Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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== Solution 3 == | == Solution 3 == | ||
We have | We have | ||
− | + | <cmath> | |
\frac{\left( 2112 - 2021 \right)^2}{169} | \frac{\left( 2112 - 2021 \right)^2}{169} | ||
& = \frac{91^2}{13^2} \\ | & = \frac{91^2}{13^2} \\ | ||
& = 7^2 \\ | & = 7^2 \\ | ||
& = 49 . | & = 49 . | ||
− | + | </cmath> | |
Therefore, the answer is \boxed{\textbf{(C) 49}}. | Therefore, the answer is \boxed{\textbf{(C) 49}}. |
Revision as of 20:46, 25 November 2021
- 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
We have
\[\frac{\left( 2112 - 2021 \right)^2}{169} & = \frac{91^2}{13^2} \\ & = 7^2 \\ & = 49 .\] (Error compiling LaTeX. Unknown error_msg)
Therefore, the answer is \boxed{\textbf{(C) 49}}.
~Steven Chen (www.professorchenedu.com)
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.