How many odd, positive divisors does 540 have?

Make a prime factorization of a number: 540=2^2*3^3*5 --> get rid of powers of 2 as they give even factors --> you'll have 3^3*5 which has (3+1)(1+1)=8 factors.

Another example: 60=2^2*3*5 it has (2+1)(1+1)(1+1)=12 factors out of which (1+1)(1+1)=4 are odd: 1, 3, 5 and 15 the same # of odd factors as 60/2^2=15 has.

Answer: B.

MUST KNOW FOR GMAT:

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.
For more on number properties check: math-number-theory-88376.html