CAT-holics


Home | Pages | Archives


A Puzzle of Knights

January 14, 2013 9:00 AM

(Note: From Mondays to Thursday, we’ll generally do maths. Akbar and Verbal will return on Friday!)

Posted by catcracker

Categories: All, Mathematics

Tags: ,

26 Responses to “A Puzzle of Knights”

  1. When there are 3 people, then how come the last remaining knight will be #3 ?
    e.g. 1,2,3
    1st pass: 1,3
    2nd pass: 1
    Please clarify.

    By nikunjbehani007 on February 18, 2013 at 7:16 PM

    1. Hi Nikunj,

      It is a continuous cycle and does not start again from the first person. So if there are 3 people, the count goes 1, 2(eliminated) 3, 1 (eliminated) and 3 is left.

      Similarly for say 5 people it would be 1, 2(eliminated) 3, 4 (eliminated) 5, 1(eliminated), 3, 5(eliminated) and so #3 would be left.

      regards
      J

      By catcracker on February 18, 2013 at 11:48 PM

      1. Thank you. I appreciate it.

        By nikunjbehani007 on February 19, 2013 at 12:03 AM

  2. once i saw a question in which they asked about second last position of the person means second last winner . how to get that ???

    By amanmalhotra09 on June 6, 2013 at 2:07 PM

    1. Again, let us see if there is a pattern. We need a minimum of 2 people of course. Let’s try….
      2 – 2
      3 – 1
      4 – 3
      5 – 5
      6 – 1
      7 – 3
      8 – 5
      9 – 7
      10 – 9
      11 – 11
      12 – 1
      13 – 3

      So do you see the pattern? Excepting the first case of 2 people, there is a clear pattern… Now I am sure you can figure out the rest from here…:)

      regards
      J

      By catcracker on June 6, 2013 at 4:57 PM

      1. Is the answer 217.. just to confirm 🙂 Thanks a lot for this technique

        By JV on October 11, 2013 at 9:15 PM

      2. what’s the pattern here….if multiples of 3 wud give 1 then why not 9.

        By Ashish on November 10, 2013 at 12:57 AM

      3. Sorry but I didn’t understand the statement “second last position of the person means second last winner”
        As far as I understood,
        If there are 3 person then 2nd person is second last winner(which is 2nd position)…So 1st will be the winner as you pointed out.
        Same way for 4 person 3rd person will be second last winner(which is 3rd position)…So here again 1st person will be the winner…A bit confused here.

        By bansalpatel on September 14, 2018 at 8:11 PM

      4. I’m sorry I could not make any sense of the comment. If we want to find the second last person, why are you finding the last one? You seem to be mixing up the two questions.

        regards
        J

        By catcracker on September 15, 2018 at 11:29 AM

      5. Okay sir…I got it now…the game remains the same just instead of the last person we will check who was the last person to get eliminated…I checked various cases and it checks out…Thank You very much for the reply…I have so much gratitude for this blog and the people who came up with this noble idea.

        By bansalpatel on September 28, 2018 at 2:13 PM

      6. What would be the general form of the pattern?
        I am not able to figure that part out…but can used this to get to answer.
        If N= Total number of knights
        So,
        1+[N-(Power of 2 X 3)]2 will be the 2nd last winner, where (Power of 2X3) has to be closest as well as less than N.
        For ex in 300,
        1+[300-(64X3)]2
        1+[300-192]2
        1+216
        217th will be the runner up

        Is the general form I have applied correct or it could be refined further?

        By Jotaro Kujo on June 29, 2022 at 3:31 AM

      7. As I said, you would have to figure out the rest, if at all you can develop pattern-recognition (if not this method is anyway useless)

        regards
        J

        By catcracker on June 29, 2022 at 8:55 PM

  3. what will happen if numbers are removed in multiples of 3 or mulitiples of 4 or 5 ?

    By nitin on June 30, 2013 at 2:51 PM

    1. No simple solution in either of these two cases. You can try for small numbers and check if a pattern emerges, but the pattern turns out to be fairly complex in both the cases you mentioned….the latter case, maybe something in bases systems would be feasible, would have to think about it.

      regards
      J

      By catcracker on June 30, 2013 at 3:29 PM

  4. also what difference it will make when they are not removed continuously but at one time then started again ? for example in a count of 2000 number every 7 number is removed in every stage , what will be the last digit to be removed ? plz explain procedure

    By nitin on June 30, 2013 at 2:57 PM

  5. Dear J, i’m able to solve a lot of other variants of this question. Thank you so much. 🙂

    By Kartik Kwatra on August 5, 2013 at 3:11 AM

  6. thanks sir, i am now able to figure out where i was going wrong when i solved a logical question of the same pattern. 😀

    By nikitasinha23 on August 20, 2013 at 9:37 AM

  7. HI Sir,

    How about if lets say every 10th was to be eliminated?

    By gaurav raj on December 7, 2014 at 1:16 PM

  8. Very nasty, Gaurav. For 3 or 4 you could still hope to derive a kind of pattern by trying it for small numbers; for 10 we would have to do a lot of trials before a pattern would emerge (if any). Thus it would be an exercise for a computer program and not for a manual effort. Consequently, you could safely guarantee that it would be unlikely to come in a test/interview.

    regards
    J

    By catcracker on December 7, 2014 at 1:40 PM

  9. sir, in 1st question in 1st pass,
    1,3,5,7,9… left
    in second pass and in third pass also you started eliminating from 2nd no. in the row, then why did you chose 9 in 4th pass, shouldn’t it again be 1,17,25.. in 4th pass? and because the trick is also based on it. i got stuck here..

    By seema on July 9, 2016 at 9:56 AM

  10. Because in the first 2 passes the last person eliminated was the last person left (#300 and #299 respectively). But in the third pass the last person eliminated was 293, so two steps further from there will be 297 (remains) and 1 (eliminated), i.e. 1 will have to be eliminated at the beginning of the 4th pass. I am not starting each pass from 1 again, it is a continuous pass as already explained in an earlier comment.

    regards
    J

    By catcracker on July 9, 2016 at 4:17 PM

  11. so can we conclude, that it is in the following a pattern of removing even terms(2nd, 4th, 6th, 8th..) in 1st three passes, odd ones in the next three passes and then evens and odds and so on in the gap of three passes?

    By seema on July 11, 2016 at 7:42 AM

    1. NO. Every different starting number of people will have a different sequence of passes.

      regards
      J

      By catcracker on July 11, 2016 at 1:31 PM

  12. sir,
    when number of people are 4 first pass
    1,2(eliminated),3,4(eliminated)
    second pass
    1(eliminated), #3 is left out?
    could you please explain me how #1 was left as per your sol.

    By gayatri on July 25, 2016 at 9:19 PM

    1. You have eliminated 4 and then 1…that is, two consecutive people! After 4 is kicked out, 1 must stay and hence 3 will get kicked out.

      regards
      J

      By catcracker on July 26, 2016 at 12:01 AM

  13. I couldn’t understand the question first

    By Agneeshwaran on September 7, 2017 at 12:02 PM

Leave a Reply



Mobile Site | Full Site


Get a free blog at WordPress.com Theme: WordPress Mobile Edition by Alex King.