Time complexity problem

#1

void function(int n)
{
int count = 0;
//executes O(n) times
for (int i=0; i<n; i++)
//Executes 0(n^2) times
for (int j=i; j< ii; j++)
if (j%i == 0)
{
//executes O(n^2) times
for (int k=0; k<j; k++)
System.out.println("");
}
}

last O(n^2) time?

#2

Yes

inner most loop runs only when j%i
means it runs i times for every i hence inner runs O(n^2) time