2021 AMC Fall 10A Problem 5

The problem promises that exactly one digit works. So you do not need to prove a number is prime. You only need to cross out the four answers that are not prime.

1. Test for 5. If $A = 5$, the number $202105$ ends in $5$, so $5$ divides it. Cross out (C).
2. Test for 3. A number is a multiple of $3$ when its digit sum is. Here the digit sum is $2 + 0 + 2 + 1 + 0 + A = 5 + A$.
- $A = 1$: digit sum $6$, a multiple of $3$. Cross out (A).
- $A = 7$: digit sum $12$, a multiple of $3$. Cross out (D).
- $A = 3$: digit sum $8$, and $A = 9$: digit sum $14$. Neither is a multiple of $3$.
3. Test for 11. Two choices are left, $3$ and $9$. The rule for $11$: starting from the last digit, add and subtract the digits in turn. If the result is a multiple of $11$ (including $0$), so is the number. For $202103$:
$$3 - 0 + 1 - 2 + 0 - 2 = 0.$$
So $11$ divides $202103$ (indeed $202103 = 11 \times 18373$). Cross out (B).
4. Only $A = 9$ is left, and the problem says one digit works, so $202109$ is the prime.

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

When a problem says "only one of these works," cross out the others instead of proving the winner. Use the cheap tests first: last digit (for $2$ and $5$), digit sum (for $3$), then the add-subtract rule (for $11$). Here those three tests remove four of the five choices.

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 $\boxed{\textbf{(E)}\ 9}$.