You must place 7 robot ants on a long rod and set each of them to move left or right starting at time 0. You can set any positive speed for each ant. When 2 or more ants meet, they turn around. When an ant reaches the end it falls off.

Show that you can place and set the ants in such a way that none of them ever falls off.

Bonus question: Can you do this for other odd numbers, too?

  • 1
    $\begingroup$ Since someone tried to wiggle out of the intended wording with the last problem, perhaps clarify that "left or right" is along the direction of the rod/parallel to its run/equivalent to moving along a long but not infinite axis $\endgroup$
    – bobble
    Commented May 29, 2022 at 4:40
  • $\begingroup$ What happens if I put two ants with the same starting position and direction and speed? $\endgroup$
    – justhalf
    Commented May 29, 2022 at 4:56
  • $\begingroup$ @justhalf Starting positions must be different. $\endgroup$
    – Eric
    Commented May 29, 2022 at 5:14
  • $\begingroup$ I have found that for 3 ants is impossible. Interesting problem. $\endgroup$
    – justhalf
    Commented May 29, 2022 at 5:32
  • 1
    $\begingroup$ @justhalf how? can you elaborate a bit $\endgroup$
    – I'm Nobody
    Commented May 29, 2022 at 6:56

1 Answer 1


Picture showing how to place 4,7 or 10 ants:

enter image description here

x-axis is time, y-axis is position. The 4 black ants alone are viable as are the blacks together with the 3 blues or all 10.

Ants here move either at speed 1 or 1/3. Except for the first and last ones, the slower ants can be left out.

Therefore this construction yields solutions for 4, 6, 7, 8, 9, 10, 11, ... ants.

P.S.: 5 is also possible:

enter image description here

  • 1
    $\begingroup$ One of those where a well presented diagram says it all. Great answer! $\endgroup$
    – hexomino
    Commented May 30, 2022 at 21:07
  • 2
    $\begingroup$ Worth noting that this involves instances of three ants meeting. (Yes, this is allowed by the rules.) (Maybe 3+ ants meeting is needed to get 7 or any odd number?) $\endgroup$
    – tehtmi
    Commented May 30, 2022 at 23:31
  • 1
    $\begingroup$ Does this mean that 5 ants is not possible? $\endgroup$ Commented May 31, 2022 at 1:20
  • $\begingroup$ @DmitryKamenetsky No, it only gives positive answers. $\endgroup$ Commented May 31, 2022 at 7:29
  • $\begingroup$ Congratulations! You've solved the most difficult 5 ants case! $\endgroup$
    – Eric
    Commented May 31, 2022 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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