How to solve system of linear diophantine equation with parameters

Question

Welcome,

I have a problem with this excercise:

A jeweler sold 60 watches of brand X last month for a total of € 1920. There are two types of this brand: one with stopwatch and one without. The watch with stopwatch is € 18 more expensive than the other type. The watch without a stopwatch costs more than € 20. How many watches of each type did the jeweler sell last month?

I tried to solve this system of eqations:

$$\begin {array} \\ x+y=60 \\ (38+k)x + (20+k)y = 1920 \end {array}$$

But I have no idea how I have to handle a linear diophantine equation with this k parameter.

I hope that you can help me.

Answer 1

We have that

$$x+y=60$$ $$(38+k)x+(20+k)y=1920$$

Multiplying the first equation by $(20+k)$ we get

$$(20+k)x+(20+k)y=60(20+k)$$ $$(38+k)x+(20+k)y=1920$$

Subtracting, we have $18x=720-60k$. Hence $x=40-\frac{10k}{3}$.

And since $x+y=60$, we have that $y=20+\frac{10k}{3}$.

The fact that $x$, $y$ are integers implies that $\frac{10k}{3}$ is an integer. Hence $3$ must divide $k$.

The fact that $x,y\ge0$, gives us that $40-\frac{10k}{3}\ge0$ and $20+\frac{10k}{3}\ge0$. Hence $-20\le\frac{10k}{3}\le40$. So $-6\le k\le12$.

Hence $k$ must be either $-6,-3,0,3,6,9,12$;

in which case $(x,y)$ would be $(60,0),(50,10),(40,20),(30,30),(20,40),(10,50),(0,60),$ respectively.

Edit: The above solution contains one or two mistakes. First, $20+k$ was the price of the watches that did not have stopwatches. Since we were given that these watches cost more than € $20$, it follows that $k>0$.

Hence the value of $k$ should really be either $3,6,9,12$;

In which case $(x,y)$ would be $(30,30),(20,40),(10,50),(0,60),$ respectively.

The other potential mistake I made was assuming that $k$ was an integer. We clearly have that $x$, $y$ are non-negative integers, since these denote the number of each type of watch that was sold. We also have that $\frac{10k}{3}$ is an integer since $x=40-\frac{10k}{3}$.

So if we drop the assumption that $k$ is an integer, then we're just left with the fact that $\frac{10k}{3}$ is an integer with $0<\frac{10k}{3}\le40$.

It follows that $\frac{10k}{3}$ must be either $1,2,3,\ldots,38,39,40$;

in which case $k$ must be either $0.3,0.6,0.9,\ldots,11.4,11.7,12$;

and $(x,y)$ would be either $(39,21),(38,22),(37,23),\ldots,(2,58),(1,59),(0,60)$ respectively.

Hence if we drop the assumption that $k$ is an integer, then the solutions are the pairs of integers $(x,y)$ with $y=60-x$ and $0\le x\le39$.