Can anyone please verify/check if there's error in this word problem?

[Math100]

Homework Statement

When Mr. Smith cashed a check at his bank, the teller mistook the number of cents for the number of dollars and vice versa. Unaware of this, Mr. Smith spent 68 cents and then noticed to his surprise that he had twice the amount of the original check. Determine the smallest value for which the check could have been written.
[Hint: If x denotes the number of dollars and y the number of cents in the check, then 100y+x-68=2(100x+y).]

Relevant Equations

None.

Proof: Let x be the number of dollars and y be the number of cents in the check.
Then we have 100y+x-68=2(100x+y).
From the hint, we get the following Diophantine equation: -199x+98y=68.
Applying the Euclidean Algorithm produces:
199=2(98)+3
98=32(3)+2
3=1(2)+1
2=2(1)+0.
Now we have gcd(199, 98)=1.
Note that 1$\mid$68.
Since 1$\mid$68, it follows that the Diophantine equation -199x+98y=68 can be solved.
Then we have 1=3-2(1)
=3-(98-32(3))
=33(3)-98
=33(199-2(98))-98
=33(199)-67(98).
This means 68=68[33(199)-67(98)]
=2244(199)-4556(98).
Thus, x\_{o}=2244 and y\_{o}=-4556.
All solutions in the integers are determined by:
x=2244+\frac{98){1}t=2244+98t for some t \in \mathbb{Z},
y=-4556-\frac{-199}{1}t=-4556+199t for some t \in \mathbb{Z}.
Therefore, x=2244+98t and y=-4556+199t.
To find the smallest non-negative integer solutions,
we solve for t:
2244+98t $\geq$ 0 $\land$ -4556+199t $\geq$ 0
t$\geq$$\frac{-1122}{49}$ $\land$ t$\geq$$\frac{4556}{199}$.
Thus, t=23, which implies that x=2244+98(23)=4498 and y=-4556+199(23)=21.

And now I am confused right here, because the answer in my textbook says "The smallest value for which the check could have been written is $10.21, but it doesn't seem like I am getting closer to this final answer/result, although the y-value is 21 does seem reasonable. Can anyone please verify/check my proof for this problem and spot any calculation error or something if anything is incorrect? Thank you.

 

[vela]

In the Diophantine equation, you have $-199x$, so I think you should have $x_o = -2248$.

 

[Math100]

In the Diophantine equation, you have $-199x$, so I think you should have $x_o = -2248$.

Then how should I set up the equation from 3-(98-32*3)=33(3)-98=33(199-2*98)-98?

 

[vela]

The rest of your work is fine as is. Comparing the Diophantine equation you started with to what you ended up with, it seems that
$$68=2244(199)-4556(98) = -\underbrace{(-2244)}_{x_o}(199)+\underbrace{(-4556)}_{y_o}(98).$$ Note that the wrong result you got, 4498, is equal to $2\times 2244+10$. Because you used the wrong sign for $x_o$, you added 2244 instead of subtracting 2244, resulting in the difference of $2\times 2244$.

 

[Math100]

The rest of your work is fine as is. Comparing the Diophantine equation you started with to what you ended up with, it seems that
$$68=2244(199)-4556(98) = -\underbrace{(-2244)}_{x_o}(199)+\underbrace{(-4556)}_{y_o}(98).$$ Note that the wrong result you got, 4498, is equal to $2\times 2244+10$. Because you used the wrong sign for $x_o$, you added 2244 instead of subtracting 2244, resulting in the difference of $2\times 2244$.

Now I got it. Thank you so much for the help!