2024 AMC 10A Problem 3

If $2$ were one of the five primes, the sum would be $2$ plus four odd numbers, which is even and larger than $2$, hence not prime. So all five primes are odd.

The smallest five odd primes give $3 + 5 + 7 + 11 + 13 = 39 = 3\cdot 13$, not prime.

Can the sum be $41$? Any five distinct odd primes other than $\{3,5,7,11,13\}$ must include a prime that is at least $17$, and the other four add to at least $3+5+7+11 = 26$, so the total is at least $43$. Thus $41$ is impossible.

And $43$ works: $3 + 5 + 7 + 11 + 17 = 43$, which is prime.

The digit sum of $43$ is $4 + 3 = 7$, so the answer is $\boxed{\textbf{(B)}\ 7}$.

Parity decides the first step: an odd target sum of five primes cannot contain the lone even prime $2$. After that, "smallest" is a bounding argument: start from the minimal candidate and show each smaller value is out of reach before exhibiting the one that works. Always finish by answering the question actually asked (the digit sum, not the prime).

• Reporting the prime 43 (or the composite 39) instead of its digit sum 7, or including 2 and getting an even sum.
• Declaring $39$ prime without checking; $39 = 3\cdot 13$.