Competition problem — Number theory

Library / /200 of 860

Number theory Difficulty 5.0 AIME Find the answer

Find an $n$ such that $n!-(n-1)!+(n-2)!-(n-3)!+\cdots \pm 1$ ! is prime. Be prepared to justify your answer for $\left\{\begin{array}{c}n, \\ {\left[\frac{n+225}{10}\right],}\end{array} n \leq 25\right.$ points, where $[N]$ is the greatest integer less than $N$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

$3,4,5,6,7,8,10,15,19,41$ (26 points), 59, 61 (28 points), 105 (33 points), 160 (38 points) are the only ones less than or equal to 335. If anyone produces an answer larger than 335, then we ask for justification to call their bluff. It is not known whether or not there are infinitely many such $n$ .

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