Difference between revisions of "2021 Fall AMC 12A Problems/Problem 1"
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− | What is the value of <math>\frac{(2112 - 2021)^2}{169} | + | {{duplicate|[[2021 Fall AMC 10A Problems#Problem 1|2021 AMC 10B #1]] and [[2021 Fall AMC 10A Problems#Problem 1|2021 AMC 12B #1]]}} |
+ | |||
+ | == Problem == | ||
+ | |||
+ | What is the value of <math>\frac{(2112-2021)^2}{169}</math>? | ||
<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 == | ||
+ | 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 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2021 Fall|ab=A|before=First Problem|num-a=2}} | ||
+ | {{AMC10 box|year=2021 Fall|ab=A|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 17:22, 23 November 2021
- The following problem is from both the 2021 AMC 10B #1 and 2021 AMC 12B #1, so both problems redirect to this page.
Problem
What is the value of ?
Solution
We have ~MRENTHUSIASM
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.