How many odd, positive divisors does 540 have?

Question

A. 6
B. 8
C. 12
D. 15
E. 24

Bunuel · Accepted Answer

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

tfincham86 · Answer

Thanks for the explanation

shrive555 · Answer

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

Bunuel · Answer

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.

shrive555 · Answer

I'll stick to this one way method, i can sense the other way will be unmanageable for me Thanks a lot B

Abhishek009 · Answer

540 = 2^2 x 3^3 x 5^1

Odd positive Divisors = ( 3 + 1 ) ( 1 + 1 ) => 8
 
Hence answer will be (B) 8

KHow · Answer

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

Cellchat · Answer

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

MHIKER · Answer

Factors of  540=2^2*3^3*5^1

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

The answer is  B

Urvasi · Answer

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?

ThatDudeKnows · Answer

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.

DanTheGMATMan · Answer

There is a standard rule for this. Just factor 540 out:

https://www.youtube.com/watch?v=QNl4yzbcPZo

How many odd, positive divisors does 540 have?

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GMAT Problem Solving (PS) Questions

How many odd, positive divisors does 540 have?

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