2024 AMC 10A Problem 3

1. Should $2$ be one of the five primes? Every prime except $2$ is odd. If $2$ is used, the other four primes are odd, and four odd numbers add to an even number. Then the total is even plus $2$, which is even. An even number bigger than $2$ is not prime. So $2$ cannot be used, and all five primes are odd.
2. Try the smallest choice. The five smallest odd primes are $3, 5, 7, 11, 13$:
$$3 + 5 + 7 + 11 + 13 = 39.$$
But $39 = 3 \cdot 13$, so it is not prime.
3. The next prime is $41$. Can we reach it? Any other set of five odd primes has to use a prime of at least $17$ (because it cannot be exactly $3, 5, 7, 11, 13$). The other four primes add to at least $3 + 5 + 7 + 11 = 26$. So the total is at least $26 + 17 = 43$. That means $41$ (and $40$, $42$) cannot be reached.
4. Try $43$. It works: $3 + 5 + 7 + 11 + 17 = 43$, and $43$ is prime.
5. Answer the question asked. The smallest such prime is $43$, and its digits add to $4 + 3 = 7$.

The answer is $\boxed{\textbf{(B)}\ 7}$.

Odd and even decide the first step: an odd total built from five primes cannot include $2$. After that, finding the "smallest" is a matter of starting at the smallest possible total and checking the next few numbers one at a time. Read the last line of the question again at the end: it wants the sum of the digits, not the prime.

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 $\boxed{\textbf{(B)}\ 7}$.