Competition problem — Number theory

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Number theory Difficulty 6.0 AIME, harder Prove it United States

Problem:

An integer is a perfect number if and only if it is equal to the sum of all of its divisors except itself. For example, $28$ is a perfect number since $28 = 1 + 2 + 4 + 7 + 14$ .

Let $n!$ denote the product $1 \cdot 2 \cdot 3 \cdots n$ , where $n$ is a positive integer. An integer is a factorial if and only if it is equal to $n!$ for some positive integer $n$ . For example, $24$ is a factorial number since $24 = 4! = 1 \cdot 2 \cdot 3 \cdot 4$ .

Find all perfect numbers greater than $1$ that are also factorials.

Solution

Solution:

The only perfect factorial is $6 = 3!$ . Certainly, $2! = 2$ is not perfect. For $n > 3$ , note that $n! = 6k$ , where $k > 1$ , and thus the factors of $n!$ will include $1, k, 2k, 3k$ . This sums to $6k + 1$ , showing that $n!$ is not perfect.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.