If k is the largest integer such that 100! is divisible by 24^k , then the value of k is

Step-by-step solution

The prime factorization of 24 is 2^3 3 . To find the highest power of 24 that divides 100! , we need to find the highest powers of 2 and 3 that divide 100! . The exponent of 2 in 100! is given by: E₂(100!) = [ 100 2 ] + [ 100 2^2 ] + [ 100 2^3 ] + = 50 + 25 + 12 + 6 + 3 + 1 = 97 Since we need the power of 2^3 , the available power is [ 97 3 ] = 32 . The exponent of 3 in 100! is given by: E₃(100!) = [ 100 3 ] + [ 100 3^2 ] + [ 100 3^3 ] + = 33 + 11 + 3 + 1 = 48 The highest power of 24 that divides 100! will be the m