Let P be the product of the first 100 odd natural numbers. If k is the largest integer such that P is divisible by 3^k , then the value of k is

Step-by-step solution

Let P be the product of the first 100 odd natural numbers. P = 1 3 5 199 Multiply and divide by the product of the first 100 even natural numbers: P = 1 2 3 4 199 200 2 4 6 200 P = 200! 2¹⁰⁰ (1 2 3 100) = 200! 2¹⁰⁰ 100! To find the highest power of 3 dividing P , we calculate the difference between the exponent of 3 in 200! and 100! . Exponent of 3 in 200! : E₃(200!) = [ 200 3 ] + [ 200 9 ] + [ 200 27 ] + [ 200 81 ] = 66 + 22 + 7 + 2 = 97 Exponent of 3 in 100! : E₃(100!) = [ 100 3 ] + [ 100 9 ] + [ 100 27 ] + [ 100