Question 1189399
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5002 = 2*41*61 = a*b*c<br>
a+b+c=104<br>
The question asks how many other positive integers are the products of exactly three distinct primes whose sum is also 104.<br>
For the sum of three distinct primes to be 104, one of them has to be 2, so the sum of the other two has to be 102.<br>
So you need to find the number of pairs of primes other than 41 and 61 that also have a sum of 102.<br>
You will learn nothing from this question if we work the whole problem for you, so I leave it to you to finish the problem.<br>
(my answer was 7....)<br>