"Rizky" Business

The second equation can be written as $y(x + z) = 37$ . Since $37$ is prime and $(x + z) \ge 2$ , (since $x,z$ are positive integers), we must have that $y = 1$ and $x + z = 37 \Longrightarrow z = 37 - x$ .

The second equation can then be written as

$x + x(37 - x) = 357 \Longrightarrow x^{2} - 38x + 357 = 0 \Longrightarrow (x - 17)(x - 21) = 0$ .

(Alternatively we can write it as $x(1 + z) = x(38 - x) = 357 = 3 \times 7 \times 17$ and then find the two factors from $1,3,7,17,21$ which add to $38$ .)

So there are $N = 2$ solution triples, namely $(17,1,20)$ and $(21,1,16)$ , and since $17 \times 20 \gt 21 \times 16$ we have $M = 17 \times 1 \times 20 = 340$ . Thus $N + M = 2 + 340 = \boxed{342}$ .

The second equation has a prime on the r. h. s., so y(x + z) = 37, so y = 1 , or y = 37. If y = 37, x + z = 1, which is impossible in positive integers x and z, so y = 1 only. We then have the equations: (1) x + xz = 357, and (2) x + z = 37. Substituting x = 37 - z into (1), z^2 - 36z + 320 = (z - 20)*(z - 16) = 0, so the two solutions are: x,y,z) =(17,1,20) and (16,1,21). Therefore N = 2 and M = 340, and N + M = 342. Ed Gray