2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 1 Easy

What is the value of \frac{(2112-2021)^{2}}{169} ?(2021 AMC Fall 10A, Question #1)

  • A.

    7

  • B.

    21

  • C.

    49

  • D.

    64

  • E.

    91

Answer:C

Solution 1 (Laws of Exponents):

We have \frac{(2112-2021)^{2}}{169}=\frac{91^{2}}{169}=\frac{91^{2}}{13^{2}}=\left(\frac{91}{13}\right)^{2}=7^{2}=(\mathbf{C}) 49

Solution 2 (Difference of Squares) :

We have \frac{(2112-2021)^{2}}{169}=\frac{91^{2}}{169}=\frac{\left(10^{2}-3^{2}\right)^{2}}{169}=\frac{(13 \cdot 7)^{2}}{169}=\frac{13^{2} \cdot 7^{2}}{13^{2}}=7^{2}=(\mathbf{C}) 49

Related Problems
Problem 2 2021 AMC 10 A Fall
Problem 3 2021 AMC 10 A Fall
Problem 4 2021 AMC 10 A Fall
Problem 5 2021 AMC 10 A Fall
Think Academy Powered by ChatGPT