→ The above iterative code is computes the GCD of two numbers using euclidean algorithm.
→ The procedure is to subtract smaller number from larger. So that we can reduce larger number then doesn’t change the value of GCD.
→ if we performing number of iterations based on condition the larger of two numbers will end up with GCD.

→ Since there is one line of output for each loop, we need to determine the number of
times the loop executes. Since i is constant, we need to see the growth of j only.
→ Let the initial value of j be denoted by J1 and the subsequent values by J_n for n= 2, 3,... so that J denotes a progression of j. We see that J_n = 2J_n-1;
→ J₁ = 2, which gives J_n = 2^n-1;
n = 1, 2, ...
→ The loop will execute as long as i/j=2.0/2^n-1 >0.001 which gives n<3log₂10+2 (or) n<11.97.
→ Thus the loop will execute 11 times which is equivalent to say there will be 11 lines of output.

Step-1: We are initialized i=0 in for loop. It means condition true because it is less than 5.
Step-2: Inside the for loop we are assigning again I value is 10 and printing value i.
Iteration-1: We are printing value 10 then increment by 1
Iteration-2: We are clearing previous value and assigning value 10. Printing value is 10;
Iteration-5: We are clearing previous value and assigning value 10. Printing value is 10
Step-3: output is 1010101010

Step-1: Take n value initially 14(any number we can take but we are taken 14)
Step-2: 14<16
Step-3: It will print 00000000 00001110