2021 Fall AMC 10A Problems/Problem 1: Difference between revisions

Latest revision as of 17:25, 23 November 2021

Redirect to:

@@ Line 1: / Line 1: @@
-== Problem ==
+#REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_1]]
-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>
-== 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