0
$\begingroup$

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 $19x+57y$

$\endgroup$
  • 1
    $\begingroup$ Just plug this in to 19x+57y $\endgroup$ – Aqua May 18 at 9:42
0
$\begingroup$

Hint:

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

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

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

$\endgroup$
0
$\begingroup$

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 $95$.

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

$17x+51y = 85\!\!\overset{\large (\ \ )/17_{\phantom{1_{1_1}}}\!\!\!\!\!\!\!\!\!}\iff x+3y= 5\!\!\overset{\large 19\,(\ \ )_{\phantom{1_{1}}}\!\!\!\!\!\!\!}\iff 19x+57y = 95$

$\endgroup$
  • $\begingroup$ 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 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 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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