Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it

I prove: For any given positive integer $n>1$ , there exist $n$ consecutive composite numbers.

Solution

1. It is easy to verify that $(n+1)!+2, (n+1)!+3, \cdots, (n+1)!+(n+1)$ are $n$ consecutive composite numbers.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.