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Number theory Difficulty 4.7 AIME Prove it Russia

Four consecutive integers greater than 100100 are given. Prove that the sum of some three of the given numbers can be represented as the product of three distinct positive integers greater than 11.

Solution

Let nn, n+1n+1, n+2n+2, n+3n+3 be the given numbers. The sum of the three smallest of them is 3n+3=3(n+1)3n + 3 = 3(n + 1), and the sum of the three largest numbers is 3(n+2)3(n + 2). But at least one of the numbers n+1n+1 and n+2n+2 is even, that is, equal to the product of 22 and kk, where k>3k > 3. Therefore, this sum can be represented as the product of three distinct natural numbers: 22, 33, and kk.

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