Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | + | We can multiply both sides by <math>2022!</math> to get rid of the fractions | |
| + | <cmath>\frac{5!x}{2022!}=\frac{20}{2021!}</cmath> | ||
| + | <cmath>5!x=20 \cdot 2022</cmath> | ||
| + | <cmath>120x=(120)(337)</cmath> | ||
| + | <cmath>x=\boxed{337}</cmath> | ||
| + | |||
| + | ~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
Contents
Problem
Let 
for all positive integers 
. Find the value of 
that satisfies ![\[\frac{5!x}{2022!}=\frac{20}{2021!}.\]](http://latex.artofproblemsolving.com/3/6/d/36df42c0f356e92ee3bcdca3abaa75bc4387fd5b.png)
Solution
We can multiply both sides by 
to get rid of the fractions
![\[\frac{5!x}{2022!}=\frac{20}{2021!}\]](http://latex.artofproblemsolving.com/7/1/1/7114489da9030ac2d999a4d742a8e91407d403bf.png)
![\[5!x=20 \cdot 2022\]](http://latex.artofproblemsolving.com/9/6/7/967dbc66c5ba348bdd938e31f827dc2865ae7b1d.png)
![\[120x=(120)(337)\]](http://latex.artofproblemsolving.com/3/5/9/359eb00ab26bc2d6cc37b3b005f10182af28b227.png)
![\[x=\boxed{337}\]](http://latex.artofproblemsolving.com/2/4/e/24ef932a425a2d016c919c1057a06b618db6bbf1.png)
~Bradygho
Solution 2
![\[\frac{120x}{2022}=20 \implies \frac{6x}{2022}=1 \implies x=337\]](http://latex.artofproblemsolving.com/1/5/4/1547cb9b041f5c361a29c02cee90174e8558c9eb.png)
- kante314 -
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
