2021 AMC Fall 10A Problem 5

A six-digit prime cannot be even and cannot end in $5$, so $A$ is one of $1, 3, 7, 9$; choice (C) is out.

Digit sum: $2 + 0 + 2 + 1 + 0 + A = 5 + A$. This is a multiple of $3$ when $A = 1$ ($6$) or $A = 7$ ($12$), so $202101$ and $202107$ are divisible by $3$. Choices (A) and (D) are out.

Now test $11$ with the alternating sum of digits from the right. For $202103$: $3 - 0 + 1 - 2 + 0 - 2 = 0$, which is divisible by $11$, so $202103 = 11 \cdot 18373$ is composite. Choice (B) is out.

The only digit left is $9$, and the problem guarantees exactly one digit works, so $202109$ is the prime.

The answer is $\boxed{\textbf{(E)}\ 9}$.

"Prime for exactly one digit" is an invitation to eliminate rather than to prove primality. Run the cheap divisibility tests in order of cost: last digit ($2$ and $5$), digit sum ($3$), alternating digit sum ($11$). Four of the five choices fall to these tests, and the problem statement certifies the survivor.

• Stopping after the rules for 2, 3 and 5 leave both 3 and 9, then guessing, instead of applying the alternating-sum test for 11.
• Miscomputing the digit sum and eliminating $A = 9$ (sum $14$, not a multiple of $3$) by mistake.