2021 AMC 12A Fall 第 4 题

The number is $202100 + A;$ even $A$ makes it even, and $A = 5$ makes it a multiple of $5$

For $A = 1$ and $A = 7$ the digit sum is a multiple of $3,$ and $202103 = 11 \cdot 18373$

The number is $202100 + A.$ Any even $A$ makes it even, and $A = 5$ makes it divisible by $5,$ so $A$ must be odd and not $5.$

For $A = 1$ the digit sum is $6$ (divisible by $3$); for $A = 7$ the digit sum is $12$ (divisible by $3$); and $202103 = 11 \cdot 18373.$ Only $202109$ survives all tests, and it is prime.

Thus, the correct answer is E.