2021 AMC 12A Fall Problem 1

Attempt Problem 1 of the 2021 AMC 12A 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 12A 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 (2112−2021)2169? \frac{(2112 - 2021)^2}{169}?
77
2121
4949
6464
9191
Answer: C
Concepts:factoringperfect square
Small hint:
2112−2021=912112 - 2021 = 91
Big hint:
91=7×1391 = 7 \times 13 and 169=132169 = 13^2
Solution:
Since 2112−2021=91=7⋅132112 - 2021 = 91 = 7 \cdot 13 and 169=132,169 = 13^2, the fraction is (7⋅13)2132=72=49.\dfrac{(7 \cdot 13)^2}{13^2} = 7^2 = 49.
Thus, the correct answer is C.
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