COUNTING DIVISORS

#1

HEY,
HOW CAN WE CALCULATE THE SUM OF DIVISORS USING SEIVE APPROACH.?

#2

Hey @CODER_JATIN
Follow this :Please explain how to calculate sum of divisors of any number

then this (if req) : How to calculate sum of all divisors

#3

In the second approach that you’ve told me, is this is the formula that i need to remember?

#4

and i’m not able to understand 1st approach, can you please again what actually we are doing?

#5

Hey ,You dont have to remember that ,We did this in 11TH 12th
Say we have 6 so prime factorisation is 2,3
So sum of all divisors is =(1+2)*(1+3)
i,e Multiplication ((summation of gp of K^P)) where K is prime factor of N and P is max times K appear in prime factorisation.

#6

Which one exactly? Please copy paste here:)

#7

Hanji okay bbhaiya, i got understand

#8

for ( int i=1; i<= sqrt (n); i++) { if (n%i == 0) { // If divisors are equal, print only one if (n/i == i) printf ( "%d " , i); else // Otherwise print both printf ( "%d %d " , i, n/i); } } Ye wala code bta do ek baar, ?

#9

So N’s maximum diviosor is sqrt(n)
So for i=1 to sqrt(n) we check if i divides n
If it does then we gets its two divisors i and n/i (note here n/i >sqrt(n) result came from this (i)*(n/i)=n)
And if i=sqrt(n) divides n then we only add it once because in this case i and n/i are same

#10

OkAY OKAY , bhaiya ek baar dobaara link bhejna jo aapne bheje they first reply mai, they are not showing now.

#11

Hey @CODER_JATIN
Follow this :Please explain how to calculate sum of divisors of any number

then this (if req) : How to calculate sum of all divisors