# Diophanthine Equation HW Question

I have this problem for homework which I am having trouble understanding..

At a cake store, the cost of a Red Cake is 11 dollars and the cost of a Blue Cake is 8 dollars. What conclusions can be made if each of the total bill's for these 3 customers happen to be these following amounts?

(a) 96

(b) 777

(c) 69

I looked for solutions for all 3, Diophantine equation solver for 8x+11y=96 , 8x+11y=777, 8x+11y=69. And the only thing that I noticed strange was that in most cases y or x was a negative number, does that mean that the bill would be impossible to create with just those two items?

EDIT: In a previous thread someone said 96 and 777 have solutions though, I can't seem to find a solution for any of these equation that don't involve x or y being negative, what am i missing here?

Let's pick one of the values, say $11$, and check the modular value of each bill:
a) $96\equiv 8 \bmod 11 \qquad$ (and $1\cdot 8 \equiv 8 \bmod 11$)
b) $777\equiv 7 \bmod 11 \qquad$ (and $5\cdot 8 \equiv 7 \bmod 11$)
c) $69\equiv 3 \bmod 11 \qquad$ (and $10\cdot 8 \equiv 3 \bmod 11$)
So now we need respectively $1$, $5$ and $10$ Blue cakes more than a multiple of $11$ to make these totals with the minimum Blue count. So the maximum Red cakes are respectively $(96-8)/11=8$, $(777-40)/11=67$, and - oh dear - $(69-80)/11 = -1$ is an impossible total for part (c).
(a) $8\cdot 11+1\cdot8=96$
(b) $67\cdot 11+5\cdot 8=777$
Now we have the option for parts (a) and (b) of converting multiples of $8$ Red cakes into $11$ Blue cakes, so part (a) has $2$ answers and part (b) has $9$ answers.