Find a solution to the equation:
29x + 30y + 31z = 366
Rem: x,y,z must be positive
(In reply to
Explanation to Puzzle Answer: Method 1 by K Sengupta)
29x+30y+31z=366
=> 29(x+y+z) +y+2z = 366
=> x+y+z +(y+2z)/29 = 12 + 18/29
Thus, x+y+z =12, and y+2z=18
Then, z-x=6 and, x+y+z=12
First equation gives z=8 corresponding to x=2, whence we have y=2
First equation gives z =7 corresponding to x=1, whence we have y=4
If z>= 9, then z-x=6 yields x>=15
So, z+x >= 24 > 12= x+y+z.
This is a contradiction.
Consequently, (x, y, z) = (2, 2, 8), (1, 4, 7) constitutes all possible positive integer solutions to the equation under reference.
Explanation to Puzzle Answer: Method 2
