Question

If $x$ and $y$ are integers, then the equation $5x + 19y = 64$ has:

• A. no solution for $x<300$ and $y<0$
• B. no solution for $x>250$ and $y>-100$
• C. a solution for $250<x<300$
• D. a solution for $-59<x<-56$

Hint:

For Diophantine equations, use modular arithmetic to find integer solutions to linear equations.

Answer 1

The Correct Option is C

The given equation is $5x + 19y = 64$. To find integer solutions for $x$ and $y$, we solve for $x$: \[ 5x = 64 - 19y \] \[ x = \frac{64 - 19y}{5} \] For $x$ to be an integer, $64 - 19y$ must be divisible by 5. Solving for $y$ modulo 5: \[ 64 \equiv 19y \ (\text{mod} \ 5) \] \[ 4 \equiv 4y \ (\text{mod} \ 5) \] \[ y \equiv 1 \ (\text{mod} \ 5) \] So, $y = 5k + 1$ for some integer $k$. Substituting this back into the equation for $x$, we get: \[ x = \frac{64 - 19(5k + 1)}{5} = \frac{64 - 95k - 19}{5} = \frac{45 - 95k}{5} = 9 - 19k \] Thus, the general solution is $x = 9 - 19k$ and $y = 5k + 1$. Checking for integer solutions within the given range for $x$, we find that the solution holds for $250<x<300$.