Let A, B, C, M, and N be integers.
If A divides both B and C, A divides NB+MC.
The Attempt at a Solution
Since we have defined A to divide both B and C, there exists an M in the integers such that B = AM, and there exists an N in the integers such that C = AN.
So, it follows that:
A = B/M and A = C/N
B/M + C/N = A + A
B/M + C/N = 2A
B + MC/N = 2AM
NB + MC = 2AMN
It is shown that 2AMN = NB + MC, and since A divides 2AMN, A divides NB+MC.
Therefore, A divides NB + MC.