Wednesday, November 2nd

Problem 
Find all possible naturals $x,y,z$ such that $xyz=4104$ and $x+y+z=77$


Solution 
First we notice that $4104=2^3 3^3 19$. There are a total of $\binom{5}{2}\binom{5}{2}\binom{3}{2}$ such triples that satisfy the first condition. For the second condition, we notice that one of the numbers $x,y,z$ has to be a multiple of 19. Going through the possibilities (19, 38, 57, etc) we can easily find that the only triples that work are $(4,54,19)$ and $(3,36,38)$ and their permutations.