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Find ALL solutions to $23x + 13y = 275$ when $x,y \in \Bbb N$.

How can I approach this exercise? do I need to use GCD?

I got to $y = 23k + 16, x = 13t + 4$, when $k,t \in \Bbb Z$

6 Answers 6

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Since $ \gcd(23,13)= 1 = 4\cdot 23 -7\cdot 13$, the general integer solution for $23x + 13y = 275$ is $x=13t+4\cdot275, y=-23t-7\cdot275$ with $t \in \mathbb Z$.

Now solve $x \ge 0, y \ge 0$ for $t$. You'll find that $t = -84, x = 8, y = 7$ is the only solution.

Indeed:

$x=13t+4\cdot275$ and $x\ge0$ imply $t \ge -1100/13 > -84.7$

$y=-23t-7\cdot275$ and $y\ge0$ imply $t \le -1925/23 < -83.6$

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    How did you get $x$ , $y$? Why did u multiply by $275$? gawd im confused.. I need to find some good material on that subject – 2017-01-09
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    @IlanAizelmanWS, see https://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations for instance. – 2017-01-09
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    I will update later, I like YouTube better. :) – 2017-01-09
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    why not $y = 23t -7 \cdot 275$? – 2017-01-09
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    @IlanAizelmanWS, because it does not work ($t$ does not cancel). – 2017-01-09
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    Still didnt manage to find $t$ from the two equations $x,y$ gawd.. – 2017-01-09
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Hint :there is a linear combination of 23 and 13 that equals to 1(why?) , can you adjust this combination such that it would yield 275?

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    I tried getting $y = 23k + 16, x = 13t + 4$, when $k,t \in Z$ – 2017-01-09
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Hint: Use the Euclidean Algorithm to express $13$ and $23$ as a linear combination of $\gcd(13,23)=1$.

Then proceed with doing Belgi's hint.

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This is called a linear Diophantine equation, if $x, y \in \mathbb{Z}$. There exists an algorithm on how to systematically solve it. And yes it involves the $\gcd$.

You can determine the solutions in $\mathbb{Z}$ and then restrict them to $\mathbb{N}_0$.

WA gives $$ x = 8 - 13 k \\ y = 23 k + 7 $$ for $k \in \mathbb{Z}$. $$ x = 8 - 13 k \ge 0 \iff k \le 8/13 \Rightarrow k \le 0 \\ y = 23 k + 7 \ge 0 \iff k \ge -7/23 \Rightarrow k \ge 0 $$ This leaves only $k=0$ and we get the solution $(x,y) = (8,7)$.

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In this case, brute-forcing isn't completely insane.

By successive subtractions, the possible values of $275-23m$ are

$$275,252,229,206,183,160,137,114,\color{green}{91},68,45,22.$$

And by successive additions, the multiples of $13$ are

$$0,13,26,39,52,65,78,\color{green}{91},107,130,143,156,169,182,195,208,221,234,247,260,273.$$

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In some cases it is possible to have an easy brute force search. Here for instance the divisibility by $5$ brings great thnigs.

$23x+13y\equiv3x+3y\equiv3(x+y)\equiv275\equiv0\pmod5$ and since $3\land5=1$ we get $x+y\equiv0\pmod 5\iff x+y=5k$.

So $23x+13y=65k+10x=275$, obviously $k$ is also odd for $10$ to divide $275-65k$

$ \begin{array}{c|cccc} k & 65k & 275-65k & x & y\\ \hline 1 & 65 & 210 & 21 & -16\\ 3 & 195 & 80 & 8 & 7\\ \end{array} $

Brute force is reduced to only 2 choices!