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If $17x+51y=85$, find the value of $19x+57y$

I know I could use substitution and figure this out but i wanted to use Diophantine equation. I'm just a little confused because I know that $\gcd(17,51)=17$ and $17|85$

I could use the extended EA to get that a particular solution is $x_0 =5$ and $y_0=0$

The complete solution is $x=5+51n$ and $y=0-17n$

Not sure where to go from here.

Also I know how to do the EA, I just didn't want to write out all the steps so I just put it into a EA calculator. I'm just a little confused on how to get the solutions for $19x+57y$

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    $\begingroup$ Just plug this in to 19x+57y $\endgroup$
    – nonuser
    May 18, 2019 at 9:42

3 Answers 3

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Expanding on Maria Mazur's comment:

Since you found one possible solution is $x_0 = 5$ and $y_0 = 0$, you can just substitute to get $17(5) + 51(0) = 85$, so it satisfies the original expression.

Now you can substitute this into $19x + 57y$ this to get $95$.

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  • $\begingroup$ you mean substitute $5$ and $0$ for x and y into $19x+57y$? $\endgroup$
    – user8358234
    May 18, 2019 at 17:22
  • $\begingroup$ also am i correct in my general solution for $17x+51y=85$? $\endgroup$
    – user8358234
    May 18, 2019 at 17:23
  • $\begingroup$ Yes, your general solution is correct and you can just substitute $5$ and $0$. $\endgroup$
    – Toby Mak
    May 19, 2019 at 0:48
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Hint:

You don't need the Euclidean algorithm to solve the problem. Note that

$19x+57y=(17x+51y)+2x+6y$.

On the other hand, $17x+51y=85\iff x+3y=5$.

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  • $\begingroup$ +1 Clever in hindsight ;) $\endgroup$
    – Toby Mak
    Oct 4, 2021 at 6:49
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$17x+51y = 85\!\!\overset{\large (\ \ )/17_{\phantom{1_{1_1}}}\!\!\!\!\!\!\!\!\!}\iff x+3y= 5\!\!\overset{\large 19\,(\ \ )_{\phantom{1_{1}}}\!\!\!\!\!\!\!}\iff 19x+57y = 95$

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  • $\begingroup$ thanks for this, I knew how to do it that way but I just wanted to use the Diophantine equation because I'm just a little confused on how to apply the solution I have for $17x+51y=85$ to get the solution for $19x+57y$ $\endgroup$
    – user8358234
    May 18, 2019 at 17:22
  • $\begingroup$ @user8358234 The above equivalences shows that they have the same set of solutions, namely $\, (x,y) = (5,0) + (3,-1)n\ $ (note the correction to your solution). $\endgroup$
    – Bill Dubuque
    May 18, 2019 at 17:34

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