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2020 AMC 10A Problem 24

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2020 AMC 10A Problem 24AMC 10 Step by Step
Name:                             Date: September 13, 2026
GCD & LCM · difficulty 4 of 5 · about 8 min · position P21-25

Let nn be the least positive integer greater than 10001000 for which gcd⁡(63,n+120)=21andgcd⁡(n+63,120)=60.\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60. What is the sum of the digits of nn ?

  1. A)

    1212

  2. B)

    1515

  3. C)

    1818

  4. D)

    2121

  5. E)

    2424

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