Competition problem — Number theory

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Number theory Difficulty 5.0 AIME, harder Prove it Ireland

For $n$ a positive integer, $n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n$ is the product of the positive integers from 1 to $n$ . Determine, with proof, all positive integers $n$ for which $n! + 3$ is a power of 3.

Solution

The only solutions are $3! + 3 = 3^2$ and $4! + 3 = 3^3$ . We can calculate directly that $5! + 3 = 123$ is not a power of 3. For $n \ge 6$ , we have $9 \mid n!$ as 3 and 6 are two of the factors defining $n!$ . But also 9 is a factor of all powers of 3 beyond $3^1$ , so we would have the contradiction that 9 divides both $n!$ and $n! + 3$ .

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