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

Revision as of 18:22, 23 November 2021

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-{{duplicate|[[2021 Fall AMC 10A Problems#Problem 1|2021 AMC 10B #1]] and [[2021 Fall AMC 10A Problems#Problem 1|2021 AMC 12B #1]]}}
+#REDIRECT [[2021 AMC 12A Problems/Problem 1]]
-== Problem ==
-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
-==See Also==
-{{AMC10 box|year=2021 Fall|ab=A|before=First Problem|num-a=2}}
-{{MAA Notice}}