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 ?