Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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− | {{duplicate|[[2021 Fall AMC 10A Problems | + | {{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 7: | Line 7: | ||
<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 == | + | == Solution 1 (Laws of Exponents) == |
− | <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> | + | 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 | ~MRENTHUSIASM | ||
− | + | == Solution 2 (Difference of Squares) == | |
+ | We have | ||
+ | <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 (Estimate)== |
+ | 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>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath> | ||
+ | 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>. | ||
+ | |||
+ | ==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== | ==See Also== |
Latest revision as of 12:23, 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.
Contents
Problem
What is the value of ?
Solution 1 (Laws of Exponents)
We have
~MRENTHUSIASM
Solution 2 (Difference of Squares)
We have
Solution 3 (Estimate)
We know that . Approximate this as
as it is pretty close to it. Also, approximate
to
. We then have
Now check the answer choices. The two closest answers are
and
. As the numerator is actually bigger than it should be, it should be the smaller answer, or
.
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
~Charles3829
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/o98vGHAUYjM
for AMC 12: https://youtu.be/jY-17W6dA3c
~IceMatrix
Video Solution
~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.