How many values of c in the equation $x^2 - 5x + c$ result in rational roots which are integers? This question was asked in the e-litmus exam.
Hint: $(x-a)(x-b) = x^2 - (a+b) x + ab$.  How many integer solutions does $a+b=5$ have?
Answer should be $6$.

In order for $x$ to be an integer, numerator must be even which would be divisible by $2$.
 $d= \sqrt{25-4c}$

$x=(5\pm\sqrt{25-4c}) /2$

Now solve inside the brackets first such that d becomes odd so that the numerator $5\pm$(any odd integer) is divisible by $2$ and and hence $x$ is an integer. 

The values of $c$ could be $24/5,-24/5,16/5,-16/5,0$ (though even)