LCM of N1 and N2 is 100, then find the possible pairs of N1 and N2?

LCM = 100

Prime factorization of 100 = 2² × 5²

                                   N = a^x × b^y

              Number of pairs = \(\frac{(2x \;+\; 1) (2y\; +\; 1) \;+\; 1}{2}\)

                                      = \(\frac{(2 × 2 + 1) (2 × 2 + 1) + 1}{2}\) = \(\frac{26}{2}\) = 13