2. Let $(a, \mathrm{~b})$ denote the greatest common divisor of $a$ and $\mathrm{b}$. Suppose $\mathrm{n}$ is an integer greater than 2021, and $(63, n+120)=21, \quad(n+63,120)=60$. What is the sum of the digits of the smallest number $n$ that satisfies the above conditions?
【Answer】 21
【Analysis】Chinese Remainder Theorem.
$n+120$ is divisible by $21, 120 \div 21$ leaves a remainder of $15, 21-15=6$, so $n \div 21$ leaves a remainder of 6; this is equivalent to: $n$ divided by 3 leaves a remainder of $0, n$ divided by 7 leaves a remainder of 6;
$n+63$ is divisible by $60, 63 \div 60$ leaves a remainder of $3, 60-3=57, n \div 60$ leaves a remainder of 57. This is equivalent to $n$ divided by 3 leaves a remainder of $0, n$ divided by 4 leaves a remainder of $1, n$ divided by 5 leaves a remainder of 2, considering $(57+60k) \div 7$ leaves a remainder of 6, the smallest number satisfying the condition is $k=3, n$ is the smallest $=237, 2, 3, 5, 7$ the least common multiple is 420, the number satisfying the condition can be written as $237+420m$, when $m=5$, 2337 satisfies the condition, but $2337+63=2400, 2400$ and 120 have a greatest common divisor of 120, discard. The next number is 2757, the sum of its digits is 21