How many even different factors does the integer P have?
(1) P = (x^2)(y^2)(z^3)(2^4), where x, y, and z are different odd prime numbers --> the number of factors of P is (2+1)(2+1)(3+1)(4+1)=180 and the number of odd factors of P is (2+1)(2+1)(3+1)=36, so the number of even factors of P is 180-36=144. Sufficient.
(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 --> basically the same here: P=3^p*5^q*7^r*2^4 --> the number of factors of P is 180=(p+1)(q+1)(r+1)(4+1) and the number of odd factors of P is 180/(4+1)=36, so the number of even factors of P is 180-36=144. Sufficient.
Finding the Number of Factors of an Integer
First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.
The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2
Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.
So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.
Hope it helps.
Is it possible to get a little bit more on this total number of even factors concept?
I realized we are finding total factors then subtracting total number of odd factors , is there no direct way for finding the total number of even factors ?
what if a number has no odd factors such as 16 or 32 etc ? In this case we are looking at the powers of 2's, to find total number of even factors, aren't we ?
But when there are both even and odd factors in a number such as 30 or 18 or 20 etc in this case why are we going other way round , is there no way of looking at the powers of 2 to get total number of even factors ?