2021 AMC 12A Fall Problem 4

Below is the professionally curated solution for Problem 4 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:prime divisibility

Difficulty rating: 1200

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 $A?$

$1$

$3$

$5$

$7$

$9$

Solution:

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.

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