Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I solve $6y + 5z = 960$ for $0 < y < z$ using the Extended Algorithm?

share|cite|improve this question
up vote 3 down vote accepted

First, you use the Euclidean algorithm to find all integer solutions.

This is very simple here. Since $\gcd(6,5)=1$, and $1$ divides $960$, there are certainly integer solutions. You can write $1 = 6-5$ (that's the Euclidean algorithm at work, though perhaps not very impressively). Thus, one integer solution is obtained by multiplying through by $960$: $960 = 6(960) - 5(960)$. That is, $y=960$, $z=-960$.

Now, since the $\gcd$ of $6$ and $5$ is $1$, every other integer solution is of the form $y=960-5k$, $z=6k -960$. The values of $k$ for which $y$ is positive are those in which $5k\lt 960$, or $k\lt \frac{960}{5} = 192$. The values of $k$ for which $z$ is greater than $y$ are those in which $960-5k \lt 6k-960$, or equivalently, those in which $1920\lt 11k$, or $k\gt\frac{1920}{11} \gt 174$. Thus, the values are those in which $174\lt k\lt 192$.

So the solutions are $y=960-5k$, $z=6k-960$, $175\leq k\leq 191$.

share|cite|improve this answer

You can see that $(y, z) = (0, 192)$ is a a solution; since you want $y, z > 0$, do not add it to your list. Since $\gcd(5,6) = 1$, we obtain the next possible solution by docking 6 from $z$ and adding 5 to $y$ to get $(5, 186)$. We keep this one since $y, z > 0$. Keep doing this until $y \ge z$ or until $y\le 0$ or $z \le 0$. You then have a list of solution-tuples.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.