If the highest power of 12 that divides N! is exactly 25 , then the sum of all possible positive integer values of N is

Step-by-step solution

The prime factorization of 12 is 2^2 3 . For 12²⁵ to be the highest power dividing N! , we must have: ( E₂(N!) 2 , E₃(N!) ) = 25 Let us first check if E₃(N!) can be exactly 25 . For N = 53 , E₃(53!) = 53 3 + 53 9 + 53 27 = 17 + 5 + 1 = 23 . For N = 54 , E₃(54!) = 54 3 + 54 9 + 54 27 = 18 + 6 + 2 = 26 . Since E₃(N!) jumps from 23 to 26 , it can never be exactly 25 . Thus, the bottleneck must be the prime 2 . We require: E₂(N!) 2 = 25 E₂(N!) = 50 or 51 We also need E₃(N!) 25 , which means E₃(N!) 26 . Let us test valu