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

(Solution)
 
(2 intermediate revisions by one other user not shown)
Line 10: Line 10:
  
 
~Bradygho
 
~Bradygho
 +
 +
== Solution 2 ==
 +
<cmath>\frac{120x}{2022}=20 \implies \frac{6x}{2022}=1 \implies x=337</cmath>
 +
 +
- kante314 -
 +
 +
==See also==
 +
#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
 +
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 10:04, 12 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

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