2021 Fall AMC 12A Problems/Problem 1

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 $\frac{(2112-2021)^2}{169}$?

$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$

Solution 1 (Laws of Exponents)

We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{\textbf{(C) } 49}.\] ~MRENTHUSIASM

Solution 2 (Difference of Squares)

We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.\]

Solution 3 (Estimate)

We know that $2112-2021 = 91$. Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$. We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$. As the numerator is actually bigger than it should be, it should be the smaller answer, or $\boxed{\textbf{(C) } 49}$.

Video Solution

https://youtu.be/sqjFA_CJNRc

~Charles3829

Video Solution (Simple and Quick)

https://youtu.be/wBf2Un_4fjA

~Education, the Study of Everything

Video Solution

https://youtu.be/jSvTHKTkod8

~savannahsolver

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/o98vGHAUYjM

for AMC 12: https://youtu.be/jY-17W6dA3c

~IceMatrix

Video Solution

https://youtu.be/3qohnl543-4

~Lucas

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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