2021 AMC Fall 10A Problem 5

P1-10~2 minPrint

Primes Divisibility & Factors Use the answer choices (mod checks, size, form) to eliminate or select Set up the equation/formula and compute; no special trick needed

The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$

Show solution without answering

Key insight

Cross out answers with quick tests: 5 ends in 5, digits 1 and 7 make the digit sum a multiple of 3, and 202103 is a multiple of 11.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)