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
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Thanks for the explanation

i remembered that, but just caught in the term " odd factors " which is cleared now. All right, this was an odd factor case in which we don't have to consider Power of 2. what if we are asked about only even +ve factors. would then we be considering only power of 2 ?
for example: 540 = 2^2*3^3*5 ----
=>(2+1) = 3
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No. One way will be to count # of all factors and subtract # of odd factors.

540=2^2*3^3*5 has (2+1)(3+1)(1+1)=24 factors out of which 8 are odd, so 24-8=16 are even.
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I'll stick to this one way method, i can sense the other way will be unmanageable for me

Thanks a lot B
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\(540 = 2^2 x 3^3 x 5^1\)

Odd positive Divisors = ( 3 + 1 ) ( 1 + 1 ) => 8

Hence answer will be (B) 8
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Hello,

Could someone help me understand the divisor formula used for these types of problems? I am not familiar with this and thought the answer was 4 ( 3*3*3*2*2*5 excluding the 2's).

Thank you!

KHow

To prove this, we first consider numbers of the form, \(n = p^a\). The divisors are \(1, p, p^2, ..., p^a\); that is, \(d(p^a)=a+1\).

Now consider \(n = p^aq^b\). The divisors would be:
1 p \(p^2\) ... \(p^a\)
q pq \(p^2q\) ... \(p^aq\)
\(q^2\) \(pq^2\) \(p^2q^2\) ... \(p^aq^2\)
... ... ... ... ...
\(q^b\) \(pq^b\) \(p^2q^b\) ... \(p^aq^b\)

Hence we prove that the function, d(n), is multiplicative, and in this particular case, \(d(p^aq^b)=(a+1)(b+1)\). It should be clear how this can be extended for any natural number which is written as a product of prime factors.

Hope it helps.

Source: mathschallenge.net

Factors of \(540=2^2*3^3*5^1\)

The Odd factors \(=(3+1)(1+1)=4*32=8\)

The answer is \(B\)

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.

For another example: 60=2^2*3*5
does this mean that # of even factors is (12-4)= 8 ??
or the # of even factors is (2+1)= 3?
What is the concept here?

If you don't know the rule cited multiple times above and you don't understand it, I have great news!! You can get to the finish line on (I think) every similar question in every OG and OG Quant.

540 factors to 1*2*2*3*3*3*5
We can't include the 2s or we will end up with an even. How many numbers can we make out of the others?
1
3
3*3
3*3*3
5
5*3
5*3*3
5*3*3

Eight.

Answer choice B.

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