2021 Fall AMC 12A Problems/Problem 1: Difference between revisions
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What is the value of <math>\frac{(2112 - 2021)^2}{169} | {{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 == | |||
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 | |||
==See Also== | |||
{{AMC12 box|year=2021 Fall|ab=A|before=First Problem|num-a=2}} | |||
{{AMC10 box|year=2021 Fall|ab=A|before=First Problem|num-a=2}} | |||
{{MAA Notice}} | |||
Revision as of 17:22, 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 
?
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
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| 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 12 Problems and Solutions | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| 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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