See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

0 1 2 n k k k k 1 2 n p 2 p ,p ....p  where pi  odd prime Total number of even divisors      0 1 2 n k k 1 k 1 .... k 1    Total number of odd divisors      1 2 n k 1 k 1 .... k 1           0 1 2 n f p k 1 k 1 k 1 .... k 1 4       k0 = 2, ki = kj = 1, p = 4.3.5 = 60 k0 = 3, ki = 1, p = 24, 40, 56 k0 = 5, p = 32.