2021 Fall AMC 10A Problems/Problem 1: Difference between revisions
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== Problem == | |||
What is the value of <math>\frac{(2112-2021)^2}{169}</math>? | What is the value of <math>\frac{(2112-2021)^2}{169}</math>? | ||
<math>\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91</math> | <math>\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91</math> | ||
== Solution == | |||
We have <cmath>\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}.</cmath> | |||
~MRENTHUSIASM | |||
Revision as of 17:36, 22 November 2021
Problem
What is the value of 
?
Solution
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
