Given P is an integer.
Considering statement 1:
P = x2 y2 z3 24, where x, y, and z are different odd prime numbers.
As we know if an integer, k = Ma × Nb × Oc × 2d, where M , N and O are different prime numbers, then number of positive factors = (a + 1)(b + 1)(c + 1)(d + 1)
and Number of odd positive factors = (a + 1)(b + 1)(c + 1)
⇒ Number of odd positive factors of P = (2 + 1)(2 + 1)(3 + 1) = 36
and total number of positive factors of P = (2 + 1)(2 + 1)(3 + 1)(4 + 1) = 180
⇒ Number of different even positive factors of P = 180 − 36 = 144
Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.
Considering statement 2:
The total number of different factors in P are 180 and P is a multiple of 16 but not a multiple of 32. Also only other prime numbers that are factors of P are 3, 5 and 7.
As P is a multiple of 16 but not 32 => Highest power of 2 present in P is 4.
From above :-
P = 24 × 3a × 5b × 7c
⇒ Number of positive factors of P = (4 + 1)(a + 1)(b + 1)(c + 1), which is given as 180.
⇒ (a + 1)(b + 1)(c + 1) = 180/5 = 36, which is same as number of positive odd factors.
Hence, number of positive even factors = 180 − 36 = 144.
Since we are getting a definite answer from above statement , statement 2 also itself is sufficient to provide the answer.
As statements 1 and 2 both individually are sufficient to provide the answer
Answer: D.