Determine all positive integers $a$ for which the equation

$$
7 a n - 3 n! = 2020
$$

has a positive integer solution $n$.

(Hint: For every positive integer $n$: $\quad n! = 1 \cdot 2 \cdot \cdots \cdot n$.)

(Richard Henner)
Answer:

$a=68$ or $a=289$.

## Solution:

The left side of the equation is divisible by $n$, so $n$ must be a divisor of 2020.

$n$ must be less than 7, because for $n \geq 7$ the left side is divisible by 7, but 2020 is not.

Therefore, the only possible values for $n$ are 1, 2, 4, and 5.

For $n=1$, we get the equation $7a=2023$, so $a=289$.

For $n=2$, we get the equation $14a=2026$, but this has no solution because 2026 is not divisible by 14.

For $n=4$, we get the equation $28a=2092$, but this has no solution because 2092 is not divisible by 28.

For $n=5$, we get the equation $35a=2380$, so $a=68$.

Thus, for $a=68$ the equation has the solution $n=5$, and for $a=289$ the solution $n=1$.

(Richard Henner)