AIME late

Number theory Difficulty 5.1 Prove it Russian Mathematical Olympiad · Russia

A student got $17$ marks during a week, each one is $2$ , $3$ , $4$ , or $5$ . The arithmetic mean of these marks is an integer. Prove that he could not get each mark at least thrice.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Suppose the contrary. Then the student received each of the marks $2$ , $3$ , $4$ , $5$ at least three times. Take three marks of each kind; the sum of these $12$ marks is $42$ . Since each of the remaining five marks is at least $2$ and at most $5$ , the sum of all $17$ marks is at least $42 + 5 \cdot 2 = 52$ and at most $42 + 5 \cdot 5 = 67$ . But none of the numbers from $52$ to $67$ is divisible by $17$ , since $3 \cdot 17 = 51$ and $4 \cdot 17 = 68$ . Therefore, the arithmetic mean of all the marks is not an integer. Contradiction.

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