2021 JMPSC Accuracy Problems/Problem 5

Revision as of 22:13, 10 July 2021 by Bradygho (talk | contribs) (Solution)

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