2021 Fall AMC 10A Problem 5

Note that $A$ cannot be even, as then the number would be divisible by $2.$ $A$ also cannot be $5,$ as that would make the number divisible by $5.$ If $A$ equaled $1$ or $7,$ then the sum of the digits of the number would be $6$ and $12$ respectively. This would make the number divisible by $3,$ so that rules out $A$ equaling either of these numbers. Finally, if $A$ equals $3,$ then the whole number becomes $202103.$ If we look at the difference of the sums of alternating digits, we get $$2 + 2 - 1 - 3 = 0,$$ which means the number is divisible by $11.$ Every choice except $9$ makes the number composite. Because the problem states that exactly one digit works, that digit must be $A=9.$ (Indeed, trial division by the primes at most $\sqrt{202109}<450$ confirms that $202109$ is prime.) Thus, E is the correct answer.