0

To find the integer values of $c$ for which the equation $$20x+22y+cz = 315$$ has integer solutions $(x,y,z)$.

One observation is that $c$ has to be odd since $315$ is odd but $20x+22y$ is even.

If $y \neq 5$, then we must have $c$ such that $\gcd(22,c) = 1$, then we can find $z$ such that $5$ divides $22y+cz$ since we already have $315-20x$ is divisible by $5$.

Can some one give some hints how to proceed with the problem?

Bill Dubuque
  • 282,220
User56
  • 197
  • Why there is a dislike ...sorry I can't understand. – User56 May 17 '22 at 16:56
  • Side note: It is possible to squeeze "For which integer $c$ does $20x+22y+cz = 315$ have integer solutions $(x, y, z)$?" into the title? Somewhat less explicit than the body, but completes the question. You'll have to decide whether that's worth it. – Brian Tung May 17 '22 at 16:56
  • 1
    Don't know why there's a down-vote. I'm afraid many people don't explain them... – Brian Tung May 17 '22 at 16:57
  • 1
    By the obvious trivariate extension of the Proof in the linked dupe, it is solvable $\iff 315$ is a multiple of $(20,22,c)= (20,2,c) = (2,c),,$ i.e. iff $c$ is odd. – Bill Dubuque May 17 '22 at 17:11

1 Answers1

0

Consider what the set $ A = \{22x+20y|x,y\in\mathbb{Z}\}$ looks like.

Once you understand that, you can reduce this problem to which integers can be added to some element of this set to get 315. In other words, what are the elements of $ B = \{315-x| x\in A\} $. Then any number which is a factor of any element of B is a possible value for $c$. And any number which is not a factor of some element of B, is not a possible value for $c$.

Moosh
  • 624