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Find number of positive integer solutions $(x,y,z)$ for the following equation:

$19x + 11y + 8z = 240$

I divided the equation by $8$ and then tried to equate remainders. It yields that $3(x + y) = 8k$ for some constant $k$ or $x + y$ is a multiple of 8. Can't choose which combinations of $x,y$ will do the work.

Solved!

For $ x + y = 8 $ All ordered pairs are allowed. Corresponding values of $z$ are within domain of positive integers. This gives 7 solutions.

For $ x + y = 16 $ Only 7 pairs of $ x,y $ allowed. They are $ (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9) $ Pairs $(8,8)$ and beyond yield negative values for $z$. Therefore total of 14 solutions.

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  • $\begingroup$ Not clear what you are asking. Do you mean integer solutions? What does nos mean? $\endgroup$ – MrYouMath Oct 16 '17 at 7:35
  • $\begingroup$ Yes, number of positive integer solution. $\endgroup$ – Ajax Oct 16 '17 at 7:36
  • $\begingroup$ Have you tried anything here? $\endgroup$ – Mark Bennet Oct 16 '17 at 7:45
  • $\begingroup$ @MarkBennet what do you mean? $\endgroup$ – Ajax Oct 16 '17 at 7:46
  • $\begingroup$ Well, have you made an attempt to solve this yourself? Where did you get stuck? Is there something about the situation you don't understand? It is a finite problem, so one way is simply to try all possible solutions. Have you tried any ways of solving this? There are tricks you can use with particular problems which might not work so well in the general case - are you looking for tricks or for general methods? Give some context so we can help. $\endgroup$ – Mark Bennet Oct 16 '17 at 7:49
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I had a very similar method, which takes advantage of $19-11=8$ and also the factor $8$ generally. Rewrite as $$11(x+y)+8(x+z)=240=8\times 30$$

$x+y$ must be divisible by $8$ and the only possibilities are $x+y=8, 16$. In the first of these cases $x+y=8$ has seven solutions. In the second we find $x+z=8 (=30-2\times 11)$ which has seven solutions. All these solutions evidently give a positive integer value for the missing number, so there are fourteen solutions.

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