2024 AMC 10A Problem 3

Suppose $2$ is one of the five primes. Then the total is even and bigger than $2,$ so it is composite. Thus all five primes must be odd. The five smallest odd primes give $3 + 5 + 7 + 11 + 13 = 39$ $= 3 \cdot 13,$ which is not prime. The next possible sum is obtained by replacing $13$ with the next prime, $17;$ changing any earlier term forces at least as large an increase to keep the primes distinct. This gives $3 + 5 + 7 + 11 + 17 = 43,$ which is prime. Its digit sum is $4 + 3 = 7.$ Thus, B is the correct answer.