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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time

Puzzle time

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  • KlausK Offline
    KlausK Offline
    Klaus
    wrote on last edited by Klaus
    #1

    Find all natural numbers x and y such that the sum of the square roots of x and y is the square root of 2009.

    No computer is needed.

    1 Reply Last reply
    • jon-nycJ Online
      jon-nycJ Online
      jon-nyc
      wrote on last edited by
      #2

      @Klaus said in Puzzle time:

      Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.

      All the square roots or just the positive ones?

      You were warned.

      KlausK 1 Reply Last reply
      • jon-nycJ Online
        jon-nycJ Online
        jon-nyc
        wrote on last edited by jon-nyc
        #3

        ||Assuming just the positive roots then itโ€™s just (1^2, 2008^2), (2^2, 2007^2), (3^2, 2006^2), ....etc.||

        You were warned.

        taiwan_girlT 1 Reply Last reply
        • jon-nycJ jon-nyc

          ||Assuming just the positive roots then itโ€™s just (1^2, 2008^2), (2^2, 2007^2), (3^2, 2006^2), ....etc.||

          taiwan_girlT Offline
          taiwan_girlT Offline
          taiwan_girl
          wrote on last edited by
          #4

          |@jon-nyc

          ||I donโ€™t think your answer is correct. For example, the square root of one is one. And the square root of 2008 is almost 45. Add those two together and only equal 46

          Not sure of correct answer however. Lol46||

          1 Reply Last reply
          • jon-nycJ Online
            jon-nycJ Online
            jon-nyc
            wrote on last edited by jon-nyc
            #5

            @taiwan_girl

            ||You misread it. I squared those numbers for ease of notation and understanding the sequence. The real numbers would be (1, 4,032,064), (4, 4,028,049), (9, 4,024,036)...||

            You were warned.

            taiwan_girlT 1 Reply Last reply
            • jon-nycJ jon-nyc

              @taiwan_girl

              ||You misread it. I squared those numbers for ease of notation and understanding the sequence. The real numbers would be (1, 4,032,064), (4, 4,028,049), (9, 4,024,036)...||

              taiwan_girlT Offline
              taiwan_girlT Offline
              taiwan_girl
              wrote on last edited by
              #6

              @jon-nyc ah okay. ๐Ÿ‘๐Ÿป

              1 Reply Last reply
              • KlausK Offline
                KlausK Offline
                Klaus
                wrote on last edited by Klaus
                #7

                Sorry, I screwed up. The sum of those numbers should be the square root of 2009. I fixed it above. The previous version is trivial, as Jon pointed out.

                1 Reply Last reply
                • jon-nycJ jon-nyc

                  @Klaus said in Puzzle time:

                  Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.

                  All the square roots or just the positive ones?

                  KlausK Offline
                  KlausK Offline
                  Klaus
                  wrote on last edited by
                  #8

                  @jon-nyc said in Puzzle time:

                  @Klaus said in Puzzle time:

                  Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.

                  All the square roots or just the positive ones?

                  Just the positive ones.

                  1 Reply Last reply
                  • AxtremusA Away
                    AxtremusA Away
                    Axtremus
                    wrote on last edited by
                    #9

                    ||Since the positive square root of 2009 is not a natural number, there is zero natural number for x and y such that they sum to the positive square root of 2009.||

                    1 Reply Last reply
                    • jon-nycJ Online
                      jon-nycJ Online
                      jon-nyc
                      wrote on last edited by
                      #10

                      Ax thatโ€™s not the problem. Iโ€™ll restate it:

                      Solve for sqrt(x) + sqrt(y) = sqrt(2009) where x,y are natural numbers

                      You were warned.

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

                        || (41 * a^2, 41 * b^2) where a,b are natural numbers with a+b=7.

                        Or to spell it out, (41,1476), (164,1025), (369,656) and then the three number pairs that have those numbers reversed||

                        You were warned.

                        1 Reply Last reply
                        • KlausK Offline
                          KlausK Offline
                          Klaus
                          wrote on last edited by
                          #12

                          That looks good, Jon. Can you show how you got that solution?

                          1 Reply Last reply
                          • jon-nycJ Online
                            jon-nycJ Online
                            jon-nyc
                            wrote on last edited by jon-nyc
                            #13

                            2009^1/2 simplifies to 7*sqrt(41)

                            So the two addends need to be in the form a * sqrt(41) and b * sqrt(41) with a and b summing to 7.

                            You were warned.

                            1 Reply Last reply
                            • KlausK Offline
                              KlausK Offline
                              Klaus
                              wrote on last edited by
                              #14

                              ๐Ÿ‘

                              1 Reply Last reply
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