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

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

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

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions