2021 JMPSC Accuracy Problems/Problem 5


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 \[\frac{5!x}{2022!}=\frac{20}{2021!}.\]


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


Solution 2

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

- kante314 -

See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

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