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Number theory Difficulty 4.7 AIME Prove it Saudi Arabia

We have n>2n > 2 nonzero integers such that every one of them is divisible by the sum of the other n1n-1 numbers. Show that the sum of the nn numbers is precisely 00.

Solution

Let these numbers be a1,a2,,ana_1, a_2, \dots, a_n and SS is the sum of these numbers. For every i{1,2,,n}i \in \{1, 2, \dots, n\}, we have SaiaiS - a_i \mid a_i, which means SaiSS - a_i \nmid S. Suppose that S0S \neq 0, without the loss of generality, S>0S > 0. We investigate two cases

* If i{1,2,,n}\exists i \in \{1, 2, \dots, n\} such that ai<0a_i < 0 then Sai>S|S - a_i| > |S| so SaiSS - a_i \nmid S.

* ai>0a_i > 0 for all i{1,2,,n}i \in \{1, 2, \dots, n\}. Let aka_k be the smallest of a1,a2,,ana_1, a_2, \dots, a_n. We have Sam>2amam=amS - a_m > 2a_m - a_m = a_m, so SamamS - a_m \nmid a_m.

Hence, S0S \neq 0 is impossible, which means S=0S = 0.

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