DRDO Multi tasking staff Model paper 6

If N is a number and,
N=p1r1p2r2p3r3.......pnrn
Where, p1,p2,p3......pn are prime numbers and r1,r2.......rn are positive integers
Then,
Number of positive factors =(r1+1)(r2+1)(r3+1).....(rn+1)
Now,
425=5×5×17=52×17
∴ Number of positive factors = (2 + 1) × (1 + 1) = 3 × 2 = 6