What is Happening at Primes here?

#1

while(p*p <=m){

if(m%p == 0){

int cnt = 0;

		while(m%p == 0){
			cnt++;
			m = m/p;
		}
		ans =ans *(cnt-1);
	}
	//Go to Next Position
	i++;
	p = primes[i];
}
#2

hello @JaveedYara
What u want to ask? pls elaborate

#3

why is m Mod p being divided and being checked?

#4

@JaveedYara
for p to be a factor of m it should divide m right?
that is why we are checking it using m%p if m%p gives 0 that means p is factor of m

#5

and what about ans variable there?

#6

and p = primes[i] what does this do?

#7

primes vector is holding prime numbers.
i++;
p=primes[i] ;
is giving use next prime number.

it is holding the overall answer

#8

so why is ans then Getting Multiplied to count-1. Last Doubt. you have been great, Thank you so Much

#9

no it should be cnt+1 in place of cnt-1.

if prime factorisation of number n is written as ->
n=p1^a * p2 ^ b * p3 ^ c … ( where p1 ,p2,p3 are prime number and a,b,c are the respecitve power)

then number of divisors will be (a+1) * (b+1) * (c+1)
that cnt variable is playing the role of a ,b,c

#10

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