Equal numbers in a circle

Source: Stanford Math Circle Sunday May 30, 2010

Problem: A circle is divided into 6 sectors. The numbers 1, 0, 1, 0, 0, 0 are written into the sectors in the counter clock-wise direction. You may increase any two neighboring numbers by 1. Is it possible to make all of the numbers equal?

Disclaimer: Think simple. Very simple problem. Solved it in less than 30 seconds. :)

Solution: Posted by Aman in comments!! He claims he took less than 10 seconds \m/ \m/. This is one thing where it feels good to win, but feels awesome to be defeated \m/ \m/

Comments

  1. Anonymous11:01 PM, September 08, 2010

    Simple problem, Me solved it in less than 10 seconds \m/
    Add all the numbers in odd places and subtract from this sum, the sum of the number at even places. The result is invariant and is = 2. The final state would demand a result = 0 and hence is not possible.:)

    ReplyDelete
  2. Pratik Poddar11:45 PM, September 08, 2010

    Great.. Nice and simple solution.. Thanks.. :)

    ReplyDelete
  3. Ankush12:13 AM, September 09, 2010

    Damn it!
    @ Aman : Awesome solution. It was very difficult to think this simple!

    ReplyDelete
  4. Shantanu1:29 PM, October 10, 2010

    ~2 mins before I read your comment on time to be taken :P
    Got it :) in 30 sec. thereafter

    ReplyDelete
  5. Pratik Poddar1:02 PM, September 09, 2026

    Ha - that is exactly the right reaction. The invariant (alternating sum of the six sectors never changes: adding 1 to two neighbors adds +1 and -1 to it) kills the problem in one line: the start has alternating sum 2, the all-equal state has 0, so it can never be reached. (replied using AI)

    ReplyDelete
  6. Pratik Poddar1:03 PM, September 09, 2026

    Nice! The hint about the time taken is basically a spoiler in itself - once you know there is a 30-second solution, you stop trying to construct the sequence and start hunting for an invariant. (replied using AI)

    ReplyDelete

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