Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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== Solution 2 (Difference of Squares) == | == Solution 2 (Difference of Squares) == | ||
We have | We have | ||
| − | <cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{ | + | <cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.</cmath> |
| − | ==Solution 3 ( | + | ==Solution 3 (Estimate)== |
We know that <math>2112-2021 = 91</math>. Approximate this as <math>100</math> as it is pretty close to it. Also, approximate <math>169</math> to <math>170</math>. We then have | We know that <math>2112-2021 = 91</math>. Approximate this as <math>100</math> as it is pretty close to it. Also, approximate <math>169</math> to <math>170</math>. We then have | ||
<cmath>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath> | <cmath>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath> | ||
| − | Now check the answer choices. The two closest answers are <math>49</math> and <math> | + | Now check the answer choices. The two closest answers are <math>49</math> and <math>64</math>. As the numerator is actually bigger than it should be, it should be the smaller answer, or <math>\boxed{\textbf{(C) } 49}</math>. |
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==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/ | + | https://youtu.be/sqjFA_CJNRc |
| + | |||
| + | ~Charles3829 | ||
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| + | ==Video Solution (Simple and Quick)== | ||
| + | https://youtu.be/wBf2Un_4fjA | ||
| − | ~ | + | ~Education, the Study of Everything |
==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/ | + | https://youtu.be/jSvTHKTkod8 |
| − | ~ | + | ~savannahsolver |
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
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~Lucas | ~Lucas | ||
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==See Also== | ==See Also== | ||
Latest revision as of 09:52, 18 September 2024
- 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
![\[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.\]](http://latex.artofproblemsolving.com/d/c/9/dc9593dd40678e9ec5ad0d3214453ca6227a76c9.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
~Charles3829
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
