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question on number of factors

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question on number of factors [#permalink]

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New post 07 Nov 2012, 03:59
Hi,

I've a question on finding out number of even and odd factors of a number. Is there any quick method to identify how many 'even' or 'odd' factors a number would have.
Eg. how many 'even' factors does 144n have, where n is a prime number greater than 2. we can quickly identify, since 144=2^4*3^2 so there would be 30 factors of 144n but, how many even and how many odd?

Thanks.
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Re: question on number of factors [#permalink]

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New post 07 Nov 2012, 05:00
Vips0000 wrote:
Hi,

I've a question on finding out number of even and odd factors of a number. Is there any quick method to identify how many 'even' or 'odd' factors a number would have.
Eg. how many 'even' factors does 144n have, where n is a prime number greater than 2. we can quickly identify, since 144=2^4*3^2 so there would be 30 factors of 144n but, how many even and how many odd?

Thanks.


Total Number of factors: (4+1)(2+1)(2) = 30
odd Factors Only = 3*2 = 6
Even facors = 30-6=24

Check for 12
factors 1,2,3,4,6,12
Total factors = 6
Even = 6-2=4
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Re: question on number of factors [#permalink]

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New post 07 Nov 2012, 21:28
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Vips0000 wrote:
Hi,

I've a question on finding out number of even and odd factors of a number. Is there any quick method to identify how many 'even' or 'odd' factors a number would have.
Eg. how many 'even' factors does 144n have, where n is a prime number greater than 2. we can quickly identify, since 144=2^4*3^2 so there would be 30 factors of 144n but, how many even and how many odd?

Thanks.


You can arrive at the answer on your own if you understand (not just 'know' but 'understand') how to calculate the total number of factors of a number.

Say, a number N = 2^a * 3*b * 7^c

Total number of factors = (a + 1)(b + 1)(c + 1)

Why? because you can choose each prime number in (power + 1) ways, the additional 1 being for the case in which you don't choose the prime number. So you can select a 2 for the factor in (a + 1) ways.

You should go through this post first if you are not comfortable with this concept: http://www.veritasprep.com/blog/2010/12 ... ly-number/

Now, say the number is N = 2^a * 3*b * 7^c

How many factors will be even? What do you need for even factors? At least one 2. In how many ways can you do it?
In a*(b + 1)*(c + 1) ways
(the 1 additional way in which you don't pick a 2 has been removed. Rest everything is the same.)

How many will be odd? Now you don't want a 2. So, you get 1*(b+1)*(c +1 ) ways

Similarly, you can pose tons of questions: How many factors will be multiples of 4/6/21 etc? How many factors will have all the primes (2, 3 and 7)? etc
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Re: question on number of factors [#permalink]

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New post 08 Nov 2012, 03:34
VeritasPrepKarishma wrote:
Vips0000 wrote:
Hi,

I've a question on finding out number of even and odd factors of a number. Is there any quick method to identify how many 'even' or 'odd' factors a number would have.
Eg. how many 'even' factors does 144n have, where n is a prime number greater than 2. we can quickly identify, since 144=2^4*3^2 so there would be 30 factors of 144n but, how many even and how many odd?

Thanks.


You can arrive at the answer on your own if you understand (not just 'know' but 'understand') how to calculate the total number of factors of a number.

Say, a number N = 2^a * 3*b * 7^c

Total number of factors = (a + 1)(b + 1)(c + 1)

Why? because you can choose each prime number in (power + 1) ways, the additional 1 being for the case in which you don't choose the prime number. So you can select a 2 for the factor in (a + 1) ways.

You should go through this post first if you are not comfortable with this concept: http://www.veritasprep.com/blog/2010/12 ... ly-number/

Now, say the number is N = 2^a * 3*b * 7^c

How many factors will be even? What do you need for even factors? At least one 2. In how many ways can you do it?
In a*(b + 1)*(c + 1) ways
(the 1 additional way in which you don't pick a 2 has been removed. Rest everything is the same.)

How many will be odd? Now you don't want a 2. So, you get 1*(b+1)*(c +1 ) ways

Similarly, you can pose tons of questions: How many factors will be multiples of 4/6/21 etc? How many factors will have all the primes (2, 3 and 7)? etc

Thanks, I got the quick method I was looking for.
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Re: question on number of factors [#permalink]

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Re: question on number of factors [#permalink]

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New post 17 Apr 2015, 23:23
Expert's post
Vips0000 wrote:
Hi,

I've a question on finding out number of even and odd factors of a number. Is there any quick method to identify how many 'even' or 'odd' factors a number would have.
Eg. how many 'even' factors does 144n have, where n is a prime number greater than 2. we can quickly identify, since 144=2^4*3^2 so there would be 30 factors of 144n but, how many even and how many odd?

Thanks.


This thread may be old, but this doubt is still relevant. Hence reviving the discussion on this one. :)

Vips' question is modeled on an easier Official question:

If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(This question is discussed in detail here)

I'll first explain the Official question briefly (to better establish the contrast with Vips' question later)

The first step to solving all such questions is to express the given number in its Prime Factorized form.

By doing that, we get: \(n=2^{2}*p^{1}\)

Had we been asked the total number of factors of n, the answer would have been: (2+1)(1+1) = 3*2 = 6 . (These 6 factors also include the cases of \(2^{0}*p\) and 2\(^{0}*p^{0}\))

But we are asked here, the number of even factors of n. Now, an even factor will always contain at least one 2. So, the cases in which the power of 2 is 0 are not allowed here. So, when asked about the number of even factors of n, we should only consider 2 powers of 2 (\(2^{1}\) and \(2^{2}\)), not the 3 powers we considered in the question of total number of factors (\(2^{1}\), \(2^{2}\) and \(2^{0}\))

So, the number of even factors of n = (2)(1+1) = 4

(Please refer to Karishma's lucid explanation above if you still have any doubt about the way we calculated the number of even factors :) )

Okay, so after having established this process of solving, let's now come to Vips' question:

How many 'even' factors does 144n have, where n is a prime number greater than 2?

Like before, our first step will be to do the prime factorization of the given number. \(144=2^{4}*3^{2}\)

So, \(144n=2^{4}*3^{2}*n\)

Now, n is a prime number greater than 2. Can n be 3? It very well can be!

So, 2 cases will arise here (this is the point where Vips' question differs from the Official question above)

I) If n = 3

Then \(144n=2^{4}*3^{3}\)

So, by applying the same logic as in the Official question above, the total number of even factors of 144n = (4)(3+1) = 16

II) If n is a prime greater than 3

Then, \(144n=2^{4}*3^{2}*n^{1}\)
So, the total number of even factors of 144n = (4)(2+1)(1+1) = 24

So, we see that in the question posed by Vips', the number of even factors of 144n may be either 16 or 24. A unique answer cannot be determined (because n may be 3 or greater than 3)

Takeaways:

1. Pay attention to the condition given in question (the question mentioned only that n was a prime greater than 2. So, the case of n = 3 did need to be considered as well)

2. Don't do calculations in the air (After prime factorizing 144, had Vips' written the next step: \(144n=2^{4}*3^{2}*n\) on paper, the possibility of n = 3 may have occurred to him. However, when we do things in the air, since our mind is already occupied by trying to remember all the numbers that are being multiplied and process the multiplication at the same time, the likelihood of us noticing something amiss decreases significantly)

Hope this discussion was helpful! :)

- Japinder
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Re: question on number of factors   [#permalink] 17 Apr 2015, 23:23
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