Q13/125) For how many positive integer values of n does the equation 2x^2+689x+n=0 have an integer solution?

let (2x+ a)(x+b) = 2x^2+689x+n

as n is +ve both a and b positive or -ve, 

or 2x^2 + (a + 2b) = 2x^2+689x+n

a and b cannot - ve as sum is positive

0 < 2b < 689 and hence 0 < b < 344.5

so there are 344 values of b and also n