Problem 19. The product of three prime numbers is 11 times their sum. What is the largest possible value of their sum? So, we know that the product of the three prime numbers is divisible by 11, since it's 11 times the sum. So this means that one of the prime numbers has to be 11. We can call the other ones p and q. Then we know that 11 times p times q is 11 times the sum, which is 11 plus p plus q. So if we divide both sides by 11, we know that pq is 11 plus p plus q. Now, we can see that the left-hand side grows much faster than the right-hand side. So as we increase p and q, the left-hand side grows by more than the right-hand side. So once we reach a point where the left-hand side is greater than the right-hand side, then by further increasing p and q, the left-hand side will only be more greater than the right-hand side, and so we'll never reach a point where the sides are equal again. So we see that already if p and q are 5, then 25 is greater than 21. And so if p is greater than or equal to 5 and q is greater than or equal to 5, it will always be that the left-hand side is greater than the right-hand side. So if this equation is true, then 1 of p and q has to be less than 5. And it doesn't matter which one we pick, let's just say p is the smaller of the two. So p has to be less than 5 and p is prime, so p is either 2 or p is 3. If p is 2, then our equation pq is 11 plus p plus q becomes 2q is 13 plus q, so q is 13. And if p is 3, then our equation becomes 3q is 14 plus q, which gives us that q is 7. Now, the problem is asking us to find the largest possible value of the sum. So let's find the sum of the prime numbers in each case. When p is 2, the sum becomes 2 plus 11 plus 13, which is 26. And if p is 3, then we have 3 plus 7 plus 11, which is 21. And so the larger value for the sum is 26, and the answer is E.

prime numbers

largest sum

product equation

mathematical problem

sum of primes