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2021 AMC Fall 10A Problem 1

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2021 AMC Fall 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$7$
B)
$21$
C)
$49$
D)
$64$
E)
$91$

Answer

Key insight

Since 2112 - 2021 = 91 = 7 * 13 and 169 = 13^2, the fraction is (91/13)^2 = 7^2 = 49.

Solution

First the subtraction: $2112 - 2021 = 91$ .

Now notice that the denominator is a perfect square, $169 = 13^2$ , and that $91 = 7 \cdot 13$ . So instead of squaring $91$ , rewrite the whole fraction as a square:
$\frac{91^2}{13^2} = \left(\frac{91}{13}\right)^2 = 7^2 = 49.$

The answer is $\boxed{\textbf{(C)}\ 49}$ .

Why this works

A fraction of squares is the square of a fraction, so simplify before squaring whenever the denominator is a perfect square. Knowing the small multiples of $13$ ( $91 = 7 \cdot 13$ , $169 = 13^2$ ) turns a two-minute division into a one-line computation.

The trap

Squaring 91 first (8281) and then long-dividing by 169, which is slow and error-prone, instead of taking the square root of 169 first.

Common mistakes

Squaring 91 first (8281) and then long-dividing by 169, which is slow and error-prone, instead of taking the square root of 169 first.
Forgetting to square after dividing, which gives $7$ , choice (A).

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Set up the equation/formula and compute; no special trick needed