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

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{{duplicate|[[2021 Fall AMC 10A Problems#Problem 1|2021 AMC 10B #1]] and [[2021 Fall AMC 10A#Problem 1|2021 AMC 12B #1]]}}
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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]]}}
  
 
== Problem ==
 
== Problem ==

Revision as of 17:19, 23 November 2021

The following problem is from both the 2021 AMC 10B #1 and 2021 AMC 12B #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

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

See Also

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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