The question is given in the following picture: enter image description here

I solved it and my answer was B, but the answer sheet said that the answer is C as you can see from the following picture:

enter image description here

I do not know why, could anyone clarify this for me please?

  • 2
    $\begingroup$ It seems as though you didn't understand what "Go to step 2" means $\endgroup$ Jul 14, 2017 at 12:42
  • $\begingroup$ @Omnomnomnom That would yield $A$ as the answer. But that raises a good point - the OP should've posted their working so we can also see where the mistake in the working was. $\endgroup$
    – Shuri2060
    Jul 14, 2017 at 12:52
  • $\begingroup$ @Shuri2060 I suspect that OP followed the steps in order (except the 4th), ignoring the "go to"s. $\endgroup$ Jul 14, 2017 at 13:09

3 Answers 3


Let us consider how $(k,i,p)$ changes. You get


It's clear $p=2$ isn't the answer. In fact, notice that the loop just doubles $i$ and increments $p$ until $i=2^{10}\ge999>2^9$ when the loop will exit and print $p$. Notice also, that $(k,i,p)=(999,2^{p},p)$.

Hence $p=10$ when the loop exits.


Have you tried actually following those instructions as you were told to do?

Initially, k= 999, i= 1 and p= 1. Since 999> 1, we do step 3: i becomes 2i= 2 and p becomes 1+ 1= 2.

Go back to step 2: i= 2< 999 so we do step 3: i becomes 2i= 4 and p becomes 2+ 1= 3.

Go back to step 2: i= 4< 999 so we do step 3: i becomes 2i= 8 and p becomes 3+ 1= 4.

That should be enough to convince you that i is being multiplied by 2 every time so after n repetitions $i= 2^n$. And p has 1 added every time so after n steps, p= 1+ n.

Now how many times do we repeat? We repeat until $2^n< 999$. You could determine n using logrithms if you have a calculator: if $2^n= 999$ then n log(2)= log(999). 0.3010n= 2.9996. n= 2.9996/0.3010= 9.9653. Since n must be an integer, n= 10. But we want $2^n< 999$, not equal to it so n= 10- 1= 9.

Or we could just do the doubling: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. The last power of 2 less than 999 is $2^{9}= 512$.

Since n= 9, p= n+ 1= 10.

  • $\begingroup$ I think in the second line the first p = 0 as u said initially. $\endgroup$
    – Emptymind
    Jul 14, 2017 at 15:52
  • $\begingroup$ I think in line 7 you mean "until 2^{n} > 999". $\endgroup$
    – Emptymind
    Jul 14, 2017 at 16:04
  • $\begingroup$ I think in this Exam no calculators are allowed. $\endgroup$
    – Emptymind
    Jul 14, 2017 at 16:05

With each execution of the loop: the value of $i$ doubles and $p$ increases by $1$. Observe that $i$ is taking values as powers of $2$ and the power is reflected in $p$. The first time $i=2^p >1000$ is when $p=10$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.