1. **Problem statement:** We need to find three distinct prime numbers whose sum is 30 and whose product is as large as possible. 2. **Recall prime numbers:** Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. 3. **Set up the problem:** Let the three distinct primes be $p_1$, $p_2$, and $p_3$ such that $$p_1 + p_2 + p_3 = 30$$ with $p_1 < p_2 < p_3$ to avoid repetition. 4. **Strategy:** We want to maximize the product $$P = p_1 \times p_2 \times p_3$$ under the sum constraint. 5. **Check possible triples:** Since 2 is the only even prime, and the sum is even (30), either all three primes are odd (sum of three odd primes is odd) or one prime is 2 (even) and the other two are odd. 6. **Try including 2:** Let $p_1 = 2$, then $$p_2 + p_3 = 28$$ Try pairs of distinct odd primes summing to 28: - 3 and 25 (25 not prime) - 5 and 23 (both prime) - 11 and 17 (both prime) - 13 and 15 (15 not prime) Calculate products: - $2 \times 5 \times 23 = 230$ - $2 \times 11 \times 17 = 374$ 7. **Try without 2:** Three odd primes summing to 30 is impossible because sum of three odd numbers is odd. 8. **Try other combinations:** Check if any other triple sums to 30: - 3, 7, 20 (20 not prime) - 3, 11, 16 (16 not prime) - 5, 7, 18 (18 not prime) No other valid triples. 9. **Conclusion:** The triple $(2, 11, 17)$ sums to 30 and has the largest product $374$. **Final answer:** $$\boxed{374}$$