Bob drew 6 letters into a circle. A, B, C, D, E, F. I don't have a graphic, but imagine those letters into a circle, with A at the top. Bob goes clockwise through the circle, writing every third letter, that he hasn't already written down. This results in the sequence A, D, B, F, C, E. Now, Alice mixes up the letters L,M,N,O,P,Q,R,S, randomly, starting with L at the top. Bob does the same thing, and goes through the circle writing every third letter. He ends up with the sequence, L, M, N, O, P, Q, R, S. What is the order that Alice put it in, starting with L at the top.

Sorry if this is vague. I read this problem earlier, but I don't know the answer.

  • 2
    $\begingroup$ Hi there, and welcome to Puzzling :) Where did you read about this problem, please? All puzzles from elsewhere need to have a source attributed to make sure the original creator is credited and avoid claims of plagiarism, etc. Thanks! $\endgroup$
    – Stiv
    Commented May 18, 2022 at 16:23
  • $\begingroup$ Read it on an assessment created by a teacher at my school. $\endgroup$
    – R3FL3CT
    Commented May 18, 2022 at 22:06

2 Answers 2


The order Alice put the letters, in clockwise order and starting from the top, in is:

L, R, O, M, S, Q, N, P

Going in order:

Put $L$ at the top
Go forwards two notches, then write $M$ on the third
Go forwards two notches again, write $N$ on the third
Go forwards two notches, we just passed $L$, so we have to add another notch, write down $O$ on the next notch
Forwards two, we passed $M$, so add another one, we can't write on the next notch because $N$ is there, so add another notch, and write $P$ on that one.
Go forwards two, add one more because we passed $L$, add as many as we need until we find an empty spot, write $Q$ on the next available.
The last two are the next available, and the last remaining.
$R$: We pass an empty spot, go forwards until we find another empty spot, and the third empty spot is going to be the first one we passed by.
$S$: It should be the last spot left.


The order is:



Repeating the same process for the order given above we get the original order.

  • $\begingroup$ I thought exactly this at first, then I tried it with the example given and understood how it really worked, if it worked like you're executing it, the example would be impossible. $\endgroup$
    – Auribouros
    Commented May 18, 2022 at 14:25
  • $\begingroup$ @Auribouros I should have cross-checked with example. $\endgroup$
    – I'm Nobody
    Commented May 18, 2022 at 14:28
  • $\begingroup$ No worries! I won't lie my heart sank a bit when I saw someone had answered while I was writing the full explanation :) $\endgroup$
    – Auribouros
    Commented May 18, 2022 at 14:30

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