Question 162 – Programming-for-Output-Problems

→ 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.