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

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,

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!

Thanks Uttara.. This was very helpful

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

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

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?

I found the formal definition at the following:

https://whyslopes.com/Number_Theory/Prim ... ctors.html

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