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Task: Two items are being sold, small and large types. The small type costs $98$ cents each, the large type costs $158$ cents. In the end, the total amount of sold items is $17028$ cents $(\$170.28)$. All in all, there have been $100$ small and $100$ large items at the beginning.*

So I wrote down an equation $98x + 158y = 17028$, where x is the amount of small types and y the amount of large types, solved for y and plotted a function. The only natural solution for that equation is $x = 48, y = 78$, with $1 \le x,y \le 100$

Any idea how this could be proved? I tried writing $xy = l$ for an $l \in \mathbb{N}$, and used $y = \dfrac{l}{x}$ in the equation above. I got $98x + 158 \dfrac{l}{x} = 17028 \iff 98x^2 - 17028x + 158l = 0$ and tried solving this through the p-q-formula, but this approach was not quite constructive so far deciding what l has to be, so that x is a natural number.

poetasis
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X3nius
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  • Yeah that's it, and in this case $17028 \ge 15229$. However, how can the solutions be constructed? As far as I could read, this theorem also proves that the number 17028 can be written as 98x+158y in this case. And another thing: the two numbers 98 and 158 are not relatively prime? – X3nius Sep 22 '21 at 19:15
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    Suppose you have two solutions $98x_1 + 158y_1 = 98x_2+158y_2=17028$ then $98(x_1-x_2)=158(y_2 - y_1)$ and $49(x_1-x_2)= 79(y_2-y_1)$. Let $m=x_1-x_2$ (wolog assume it is positive) and $n=y_2-y_1$ then $49m = 79n$. That mean $49|79m$. But $49$ has no factors in common with $79$ so $49|m$ and likewise 79$n$. So the difference between $x_1$ and $x_2$ must be a multiple of $79$. As you found $x=48$ is one possible value, there can't be any onther values less or equal to $100$. – fleablood Sep 22 '21 at 19:34
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    " And another thing: the two numbers 98 and 158 are not relatively prime? " Just reduce. Divide everything by the $\gcd(98,158)$ and the resulting problem will have the exact same solutions. – fleablood Sep 22 '21 at 19:35
  • I didn't quite get that. Okay the equality $49m = 79n$ holds. But why does this mean 49|79m? I think, because there exists $m \in \mathbb{N}$ with $49m = 79n$, we have 49|79n? – X3nius Sep 22 '21 at 19:52
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    Yeah... it is not true that every $n \ge 15299 =(98-1)(158-1)$ can be created (no odd number can) but it is true that every even number $2n \ge 7488 = 2(49-1)(79-1)$ can. – fleablood Sep 22 '21 at 19:55
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    Yes $49|79n$ so $49|n$. If I typed $m$ it was typo. – fleablood Sep 22 '21 at 19:56
  • Okay then we have $49|n$ and likewise $79|m$. So the difference between $x_1$ and $x_2$ must be a multiple of 79. How does this lead to x = 48? – X3nius Sep 22 '21 at 20:00
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    Here is one link to the Frobenius coin problem. – Dietrich Burde Sep 22 '21 at 20:40

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