Let n be a natural number. The number of consecutive zeros at the end of the expansion of n! is exactly 2. How many values of n are possible?

Step-by-step solution

The number of consecutive zeros at the end of n! is determined by the exponent of 5 in the prime factorization of n! . Using Legendre's formula, the exponent of 5 in n! is given by: E₅(n!) = n 5 + n 5^2 + n 5^3 + Given that E₅(n!) = 2 , n must be less than 25 , which means n 25 = 0 and all higher terms are also 0 . Thus, the equation simplifies to: n 5 = 2 This inequality holds for: 2 n 5 10 n Since n is a natural number, the possible values of n are 10, 11, 12, 13, 14 . There are exactly 5 possible values for n .