AoPS - Problem

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Tags: number theory, systems of equations, Digits, Diophantine Equations, diophantine

parmenides51

23.09.2021 02:29

The cost of $1000$ grams of chocolate is $x$ dollars and the cost of $1000$ grams of potatoes is $y$ dollars, the numbers $x$ and $y$ are positive integers and have not more than $2$ digits. Mother said to Maria to buy $200$ grams of chocolate and $1000$ grams of potatoes that cost exactly $N$ dollars. Maria got confused and bought $1000$ grams of chocolate and $200$ grams of potatoes that cost exactly $M$ dollars ($M >N$). It turned out that the numbers $M$ and $N$ have no more than two digits and are formed of the same digits but in a different order. Find $x$ and $y$.

$\dfrac{1}{5}x +y=N=10k+h$ $x+ \dfrac{1}{5}y=M = 10h+k$ $100>M>N$ $10>h$ and $10>k$. We can say that $10h+k>10k+h$, where $9h>9k$ or $h>k$ is evident. Note that $x+5y=50k+5h$ and $5x+y=50h+k$. $6x+6y=55k+55h$ and $4x-4y=45h-45k$. Let us say that $12x+12y=165k+165h$ and $12x-12y=135h-135k$. Add the two to get that $24x=200h+30k$. This simplified yields $12x=100h+15k$. We can also derive the fact that $24y=30h+200k$, or simplified $12y=15h+100k$. $12x-12y$ is apparently equal to $100h+15k-15h-100k$, or simply $85h-85k$. Note that $h-k>0$, as $h>k$. Hence, $85h-85k>0$. But wait... I give up I'm stuck.