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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time - integers

Puzzle time - integers

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  • jon-nycJ Offline
    jon-nycJ Offline
    jon-nyc
    wrote on last edited by
    #9

    :::

    On the right track but not quite there

    :::

    Thank you for your attention to this matter.

    1 Reply Last reply
    • KlausK Online
      KlausK Online
      Klaus
      wrote on last edited by
      #10

      So you are saying my last statement is wrong, or are you saying it's not precise enough?

      1 Reply Last reply
      • jon-nycJ Offline
        jon-nycJ Offline
        jon-nyc
        wrote on last edited by
        #11

        Depends on how one defines ‘noise‘. But what I really mean is “from what I infer from your words you’re still missing an insight here”

        Thank you for your attention to this matter.

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        • KlausK Online
          KlausK Online
          Klaus
          wrote on last edited by
          #12

          OK, here's a precise version of the statement:

          :::

          There is a number N, such that for all n >N, n is not in S if and only if the last digit of n is 0 or 5.

          :::

          Is that correct?

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          • jon-nycJ Offline
            jon-nycJ Offline
            jon-nyc
            wrote on last edited by jon-nyc
            #13

            Yes but tell me N. You’re missing something or you would know what N is.

            Thank you for your attention to this matter.

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            • KlausK Online
              KlausK Online
              Klaus
              wrote on last edited by Klaus
              #14

              N is smaller than or equal to 2915. Now don't tell me you want me to worry about selecting a particular number between 1 and 2915!!!

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              • jon-nycJ Offline
                jon-nycJ Offline
                jon-nyc
                wrote on last edited by
                #15

                Yes I do.

                Thank you for your attention to this matter.

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                • jon-nycJ Offline
                  jon-nycJ Offline
                  jon-nyc
                  wrote on last edited by
                  #16

                  What Klaus missed:

                  :::

                  The only ‘noise’ (besides all multiples of 5) is the number 1.

                  • 2 is granted which gets you all numbers ending in 2 or 7.
                  • 7^2 is 49 which gets you all the numbers ending in 9 and 4 above that
                  • after 49 is 54. 54^2 is 3136 which gets you all the numbers ending in 6 or 1 above it.
                    BUT
                  • once you have the *6s, you’ll get to 6^8 which gets you back to 6 and 11, etc.
                  • that gets you to 16 which gets you back to 4 and 9
                  • that 9 gets you back to 3 and 8

                  So we have 2,3,4,6,7,8,9 covered plus any number that is a multiple of 5 above them.

                  So only 1 is missing, along with all multiples of 5

                  :::

                  Thank you for your attention to this matter.

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                  • KlausK Online
                    KlausK Online
                    Klaus
                    wrote on last edited by
                    #17

                    Nice!

                    54^2 is 2916 and not 3136, though - that was the source of the 2915 bound I was giving above. So my bound was pointing in the right direction 🙂

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                    • jon-nycJ Offline
                      jon-nycJ Offline
                      jon-nyc
                      wrote on last edited by
                      #18

                      My math buddy at CS pointed out that Fermat’s Little Theorem could help here too rather than finding actual paths back to the lower numbers.

                      Thank you for your attention to this matter.

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