B) 72 c) 73If d(N) represents number of divisors of N then d(d(18)) isa)72ERALD PACKAGEa) 4b) 5c) 6The product of divisors of 512 isa) 245b) 254c) 2100c) 3The number of prime factors of 54 isa) 8b) 2The product of divisors of N isId(N)a)Nb) N(N)c) Nwan

B) 72 c) 73If d(N) represents number of divisors of N then d(d(18))

1.	B) 72 c) 73If d(N) represents number of divisors of N then d(d(18)) isa)72ERALD PACKAGEa) 4b) 5c) 6The product of divisors of 512 isa) 245b) 254c) 2100c) 3The number of prime factors of 54 isa) 8b) 2The product of divisors of N isId(N)a)Nb) N(N)c) Nwan
Answer» Step-by-step explanation:d(18) = NUMBER of DIVISORS of 18 The of divisors are 1, 2, 3, 6, 9, 18 Therefore, Number of divisors of 18 = 6 Therefore, d(18)=6 --------------(1) Now, As we have to fine d(d(18)) From result 1 we substitute d(18)=6 in d(d(18)) therefore, Now we have to find d(d(18)) = d(6) the divisors of 6 = 1, 2, 3, 6 therefore d(6) = 4 Answer:- d(d(18)) = 4