# Solve equation with constrain on number of decimals?

Given this equation:

11 x + 102 y = 100


How can I find all possible solutions for which the following condition is satisfied:

x and y rounded up to two decimal places will be exactly the same as non-rounded x and y, i.e. all decimal places from the third one onwards are 0.

Edit:

Both x and y must be > 0 and are positive real numbers.

Practical example

A trader had a deal with a client to deliver him 11 breads and 102 lemons. The price of the whole shipment was $100 and the invoice only showed the total price. However, the Tax services halted the transaction because the trader is obliged to list the prices of individual breads and lemons on the invoice. Given that the price cannot change and must remain exactly$100, what prices of individual breads and lemons can the trader put on the invoice?

Let $$u=100x$$ and $$v=100y$$ and find an equation for these integers. In other words, work with cents instead of dollars. So,
$$11u+102v=10000.$$
As \begin{align}10000 &= 102\cdot 98+4\\ &=102\cdot 97+99+7\\ &=102\cdot 96+2\cdot 99+10\\ &=102\cdot 95+3\cdot 99+13\\ &=102\cdot 94+4\cdot 99+16\\ &=102\cdot 93+5\cdot 99+19\\ &=102\cdot 92+6\cdot 99+22\\ &=102\cdot 92+11\cdot 56\\ \end{align} apparently the maximum valid price for lemons is $$92$$ cents, in combination with a price of $$56$$ cents per bread. Since $$102$$ and $$11$$ have no factor in common, we can only find other solutions by (perhaps repeatedly) making the lemons $$11$$ cents cheaper and the bread $$102$$ cents more expensive. So lemons @ $$81$$ cents with bread @ $$158$$ cents is also possible, as is lemons @ $$70$$ cents with bread @ $$260$$ cents, etc.
From $$11 x =100- 102 y \implies 11\mid 100-102 y$$
Since $$11\mid 99-99y$$ we have $$11\mid (100-102y)-(99-99y)=3y+1$$
So $$11 \mid 4(3y+1)-11y = y+4$$ and thus $$y=11t-4$$ for some integer $$t$$ and now you can express $$x$$ with $$t$$ also...