Playing with divisors is fun

#1

I don’t understand the logic and the math behind the problem

#2

hey @shivanshu12800
It’s well known fact that the number of divisors of n=∏piαi equals ∏(αi+1) where pi are primes in the factorization of n . Lets define P as the product of all divisors of n . If we find βi where P=∏piβi we will solve the problem.

All divisors of n=∏piαi can be represented as d=∏piki where 0≤ki≤αi. Lets fix some i0 and 0≤k0≤αi. There are ∏i≠i0(αi+1) divisors that ki0=k0. So βi=(0+1+…+αi)⋅∏i≠i0(αi+1)=αi(αi+1)2⋅∏i≠i0(αi+1)=αi(αi+1)2⋅allProdαi+1=αi2⋅allProd, where allProd=∏(αi+1).

Now the can calculate the number of divisors of P using βi.

#3

@shivanshu12800 you can also refer this post for more understanding .

#4

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