B/M + C/N = 2AHow does this proof show that A divides NB+MC?

[1MileCrash]

Homework Statement

Theorem:

Let A, B, C, M, and N be integers.

If A divides both B and C, A divides NB+MC.

Homework Equations

The Attempt at a Solution

Proof:

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.

 

[Mark44]

Homework Statement

Theorem:

Let A, B, C, M, and N be integers.

If A divides both B and C, A divides NB+MC.

Homework Equations

The Attempt at a Solution

Proof:

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.

I think that this is not as direct as it could be.

You are given that A|B and A|C, which means that B = rA and C = sA for integers r and s.
Then NB + MC = N*rA + M*sA = A(Nr + Ms).

 

[1MileCrash]

I see... for some reason I felt compelled not to introduce any new integers, but I like what you did.

 

[Mark44]

B/M + C/N = A + A

B/M + C/N = 2A

You should avoid doing stuff like this, with a whole separate equation that does nothing more than show that A + A = 2A.

Instead, you could do this:
B/M + C/N = A + A = 2A