Number system- Divisors/Factors

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NehaBhandari: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Mon Nov 17, 2008 12:08 pm; Thanked: 1 times

by NehaBhandari » Wed Nov 19, 2008 10:46 am

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

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Source: Beat The GMAT — Problem Solving | Wed Nov 19, 2008 10:46 am

uttara: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sun Sep 07, 2008 7:22 am; Location: India; Thanked: 4 times; GMAT Score:700

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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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

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
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

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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NehaBhandari: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Mon Nov 17, 2008 12:08 pm; Thanked: 1 times

by NehaBhandari » Fri Nov 21, 2008 3:05 pm

Thanks Uttara.. This was very helpful

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orel: Master | Next Rank: 500 Posts; Posts: 247; Joined: Sun Jul 27, 2008 1:39 am; Thanked: 2 times; GMAT Score:660

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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vishubn: Legendary Member; Posts: 574; Joined: Sun Jun 01, 2008 8:48 am; Location: Bangalore; Thanked: 28 times

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

KILL !! DIE !! or BEAT my FEAR !!! de@D END!!

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GMATCHPOINT: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Wed Oct 08, 2008 12:50 pm

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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toshiblue: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Tue Nov 25, 2008 7:44 am; Location: Chicago, IL; Thanked: 1 times

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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resepulv: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Sun Aug 03, 2008 4:26 pm

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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