Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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Revision as of 13:10, 13 June 2023
- 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
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{169}=\frac{(13\cdot7)^2}{169}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.\]](http://latex.artofproblemsolving.com/7/2/4/724decb94a7c9bf53c286a088c65ed7d2c9b8099.png)
Solution 3 (Estimate)
We know that 
. Approximate this as 
as it is pretty close to it. Also, approximate 
to 
. We then have
![\[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\]](http://latex.artofproblemsolving.com/0/5/9/059ba0d910711e20cd3185bc352538e607caec13.png)
Now check the answer choices. The two closest answers are 
and 
. As the numerator is actually bigger than it should be, it should be the smaller answer, or 
.
Video Solution
~savannahsolver
Video Solution
~Charles3829
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/o98vGHAUYjM
for AMC 12: https://youtu.be/jY-17W6dA3c
~IceMatrix
Video Solution
~Lucas
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. 
