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Counting factors of a number

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Director
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Joined: 08 Jul 2004
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Counting factors of a number [#permalink] New post 22 Jul 2004, 08:25
Hi all
How does one count the factors of a given number consdiering 1 and number itself.
Thanks
CIO
CIO
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 [#permalink] New post 22 Jul 2004, 11:08
Be extremely systematic. I make couples, starting with one, and counting up. You'll soon cover everything:

60:
1&60
2&30
3&20
4&15
5&12
6&10

By counting up on the low end, you cover everything on the high end as well. In this example, the next factor up from 6 is 10, and that's already there, and so is everything else above 10. This is better than randomly throwing out numbers until you hit them all.

Try it with a couple of examples that have lots of factors, like 24, 72, 36, 105.

Try it also with some variables: Name all the factors of 4p if p is a prime number.
4p:
1&4p
2&2p
4&p

That's it!
Manager
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Re: Counting factors of a number [#permalink] New post 22 Jul 2004, 11:55
saurya_s wrote:
Hi all
How does one count the factors of a given number consdiering 1 and number itself.
Thanks


Dear saurya_s, we(me and gravedigger) are working on a system of automated training which will contain easy and quick formulas for this and other things.
Joined: 31 Dec 1969
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CAT Tests
Combinations [#permalink] New post 22 Jul 2004, 12:55
I know it is done using combinations and using binomial coefficients. For example for 6 (excluding 1 and 6) ie 2x3 is 2C0+2C1 or something like this but I am not sure exactly. There might be some one of us knowing it. Let's take an example how many salads can one make using- lettuce, cucumber, cabbage. Now salad can be made by selecting one of these at at time or 2 at a time or 3 at a time. So, it is 3C1+3C2+3c3= 2^3-1=7
as 3C0+3C1+3C2+3c3=(1+1)^3. Now 3C0 is one so, the answer is 2^3-1=7
Now can some one extend this for factors of number.
Best
S
Combinations   [#permalink] 22 Jul 2004, 12:55
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