2021 AMC 10A Fall Problem 1

Attempt Problem 1 of the 2021 AMC 10A Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Fall solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

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

77

2121

4949

6464

9191

Answer: C
Concepts:fractionexponent
Difficulty rating: 450
Solution:

Since 21122021=91=713,2112-2021=91=7\cdot13, (21122021)2169=(713)2132=72=49. \begin{aligned} \frac{(2112-2021)^2}{169} &=\frac{(7\cdot13)^2}{13^2}\\ &=7^2=49. \end{aligned}

Thus, C is the correct answer.

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