Pattern Recognition Problem: If $7,24 to 25 ; 12,35 to 37;$ … , then M=?












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$begingroup$


A friend has given me a puzzle to solve, the puzzle is as follows:
$$color{red}{7 , , } 24 to color{blue}{25}$$
$$color{red}{12 , , } 35 to color{blue}{37}$$
$$color{red}{11 , , } 60 to color{blue}{61}$$
$$color{red}{8 , , } 15 to color{blue}{17}$$
$$color{red}{9 , , } 40 to color{blue}{text{M}}$$
then $color{blue}{text{M}}$=?






pattern







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    0












    $begingroup$


    A friend has given me a puzzle to solve, the puzzle is as follows:
    $$color{red}{7 , , } 24 to color{blue}{25}$$
    $$color{red}{12 , , } 35 to color{blue}{37}$$
    $$color{red}{11 , , } 60 to color{blue}{61}$$
    $$color{red}{8 , , } 15 to color{blue}{17}$$
    $$color{red}{9 , , } 40 to color{blue}{text{M}}$$
    then $color{blue}{text{M}}$=?






    pattern







    share|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      A friend has given me a puzzle to solve, the puzzle is as follows:
      $$color{red}{7 , , } 24 to color{blue}{25}$$
      $$color{red}{12 , , } 35 to color{blue}{37}$$
      $$color{red}{11 , , } 60 to color{blue}{61}$$
      $$color{red}{8 , , } 15 to color{blue}{17}$$
      $$color{red}{9 , , } 40 to color{blue}{text{M}}$$
      then $color{blue}{text{M}}$=?






      pattern







      share|improve this question











      $endgroup$




      A friend has given me a puzzle to solve, the puzzle is as follows:
      $$color{red}{7 , , } 24 to color{blue}{25}$$
      $$color{red}{12 , , } 35 to color{blue}{37}$$
      $$color{red}{11 , , } 60 to color{blue}{61}$$
      $$color{red}{8 , , } 15 to color{blue}{17}$$
      $$color{red}{9 , , } 40 to color{blue}{text{M}}$$
      then $color{blue}{text{M}}$=?






      pattern




      pattern






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      share|improve this question








      edited 4 hours ago









      Rubio♦

      29.5k566180




      29.5k566180










      asked 4 hours ago









      SureshSuresh

      1335




      1335






















          2 Answers
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          10












          $begingroup$

          The answer is




          41 because $red^2 + black^2 = blue^2$.




          These are all




          Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $sqrt{1681} = 41 = M$.







          share|improve this answer









          $endgroup$





















            8












            $begingroup$

            Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:




            if red is uneven then black + 1
            else if red is even then black + 2







            share|improve this answer








            New contributor




            user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






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            • $begingroup$
              Huh. I do think that @El-Guest is correct, but this is very interesting.
              $endgroup$
              – Brandon_J
              2 hours ago











            Your Answer





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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            The answer is




            41 because $red^2 + black^2 = blue^2$.




            These are all




            Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $sqrt{1681} = 41 = M$.







            share|improve this answer









            $endgroup$


















              10












              $begingroup$

              The answer is




              41 because $red^2 + black^2 = blue^2$.




              These are all




              Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $sqrt{1681} = 41 = M$.







              share|improve this answer









              $endgroup$
















                10












                10








                10





                $begingroup$

                The answer is




                41 because $red^2 + black^2 = blue^2$.




                These are all




                Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $sqrt{1681} = 41 = M$.







                share|improve this answer









                $endgroup$



                The answer is




                41 because $red^2 + black^2 = blue^2$.




                These are all




                Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $sqrt{1681} = 41 = M$.








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 4 hours ago









                El-GuestEl-Guest

                20.2k24489




                20.2k24489























                    8












                    $begingroup$

                    Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:




                    if red is uneven then black + 1
                    else if red is even then black + 2







                    share|improve this answer








                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      Huh. I do think that @El-Guest is correct, but this is very interesting.
                      $endgroup$
                      – Brandon_J
                      2 hours ago
















                    8












                    $begingroup$

                    Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:




                    if red is uneven then black + 1
                    else if red is even then black + 2







                    share|improve this answer








                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      Huh. I do think that @El-Guest is correct, but this is very interesting.
                      $endgroup$
                      – Brandon_J
                      2 hours ago














                    8












                    8








                    8





                    $begingroup$

                    Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:




                    if red is uneven then black + 1
                    else if red is even then black + 2







                    share|improve this answer








                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:




                    if red is uneven then black + 1
                    else if red is even then black + 2








                    share|improve this answer








                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|improve this answer



                    share|improve this answer






                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 2 hours ago









                    user57867user57867

                    811




                    811




                    New contributor




                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    user57867 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.












                    • $begingroup$
                      Huh. I do think that @El-Guest is correct, but this is very interesting.
                      $endgroup$
                      – Brandon_J
                      2 hours ago


















                    • $begingroup$
                      Huh. I do think that @El-Guest is correct, but this is very interesting.
                      $endgroup$
                      – Brandon_J
                      2 hours ago
















                    $begingroup$
                    Huh. I do think that @El-Guest is correct, but this is very interesting.
                    $endgroup$
                    – Brandon_J
                    2 hours ago




                    $begingroup$
                    Huh. I do think that @El-Guest is correct, but this is very interesting.
                    $endgroup$
                    – Brandon_J
                    2 hours ago


















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