MaxymDvali_2012

We use $\text{choices}[i]$ to represent the choices for digit $i$ , let's consider digit-by-digit

$0$ : we only can have it as $0$ so we've only single choice

$1$ : we can multiply by $2$ three times then we've $\{1,2,4,8\}$ which means four choices.

$2$ : we can multiply by $2$ two times and divide once then we've $\{1,2,4,8\}$ which means four choices.

$3$ : we can multiply by $2$ once then $\{3,6\}$ which means we've two choices

$4$ : we can multiply by $2$ once and divide twice then we've $\{1,2,4,8\}$ which means four choices.

$5$ : we only have it as a choice

$6$ : we can divide once by $2$ then we've $\{6,3\}$ which means two choices

$7$ : we only have it as a choice

$8$ : we can divide three times

$9$ : we only have it as a choice

putting it all gives

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