Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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\begin{align*} | \begin{align*} | ||
\frac{\left( 2112 - 2021 \right)^2}{169} | \frac{\left( 2112 - 2021 \right)^2}{169} | ||
| − | = \frac{91^2}{13^2} \\ | + | & = \frac{91^2}{13^2} \\ |
| − | = 7^2 \\ | + | & = 7^2 \\ |
| − | = 49 . | + | & = 49 . |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Revision as of 19:49, 25 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.
Contents
Problem
What is the value of 
?
Solution 1 (Laws of Exponents)
We have
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{\textbf{(C) } 49}.\]](http://latex.artofproblemsolving.com/7/0/4/704a7253fd5b1d77107061770699bfe555788438.png)
~MRENTHUSIASM
Solution 2 (Difference of Squares)
We have
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{169}=\frac{(10+3)^2(10-3)^2}{169}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.\]](http://latex.artofproblemsolving.com/9/b/6/9b64028dbd1ab61b284f7b50a9d9b653caab91bc.png)
Solution 3
We have
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
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. 
