What is the value of ((2112-2021)^2)/(169)?

2021 AMC 10A Fall Problem 1 — Solution

2021 AMC 10A Fall Problem 1

Below is the professionally curated solution for Problem 1 of the 2021 AMC 10A Fall, 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.

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Concepts:fraction exponent

Difficulty rating: 450

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

$7$

$21$

$49$

$64$

$91$

Solution:

We can simplify the expression as follows: $\begin{align*} \dfrac{(2112 - 2021)^2}{169} &= \dfrac{91^2}{169} \\ &= \dfrac{91^2}{13^2} \\ &= \left(\dfrac{91}{13}\right)^2 \\ &= 7^2 \\ &= 49. \end{align*}$

Thus, C is the correct answer.

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