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2021 Fall AMC 12A Problems/Problem 1: Difference between revisions

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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]]}}
{{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 ==
Line 14: Line 14:
== Solution 2 (Difference of Squares) ==
== Solution 2 (Difference of Squares) ==
We have
We have
<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>
<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>


== Solution 3 ==
==Solution 3 (Estimate)==
We have
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  
<cmath>
<cmath>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath>
\begin{array*}
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>.
\frac{\left( 2112 - 2021 \right)^2}{169}
 
= \frac{91^2}{13^2} \\
==Video Solution==
= 7^2 \\
https://youtu.be/sqjFA_CJNRc
= 49 .
 
\end{array*}
~Charles3829
</cmath>
 
==Video Solution (Simple and Quick)==
https://youtu.be/wBf2Un_4fjA
 
~Education, the Study of Everything
 
==Video Solution==
https://youtu.be/jSvTHKTkod8
 
~savannahsolver
 
==Video Solution by TheBeautyofMath==
for AMC 10: https://youtu.be/o98vGHAUYjM
 
for AMC 12: https://youtu.be/jY-17W6dA3c
 
~IceMatrix


Therefore, the answer is <math>\boxed{\textbf{(C) 49}}</math>.
==Video Solution==
https://youtu.be/3qohnl543-4


~Steven Chen (www.professorchenedu.com)
~Lucas


==See Also==
==See Also==

Latest revision as of 09:52, 18 September 2024

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

https://youtu.be/sqjFA_CJNRc

~Charles3829

Video Solution (Simple and Quick)

https://youtu.be/wBf2Un_4fjA

~Education, the Study of Everything

Video Solution

https://youtu.be/jSvTHKTkod8

~savannahsolver

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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