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 $\frac{5!x}{2022!}=\frac{20}{2021!}$ $5!x=20 \cdot 2022$ $120x=(120)(337)$ $x=\boxed{337}$

~Bradygho

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2021 JMPSC Accuracy Problems/Problem 5

Problem

Solution