Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | + | <cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{\textbf{(C) } 49}</cmath> | |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | Alternatively, | ||
| + | |||
| + | <cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{169}=\frac{(10+3)^2(10-3)^2}{169}=\frac{(13)^2(7)^2}{13^2}=7^2\boxed{\textbf{(C) } 49}</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 21:31, 23 November 2021
- The following problem is from both the 2021 Fall AMC 10A #1 and 2021 Fall AMC 12A #1, so both problems redirect to this page.
Problem
What is the value of 
?
Solution
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{\textbf{(C) } 49}\]](http://latex.artofproblemsolving.com/4/0/7/407cd9459ae95627aa58757ee1ca3b069e4cb1c0.png)
~MRENTHUSIASM
Alternatively,
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{169}=\frac{(10+3)^2(10-3)^2}{169}=\frac{(13)^2(7)^2}{13^2}=7^2\boxed{\textbf{(C) } 49}\]](http://latex.artofproblemsolving.com/0/0/c/00c7a8954dc440b7c92101868723092e56ae2b70.png)
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
