tolikshift Thanks ! , Fixed.

Dyadka_Dimka Your welcome , Happy I'm helped.

Dyadka_Dimka

regardless of $k$ , We're asking how many $k$ that's $f(n,k)$ is a multiple of $k$

this is equivalent to count of divisors of $1+2+3+....+n$

I cannot write more because it's already covered in the editorial.

Have a nice day.

Dyadka_Dimka Read Statement Carefully Please

Dyadka_Dimka There is no division .

This maybe useful , Let $f(n)=\{d:d\mid n\}$ i.e. $f(n)$ is the set of divisors of $n$.

We claim that $f(n)\cap f(n+1)=\{\}$ , This is because $\gcd (n,n+1)=1$ according to Euclidean Algorithm , which shows that no divisors are common expert trivial $1$ , Dividing by $2$ isn't affecting the claim because it only affect power of $2$ and no consecutive integers can be even.

And thus answer is $|f(\frac{n}{2})|\times |f(n+1)|$ when $n$ is even and $|f(\frac{n+1}{2})|\times |f(n)|$ when $n$ is odd.

Excited to Participate !!