Diophantine Equation (Hint or Help)

Question

Solve the equation below:

$19x+29y=1000$

My try:

$19x=1000-29y \rightarrow x=\frac {1000-29y}{19} , x \in N \rightarrow 1000-29y\equiv 0\pmod {19} \rightarrow 29y\equiv 1000\pmod{19} \rightarrow 10y\equiv 1000\pmod{19} \rightarrow y\equiv 100\pmod{19} \rightarrow y\equiv 5\pmod{19} \rightarrow y-5=19k \rightarrow y=19k+5$

Also from the first equation we can conclude that:

$29y\le 1000 \rightarrow y\le 34$

From the equation $y=19k+5$ and the inequality above, we conclude that $19k+5\le 34 \rightarrow k=1 $

For $k=1$ :

$y=19+5=24 \rightarrow x=16$ which is a correct answer

So I've found an answer but I wanted to know if there's any other answer or any point or anything (!) I might have missed.Thanks in advance.

EDIT:

Guys I'm so sorry but I forgot to say that $x,y \in N$.

Answer 1

Diophantine Equation implies solutions in integers, not necessarily positive

$$19x+29y=1000=100(29-19)\iff 19(x+100)=29(100-y)$$

$$\iff \frac{29(100-y)}{19}=x+100$$ which must be an integer

$$\implies19|29(100-y)\iff19|(100-y)\text{ as }(19,29)=1$$

$$\implies y\equiv100\pmod{19}\equiv5$$

Similarly, $$x+100\equiv0\pmod{29}\iff x\equiv-100\equiv16\pmod{29}$$

Answer 2

You may want to look into the general idea of solving $ax+by=c$ (with $a,b,c \in \mathbb{Z}$) for integer solutions. This equation will have solutions if and only if $\gcd(a,b)$ divides $c$.

In the present scenario, $\gcd(19,29)=1$ and $1$ divides $1000$ so for sure there will be a solution. To find that all you need to do is to first find the solutions to $$19x+29y=1.$$ This can be achieved by using Euclidean algorithm. Once you have a solution say $(x_0,y_0)$, then you can get the solutions to $$19x+29y=N.$$ by $$19(x_0N)+29(y_0N)=N.$$

Answer 3

It is clear that solutions of the equation: $19x+29y=1000$

Have the form:

$y=19k+5$

$x=45-29k$

It is seen that two positive solutions, $k=0$ ; $k=1$

If the numbers are different sign infinitely many.

Answer 4

Look at it mod 19: 19x +29y = 1000 mod 19 becomes 29y = 1000 mod 19 becomes 10y = 1000 mod 19 becomes y = 100 = 5 mod 19

if y = 5, x = 45

so if y = 5 + 19k x = 45 - 29k.

Answer 5

Select the lower coëfficient and isolate its term.
$19x+29y=1000\to 19x=1000-29y$
$x={{1000-29y}\over 19}\to{53-y+{{-7-10y}\over 19}}$
New variable $a={{-7-10y}\over 19}\to{-7-10y=19a}$
Select the lower coëfficient and isolate its term.
$10y=-7-19a\to{y={{-7-19a}\over 10}}=-1-2a+{{3+a}\over 10}$
New variable $b={{3+a}\over 10}\to{3+a=10b}$
$a=-3+10b$
$y={{-7-19(-3+10b)}\over 10}={{-7+57-190b}\over 10}=5-19b$
$x={{1000-29(5-19b)}\over 19}={{855+551b}\over 19}=45+29b$