for(i=1;i<=n;i++)
{ for(j=1;j<=i*i;j++)
if((j%i)==0)
{ for(k=1;k<=j;k++);
// some text
}
}
Execution ;
Step 1- Outermost for loop is executes O(n) time.
Check for inner for loop
if i=1 and j=1 [((j%i)==0) is true] then k=1 time executed.
if i=2 and j=1;j<=4
[(j%i)==0] is true only for half of values of j. i.e)j=2,j=4.then k is executed 2 times.
if i=3 and j=1:j<=9
[(j%i)==0] is true for j=3,6,9 [i.e i times] but j is check condition for j=1 to 9 in which only 3 times k is executed.
..
...
if i=n and j is executed until n*n and only $\frac{n^{2}}{n}$=n is succesful.This 'n' time innermost k loop is executed.
Hence
Time for first for loop=O(n)
and Second for loop=O($n^{2}$)
Third for loop =O(n) // only n values are true in total of $n^{2}$.
Total time complexity=O(n*$n^{2}$*n)=O($n^{4}$)
