Let P be the product of the first 50 even natural numbers. The maximum integer k such that 24^ k divides P is

Step-by-step solution

The product of the first 50 even natural numbers is given by P = 2 4 6 100 P = (2 1) (2 2) (2 3) (2 50) P = 2⁵⁰ (1 2 3 50) = 2⁵⁰ 50! We need to find the highest power of 24 that divides P . The prime factorization of 24 is 2^3 3 . First, we find the exponent of the prime 2 in P . The exponent of 2 in 50! is: E₂(50!) = 50 2 + 50 4 + 50 8 + 50 16 + 50 32 E₂(50!) = 25 + 12 + 6 + 3 + 1 = 47 Thus, the total exponent of 2 in P is 50 + 47 = 97 . Next, we find the exponent of the prime 3 in P . Since the factor 2⁵⁰ does no