+82 If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+3xz+8yz+15x+40y+4z=127$, find $x+y+z$.
+128732 3xz + 6xyz + 15x + 30xy + 8yz + 40y + 4z = 127
3xz ( 1 + 2y) + 15x ( 1 + 2y) + 4z (1 + 2y) + 40y = 127
(1 + 2y) ( 3xz + 15x + 4z) + 40y = 127
Let y = 1
(3) (3xz + 15x + 4z) + 40 = 127
(3) (3xz +15x + 4z) = 87
3xz + 15x + 4z = 29
Let x = 1 and z = 2
3(1)(2) + 15(1) + 4(2) = 29
6 + 15 + 8 = 29
29 = 29
x + y + z = 4
