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 $-59<y<-56$

Answer 1

Given equation is $5x+19y=64$

We need to look for integers $x$ and $y$ that satisfy the above equation.

Let us find a solution for $5x+19y=1$

We are doing this because when we will multiply that equation by 64 we will get as follows:

$64\times 5x+64\times 19y=64\times 1$

Which can be written as $5\times \left(64x\right)+19\times \left(64y\right)=64$ thus giving two integer $64x$ and $64y$ as solution for our required equation.

So let us find a solution for $5x+19y=1$.So with basic idea we observe $x=4,y=-1$ works.

So we have a solution for our given equation $5x+19y=64$ as $64\times 4$ and $64\times -1$.

With this solution we see that options A,B are incorrect.

Now let us look at Option D.

$y$ should be multiplied with such a value which when added to RHS $64$ leads to its last digit as $5$ or $0$ so that it can be divided further by 5 to find value of x.So neither $-58$ nor $-57$ will work.

Hence option D is incorrect.

And we have already found $x=64\times 4$ gives us integer solution hence correct option is $C$.