NDA Mathematics Question (2025) — Solution

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: E_5(n!) = n 5 + n 5^2 + n 5^3 + Given that E_5(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. Answer: 5