2021 JMPSC Accuracy Problems/Problem 5

Revision as of 10:04, 12 July 2021 by Kante314 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Invalid username
Login to AoPS