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

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==See also==
 
==See also==
#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Sprint Answer Key]]
+
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
{{JMPSC Notice}}
 
{{JMPSC Notice}}

Revision as of 16:23, 11 July 2021

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


See also

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

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