# Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 5"

## Problem

Let $n!=n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$ for all positive integers $n$. Find the value of $x$ that satisfies $$\frac{5!x}{2022!}=\frac{20}{2021!}.$$

## Solution

We can multiply both sides by $2022!$ to get rid of the fractions $$\frac{5!x}{2022!}=\frac{20}{2021!}$$ $$5!x=20 \cdot 2022$$ $$120x=(120)(337)$$ $$x=\boxed{337}$$

$$\frac{120x}{2022}=20 \implies \frac{6x}{2022}=1 \implies x=337$$