Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Ukraine

Determine all possible pairs of positive integers $(m, n)$ that satisfy the following:
$m^{10} + 23mn + n! = 2020,$
where $n!$ denotes the product of all positive integers from $1$ to $n$ .

Solution

When $m \ge 3$ , $m^{10} \ge 3^{10} > 3^7 = 2187 > 2020$ , so the given equality doesn't hold. Thus, there are two possible values for $m$ : $m=1$ or $m=2$ .

If $m=1$ , the condition to be satisfied is $23n + n! = 2019$ . Since $7! = 5040 > 2019$ , $n \le 6$ . For $n \le 6$ , $23n + n! \le 138 + 720 = 850 < 2019$ , so, considered case has no solutions.

If $m=2$ , the condition to be satisfied is $46n + n! = 996$ . Similarly, $n \le 6$ . If $n=6$ , then $46n + n! = 276 + 720 = 996$ , so the equation holds. Clearly, if $n$ is smaller, then so is the left side of equation, and thus, there are no more solutions.

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