diophantine equation $ 5x + 19y = 100 $ with natural numbers

Question

5x + 19y = 100

given the constraint that the two variables belong to N

I get the answer:

x = 400 y = (-100)

I dont know if this is the right asnwer to the question

Whenever you have an algebra problem and you don't know if an answer is correct , you only need to plug the numbers in and see what happens. What result do you get when you plug the values in? — Matti P., Jan 31 '20 at 12:47
my numbers works but do they meat the requirment (given the constraint that the two variables belong to N) — zellez11, Jan 31 '20 at 12:49
No, your solution does not work since $-100$ is not a natural number. — lulu, Jan 31 '20 at 12:51
Does this answer your question? Linear diophantine equation $100x - 23y = -19$ — Sophie, Feb 01 '20 at 14:34

score 1 · Answer 1 · answered Jan 31 '20 at 12:50

1

Since $5x$ and $100$ are divisible by $5$, so must $19y$. Since $19$ is not divisible by $5$, then $y$ must be. So $y = 5, 10, 15 \ldots$.

Now $5x= 100 - 19y$ has to be positive, so $100 > 19 y$, so $y < 100/19 < 6$. There for the only possibility is $y=5.$

answered Jan 31 '20 at 12:50

B. Goddard

1

can it not be x,y (20, 0) – zellez11 Jan 31 '20 at 13:00
Most people use $N$ for ${1, 2, 3, \ldots}.$ If $0$ is included, they call it $N_0.$ – B. Goddard Jan 31 '20 at 13:01

lulu · Answer 2 · 2020-01-31T13:26:33.190

1

Note first that we must have $5\,|\,y$, so write $y=5Y$. Then we can divide by $5$ to obtain $$x+19Y=20$$

Thus the integer solutions (without attention to signs) are the pairs $(x, 5Y)$ with $x=20-19Y$.

For instance, taking $Y=1$ we get $x=20-19\times 1=1$ hence the solution $(1,5)$.

Taking $Y=2$ we get $x=20-19\times 2=-18$ hence the solution $(-18,10)$.

And so on.

If you require both $x,Y$ to be positive we see that the only solution is $(1,5)$, since $Y>1\implies x= 20-19Y<0$.

To be clear: the question asked about "natural numbers". Some people define natural numbers as integers $≥0$, others define it as integers $≥1$. If you allow $0$ to be a natural number then you must also consider the solution with $Y=0$, hence $(20,0)$.

edited Jan 31 '20 at 13:26

answered Jan 31 '20 at 12:52

lulu

70,402

can it not be x,y (20, 0) – zellez11 Jan 31 '20 at 12:58
This depends on your definition of natural numbers. For me, a natural number must be $≥1$, so it excludes $0$. If you allow $0$ then your solution works. People disagree as to the definition. I have edited my post to stress this point. – lulu Jan 31 '20 at 13:06

score 1 · Accepted Answer · answered Jan 31 '20 at 13:19

5 divides into 19 3 times with remainder 4: 19- 3(5)= 4

4 divides into 5 once with remainder 1: 5- 4= 1

So 5- (19- 3(5))= 4(5)- 1(19)= 1

Multiply by 100: 5(400)- 19(100)= 100.

One solution to 5x + 19y = 100 is x= 400, y= -100.

But x= 400- 19n, y= -100+ 5n is also a solution for any n: 5(400- 19n)+ 19(-100+ 5n)= 5(400)- 19(100)- 95n+ 95n= 100.

y will be positive for any n greater than 20. In particular, for n= 21, y= -100+ 105= 5. If n= 21, x= 400- 19(21)= 400- 399= 1.

The only solution with both x and y positive is x= 1, y= 5.

score 0 · Answer 4 · answered Jan 31 '20 at 13:06

In general for the equation $ax+by=c$ if you have a particular solution $(x_0,y_0)$ then you can do $$ax+by=c\\ax_0+by_0=c$$ so you get $$a(x-x_0)+b(y-y_0)=0$$ from which making $x-x_0=bt$ and $y-y_0=-at$ you get the identity $abt-abt=0$ from which all the solutions (there are infinitely many) $$x=x_0+bt\\y=y_0-at$$ where $t$ is a parameter taking all integer value. You can discriminate if you want to have only natural numbers (which obviously are in finite number of them).

diophantine equation $ 5x + 19y = 100 $ with natural numbers

4 Answers4