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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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{{duplicate|[[2021 Fall AMC 10A Problems#Problem 1|2021 Fall AMC 10A #1]] and [[2021 Fall AMC 10A Problems#Problem 1|2021 Fall AMC 12A #1]]}}
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{{duplicate|[[2021 Fall AMC 10A Problems/Problem 1|2021 Fall AMC 10A #1]] and [[2021 Fall AMC 12A Problems/Problem 1|2021 Fall AMC 12A #1]]}}
  
 
== Problem ==
 
== Problem ==
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== Solution 2 (Difference of Squares) ==
 
== Solution 2 (Difference of Squares) ==
 
We have
 
We have
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<cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{169}=\frac{(10+3)^2(10-3)^2}{169}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.</cmath>
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<cmath>\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.</cmath>
  
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== Solution 3 ==
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==Solution 3 (Estimate)==
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We have
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We know that <math>2112-2021 = 91</math>. Approximate this as <math>100</math> as it is pretty close to it. Also, approximate <math>169</math> to <math>170</math>. We then have  
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<cmath>
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<cmath>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath>
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\frac{\left( 2112 - 2021 \right)^2}{169}
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Now check the answer choices. The two closest answers are <math>49</math> and <math>64</math>. As the numerator is actually bigger than it should be, it should be the smaller answer, or <math>\boxed{\textbf{(C) } 49}</math>.
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= \frac{91^2}{13^2} \\
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= 7^2 \\
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==Video Solution (Simple and Quick)==
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= 49 .
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https://youtu.be/wBf2Un_4fjA
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</cmath>
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 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/jSvTHKTkod8
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://youtu.be/sqjFA_CJNRc
 +
 
 +
~Charles3829
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
for AMC 10: https://youtu.be/o98vGHAUYjM
 +
 
 +
for AMC 12: https://youtu.be/jY-17W6dA3c
 +
 
 +
~IceMatrix
  
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Therefore, the answer is <math>\boxed{\textbf{(C) 49}}</math>.
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==Video Solution==
 +
https://youtu.be/3qohnl543-4
  
−
~Steven Chen (www.professorchenedu.com)
+
~Lucas
  
 
==See Also==
 
==See Also==

Revision as of 11:23, 21 July 2023

The following problem is from both the 2021 Fall AMC 10A #1 and 2021 Fall AMC 12A #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 1 (Laws of Exponents)

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

Solution 2 (Difference of Squares)

We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{\textbf{(C) } 49}.\]

Solution 3 (Estimate)

We know that $2112-2021 = 91$. Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$. We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$. As the numerator is actually bigger than it should be, it should be the smaller answer, or $\boxed{\textbf{(C) } 49}$.

Video Solution (Simple and Quick)

https://youtu.be/wBf2Un_4fjA

~Education, the Study of Everything

Video Solution

https://youtu.be/jSvTHKTkod8

~savannahsolver

Video Solution

https://youtu.be/sqjFA_CJNRc

~Charles3829

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/o98vGHAUYjM

for AMC 12: https://youtu.be/jY-17W6dA3c

~IceMatrix

Video Solution

https://youtu.be/3qohnl543-4

~Lucas

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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