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) =
16II) 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) =
24So, 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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