Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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== 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}{ | + | <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)== | ==Solution 3 (Estimate)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== |
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.