2020 AMC 10A Problem 24

Attempt Problem 24 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

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

1212

1515

1818

2121

2424

Answer: C
Concepts:greatest common divisorChinese Remainder Theoremmodular arithmetic
Difficulty rating: 1820
Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

Written solution:

The first gcd condition gives n+1200(mod21)n+120\equiv0\pmod{21}, so n6(mod21)n\equiv6\pmod{21}, but n+120n+120 must not be divisible by 6363. The second gives n+630(mod60)n+63\equiv0\pmod{60}, so n57(mod60)n\equiv57\pmod{60}, but n+63n+63 must not be divisible by 120120.

Solving n6(mod21)n\equiv6\pmod{21} and n57(mod60)n\equiv57\pmod{60} gives n237(mod420)n\equiv237\pmod{420}. The candidates above 10001000 are 1077,1497,1917,1077,1497,1917,\ldots. The first fails the first gcd condition, the second fails the second gcd condition, and 19171917 works. The digit sum is 1818. Thus, C is the correct answer.

← Problem 23#23
Full Exam

Problem 24 in Other Years