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Number system- Divisors/Factors

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Re: Number system- Divisors/Factors

by uttara » Wed Nov 19, 2008 11:02 am
NehaBhandari wrote:Hey, can someone tell me a shortest way to calculate the no of factors for a number
Ex How many divisors/factors are there for the number 220..

Thanks
First break it down to its prime factors
220 = 2^2 * 5^1 *11^1

Total no of divisors/factors for 220 = (2+1)*(1+1)*(1+1) = 3*2*2=12
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Re: Number system- Divisors/Factors

by logitech » Wed Nov 19, 2008 3:58 pm
uttara wrote:
NehaBhandari wrote:Hey, can someone tell me a shortest way to calculate the no of factors for a number
Ex How many divisors/factors are there for the number 220..

Thanks
First break it down to its prime factors
220 = 2^2 * 5^1 *11^1

Total no of divisors/factors for 220 = (2+1)*(1+1)*(1+1) = 3*2*2=12
I know that rule and it is very handy. Do you know how you drive this equation ?

Thanks,
LGTCH
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Re: Number system- Divisors/Factors

by dmateer25 » Wed Nov 19, 2008 6:13 pm
uttara wrote:
NehaBhandari wrote:Hey, can someone tell me a shortest way to calculate the no of factors for a number
Ex How many divisors/factors are there for the number 220..

Thanks
First break it down to its prime factors
220 = 2^2 * 5^1 *11^1

Total no of divisors/factors for 220 = (2+1)*(1+1)*(1+1) = 3*2*2=12
I didn't know this rule. One more for the flashcards! Thanks uttara!
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by NehaBhandari » Fri Nov 21, 2008 3:05 pm
Thanks Uttara.. This was very helpful :P
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Re: Number system- Divisors/Factors

by orel » Sat Nov 22, 2008 1:46 am
uttara wrote:
NehaBhandari wrote:Hey, can someone tell me a shortest way to calculate the no of factors for a number
Ex How many divisors/factors are there for the number 220..

Thanks
First break it down to its prime factors
220 = 2^2 * 5^1 *11^1

Total no of divisors/factors for 220 = (2+1)*(1+1)*(1+1) = 3*2*2=12
Hi Uttara,

can you please explain how the total number of divisors can be 12?
i see it like this: there are two 2s, one 5 and one 11.
thanks
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by vishubn » Sat Nov 22, 2008 5:19 am
Hi Uttara,

can you please explain how the total number of divisors can be 12?
i see it like this: there are two 2s, one 5 and one 11.
thats the numer for factors is calcualetd !

u break down to factors and add the powers with one each !!

so as u said

2 two's and one 5s and one 11

add the powers with one each

(2+1)*(2)*(2)=24 is the answer

And also many a times facotrs is disccused on the forum !! pelase do search more if u stil hve doubts ! :)

And also do let me know if u have mroe q

vishu
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by GMATCHPOINT » Mon Nov 24, 2008 1:15 pm
Let me see if I understood the rule. For instance, 420 = (2)(2)(3)(5)(7), then the rule would be (2+1)(1+1)(1+1)(1+1), meaning (two 2s) (one 3) (one 5) (one 7), thus 24 is the total number of divisor/factors of 420. am I right?
Could someone explain? I would love to hv this on my flashcards. Logitech? Uttara?
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formal rule

by toshiblue » Tue Nov 25, 2008 3:21 pm
I found the formal definition at the following:

https://whyslopes.com/Number_Theory/Prim ... ctors.html
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by resepulv » Wed Nov 26, 2008 12:21 pm
Is really intuitive
if u have N=a^2*b^3*c^2
any factor of N must be F=a^i*b^j*c^k
where
i=0,1,2
j=0,1,2,3
k=0,1,2
then u have a combinatory problem so just multiply the number of options for each prime factor

The solution is 3*4*3
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