2021 AMC 12A Fall Problem 1

Below is the professionally curated solution for Problem 1 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:factoringperfect square

Difficulty rating: 890

1.

What is the value of (21122021)2169? \frac{(2112 - 2021)^2}{169}?

77

2121

4949

6464

9191

Solution:

Since 21122021=91=7132112 - 2021 = 91 = 7 \cdot 13 and 169=132,169 = 13^2, the fraction is (713)2132=72=49.\dfrac{(7 \cdot 13)^2}{13^2} = 7^2 = 49.

Thus, the correct answer is C.

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