Plz explain it's time complexity

#1

void function(int n)
{
int count = 0;

     for (int i=0; i<n; i++) 
     for (int j=i; j< i*i; j++) 
         if (j%i == 0) 
         { 
            for (int k=0; k<j; k++) 
                 printf("*"); 
          } 
    }

??

#2

the answer would be between theta(n^4) to bigOh(n^5)
for every i , j will take value from i to i * i
and j%i will be equal to 0 i times that is when j equals to (1 * i ,2 * i,3 * i,…i * i)
so for every i k will run (i+2i+3i+…+ii) i.e i(i+1)*i can be aproximated to be i^3
and i takes values from 1 to n (using 1 index for simplification)
so total number of operations will be (1^3 + 2^3 + 3^3 + 4^3+…+n^3) = (n^2(n+1)^2)/2
which is theta(n^4) to bigOh(n^5)

so O(n^5) should be the answer in case of mcq

#3

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