Loading ...
Sorry, an error occurred while loading the content.

13th root of 100 digits

Expand Messages
  • Paul Reed
    Dear all I ve spent some weeks now looking at 13th root of 100 digits. My main motivation for this was the way this category of mental calculation was put on a
    Message 1 of 17 , Aug 5, 2003
    • 0 Attachment
      Dear all

      I've spent some weeks now looking at 13th root of 100 digits. My main motivation for this was the way this category of mental calculation was put on a pedestal as the ultimate category. I have yet to really start practicing in earnest but already believe I have derived a method for 13th root of odd powers that does not require an incredibly high IQ as has been suggested, nor does it require memorisation of log tables, or great ability at factoring numbers a la Kleins method. I have not yet finalised the method for the last 4 digits of even powers. Also, odd powers ending in 5 are a bit tricky still - I will work on these next. However, my method for odd powers (except 5) is very rapid indeed.

      My method uses mnemonics for digits 1 to 4 (based on the first 4 digits of the power) and exceedingly simple calculations and for digits 5 to 8...... I have to remember less than 1000 numbers using mnemonics and only simple division by three or simple subtraction for the mental calculations. Surely this is easier than learning up to 100 x 100 table yet 13th root of 100 digits was talked about as if it was some kind of holy grail.

      13.5 seconds is Alexis record I believe - well, with digit 8 being immediately recalled, and digits 1 to 4 being recalled within a couple of seconds through my mnemonic system, this would then leave approx 10 seconds to make three further small calculations......hmm, seems imminently possible! A nice power (ending in 1 for example) could be done in 6 or 7 seconds I believe......

      So, what are the parameters for setting world records? How many attempts in what timeframe? Where does this take place??? Are you able to pass on some attempts to wait for a 'nicer' calculation?

      I believe I will be <20 seconds within 1 month - I shall keep you all informed of my progress.

      Paul


      ---------------------------------
      Want to chat instantly with your online friends? Get the FREE Yahoo!Messenger

      [Non-text portions of this message have been removed]
    • ralf_laue
      ... attempts in what timeframe? Where does this take place??? Are you able to pass on some attempts to wait for a nicer calculation? The suggested rules for
      Message 2 of 17 , Aug 5, 2003
      • 0 Attachment
        --- In MentalCalculation@yahoogroups.com, Paul Reed <t_paulr@y...> wrote:

        > So, what are the parameters for setting world records? How many
        attempts in what timeframe? Where does this take place??? Are you able
        to pass on some attempts to wait for a 'nicer' calculation?

        The suggested rules for the record list at
        http://www.recordholders.org/en/list/memory.html
        can be found at
        http://www.recordholders.org/en/rules/calculating.html.

        The number of attempts per day / per year must be limited in order
        to avoid good luck-"records".

        Ralf
      • A.W.A.P. Bouman
        Dear Calculator friends, What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea? Eg I
        Message 3 of 17 , Jun 15, 2008
        • 0 Attachment
          Dear Calculator friends,

          What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

          Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

          And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

          But why only 13th??

          Regards,

          Willem Bouman



          [Non-text portions of this message have been removed]
        • Andy Robertshaw
          Hi Willem, I think one reason for this is because prime roots are more difficult than composite roots. For example if we have the 15th root then we can take
          Message 4 of 17 , Jun 15, 2008
          • 0 Attachment
            Hi Willem,
            I think one reason for this is because prime roots are more difficult than composite roots.
            For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
            So we've got the 7th, 11th, 13th and 17th roots to contend with.
            I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
            But certainly the 7th and 11th roots would be tricky enough!
            Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
            Andy


            ----- Original Message ----
            From: A.W.A.P. Bouman <awap.bouman@...>
            To: mental calculation <MentalCalculation@yahoogroups.com>
            Sent: Sunday, 15 June, 2008 10:40:11 AM
            Subject: [Mental Calculation] roots


            Dear Calculator friends,

            What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

            Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

            And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

            But why only 13th??

            Regards,

            Willem Bouman

            [Non-text portions of this message have been removed]




            __________________________________________________________
            Sent from Yahoo! Mail.
            A Smarter Email http://uk.docs.yahoo.com/nowyoucan.html

            [Non-text portions of this message have been removed]
          • jsh.flynn
            The nth root challenge requires n to be prime because the fourth root for example could involve taking the square root twice and so may be less challenging
            Message 5 of 17 , Jun 15, 2008
            • 0 Attachment
              The nth root challenge requires n to be prime because the fourth root
              for example could involve taking the square root twice and so may be
              less challenging than taking cube roots. This is likely only part of
              the answer. This also applies to integer exponentiation. Raising a
              number like 57 to the power of 8 is easier than raising it to the power
              of 7 because one can square it 3 times. E.g.

              57^2 = 3249 (Known before hand)

              3249^2 = (3200^2) + (here comes the difficult part) 2*3200*49 (This is
              6400 * 49 (hey both start with squares! a shortcut) = 560^2) + (49^2)
              =10556001

              As you can see the only problem now is squaring that number which I
              would choose over 57*57*57*57*57*57*57 anyday.

              From Jsh
            • Oleg Stepanov
              It was 23 and 73 too... http://stepanov.lk.net/mnemo/smith13e.html AWAPB Dear Calculator friends, AWAPB What is the special reason that the 13th root - of an
              Message 6 of 17 , Jun 15, 2008
              • 0 Attachment
                It was 23 and 73 too...
                http://stepanov.lk.net/mnemo/smith13e.html

                AWAPB> Dear Calculator friends,

                AWAPB> What is the special reason that the 13th root - of an
                AWAPB> huge number- seems to be the magic standard? Does one of you
                AWAPB> have an idea?

                AWAPB> Eg I never see something of 5th root, a11 th or a
                AWAPB> 15th or a 17th. Why is that?

                AWAPB> And then I renounce of even roots. I know: a 4th root
                AWAPB> gives heaps of possibilities and an 8th still much more! So I
                AWAPB> can imagine this is almost impossible.

                AWAPB> But why only 13th??

                AWAPB> Regards,

                AWAPB> Willem Bouman




                Sincerely Yours, Oleg Stepanov.
                http://stepanov.lk.net/
              • A.W.A.P. Bouman
                Dear Mr. Stepanov, Thank you very much for your interesting article, which I printed immediately. I ll tell you something very interesting. When I met Wim
                Message 7 of 17 , Jun 15, 2008
                • 0 Attachment
                  Dear Mr. Stepanov,

                  Thank you very much for your interesting article, which I printed immediately.

                  I'll tell you something very interesting. When I met Wim Klein in December 1959 he examined me and asked me which problems I could lose. And among others I mentioned cubis roots. "impossible" he reacted. Even roots can be done, odd ones cannot. "Make me please an excersize of an integer cubic root of 15 digits". Klein did, and asked of course how I did this. In a flash he understood and later on he worked my system more and more out.

                  He gave me the advise not do calendar calculation, it is a banality, he said. And now I pay the price: now I have to do in the competition, and although I make some progress, I'll never be a star in it.

                  In August 1986 I wanted to renew the contact, but unfortunately he was murdered.

                  Regards,

                  Willem Bouman








                  ----- Oorspronkelijk bericht -----
                  Van: Oleg Stepanov
                  Aan: A.W.A.P. Bouman
                  Verzonden: zondag 15 juni 2008 14:48
                  Onderwerp: Re: [Mental Calculation] roots


                  It was 23 and 73 too...
                  http://stepanov.lk.net/mnemo/smith13e.html

                  AWAPB> Dear Calculator friends,

                  AWAPB> What is the special reason that the 13th root - of an
                  AWAPB> huge number- seems to be the magic standard? Does one of you
                  AWAPB> have an idea?

                  AWAPB> Eg I never see something of 5th root, a11 th or a
                  AWAPB> 15th or a 17th. Why is that?

                  AWAPB> And then I renounce of even roots. I know: a 4th root
                  AWAPB> gives heaps of possibilities and an 8th still much more! So I
                  AWAPB> can imagine this is almost impossible.

                  AWAPB> But why only 13th??

                  AWAPB> Regards,

                  AWAPB> Willem Bouman

                  Sincerely Yours, Oleg Stepanov.
                  http://stepanov.lk.net/





                  [Non-text portions of this message have been removed]
                • A.W.A.P. Bouman
                  Dear Mr. Flynn, With all respect: I do not agree with you. An integer cubic root is far less challenging: when you work on the right way in combination with
                  Message 8 of 17 , Jun 15, 2008
                  • 0 Attachment
                    Dear Mr. Flynn,

                    With all respect: I do not agree with you. An integer cubic root is far less challenging: when you work on the right way in combination with modulo calculation, there is always 1 possibility. 4 th roots can surely be more tricky because of the number of possibilities.

                    I agree with your argument in multiplication, my question nevertheless is the opposite: the roots.

                    Regards,

                    Willem Bouman


                    ----- Oorspronkelijk bericht -----
                    Van: jsh.flynn
                    Aan: MentalCalculation@yahoogroups.com
                    Verzonden: zondag 15 juni 2008 14:23
                    Onderwerp: [Mental Calculation] Re: roots


                    The nth root challenge requires n to be prime because the fourth root
                    for example could involve taking the square root twice and so may be
                    less challenging than taking cube roots. This is likely only part of
                    the answer. This also applies to integer exponentiation. Raising a
                    number like 57 to the power of 8 is easier than raising it to the power
                    of 7 because one can square it 3 times. E.g.

                    57^2 = 3249 (Known before hand)

                    3249^2 = (3200^2) + (here comes the difficult part) 2*3200*49 (This is
                    6400 * 49 (hey both start with squares! a shortcut) = 560^2) + (49^2)
                    =10556001

                    As you can see the only problem now is squaring that number which I
                    would choose over 57*57*57*57*57*57*57 anyday.

                    From Jsh





                    [Non-text portions of this message have been removed]
                  • A.W.A.P. Bouman
                    Ha, Andy, Thank you for your reaction. you are right when you speak about the number of digits of the answer. If I remember wll it was Wim Klein who said: The
                    Message 9 of 17 , Jun 15, 2008
                    • 0 Attachment
                      Ha, Andy,

                      Thank you for your reaction. you are right when you speak about the number of digits of the answer. If I remember wll it was Wim Klein who said: "The trickyness of a root is not in the exponent, but in the number of digits of the answer". From this point of view a 17 th root of 100 digits would be easier than a 13 th root of 100 digit. With an answer of 8 digits a 17 th power couild have between 120 and 136 digits, as you know.

                      29 June there will be a Dutch performance in Leipzig: a certain Mr. Willem Bouman will give then a presentation with exersizes followed by an explanation. You can ask Ralf Laue if there is a place for you, the spoken language will be German.


                      Till then!

                      Willem Bouman

                      ----- Oorspronkelijk bericht -----
                      Van: Andy Robertshaw
                      Aan: MentalCalculation@yahoogroups.com
                      Verzonden: zondag 15 juni 2008 13:22
                      Onderwerp: Re: [Mental Calculation] roots


                      Hi Willem,
                      I think one reason for this is because prime roots are more difficult than composite roots.
                      For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
                      So we've got the 7th, 11th, 13th and 17th roots to contend with.
                      I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
                      But certainly the 7th and 11th roots would be tricky enough!
                      Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
                      Andy

                      ----- Original Message ----
                      From: A.W.A.P. Bouman <awap.bouman@...>
                      To: mental calculation <MentalCalculation@yahoogroups.com>
                      Sent: Sunday, 15 June, 2008 10:40:11 AM
                      Subject: [Mental Calculation] roots

                      Dear Calculator friends,

                      What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

                      Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

                      And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

                      But why only 13th??

                      Regards,

                      Willem Bouman

                      [Non-text portions of this message have been removed]

                      __________________________________________________________
                      Sent from Yahoo! Mail.
                      A Smarter Email http://uk.docs.yahoo.com/nowyoucan.html

                      [Non-text portions of this message have been removed]





                      [Non-text portions of this message have been removed]
                    • Oleg Stepanov
                      AWAPB Dear Mr. Stepanov, AWAPB Thank you very much for your interesting article, which I printed immediately. This is not article. This is chapter from book.
                      Message 10 of 17 , Jun 15, 2008
                      • 0 Attachment
                        AWAPB> Dear Mr. Stepanov,

                        AWAPB> Thank you very much for your interesting article, which I printed immediately.
                        This is not article. This is chapter from book.

                        Sincerely Yours, Oleg Stepanov.
                        http://stepanov.lk.net/
                      • A.W.A.P. Bouman
                        Dear Mr. Stepanov, Thanks again for your interesting reaction and the link it contents. My problem: it is more than 50 years ago that I left the secundary
                        Message 11 of 17 , Jun 15, 2008
                        • 0 Attachment
                          Dear Mr. Stepanov,

                          Thanks again for your interesting reaction and the link it contents.

                          My problem: it is more than 50 years ago that I left the secundary school, I have to study the log questions and mantissas thoroughly.

                          My solution: not so long ago I found a formula for finding end figures of roots.
                          BN = basic number, in this case 11, I speak about the second problem in your link, the 13 th rot of 75185.....85831.
                          N= exponent, or power, in this case 13.
                          Jump: the number which with the end figures change. 11^13 end on 3931, the problem ends on 3931.
                          My formula - on which I am a little proud- is

                          Jump = BN^(n-1) × power × 100. In this case:

                          11^12×13×100= 721×13×100= ....7300.

                          Now we subtract 5831-3931= 1900. The question is yet how many ...7300's go in ...1900. It is 3, so the answer ends on 0311.

                          Regards,

                          Willem Bouman





                          ----- Oorspronkelijk bericht -----
                          Van: Oleg Stepanov
                          Aan: A.W.A.P. Bouman
                          Verzonden: zondag 15 juni 2008 14:48
                          Onderwerp: Re: [Mental Calculation] roots


                          It was 23 and 73 too...
                          http://stepanov.lk.net/mnemo/smith13e.html

                          AWAPB> Dear Calculator friends,

                          AWAPB> What is the special reason that the 13th root - of an
                          AWAPB> huge number- seems to be the magic standard? Does one of you
                          AWAPB> have an idea?

                          AWAPB> Eg I never see something of 5th root, a11 th or a
                          AWAPB> 15th or a 17th. Why is that?

                          AWAPB> And then I renounce of even roots. I know: a 4th root
                          AWAPB> gives heaps of possibilities and an 8th still much more! So I
                          AWAPB> can imagine this is almost impossible.

                          AWAPB> But why only 13th??

                          AWAPB> Regards,

                          AWAPB> Willem Bouman

                          Sincerely Yours, Oleg Stepanov.
                          http://stepanov.lk.net/





                          [Non-text portions of this message have been removed]
                        • Oleg Stepanov
                          Dear Mr. Bouman. I do not know is it interesting. But here one more chapter from that book: http://stepanov.lk.net/mnemo/sm16.html AWAPB Dear Mr. Stepanov,
                          Message 12 of 17 , Jun 16, 2008
                          • 0 Attachment
                            Dear Mr. Bouman.
                            I do not know is it interesting. But here one more chapter from
                            that book:
                            http://stepanov.lk.net/mnemo/sm16.html


                            AWAPB> Dear Mr. Stepanov,

                            AWAPB> Thanks again for your interesting reaction and the link it contents.

                            AWAPB> My problem: it is more than 50 years ago that I left
                            AWAPB> the secundary school, I have to study the log questions and
                            AWAPB> mantissas thoroughly.

                            AWAPB> My solution: not so long ago I found a formula for
                            AWAPB> finding end figures of roots.
                            AWAPB> BN = basic number, in this case 11, I speak about the
                            AWAPB> second problem in your link, the 13 th rot of 75185.....85831.
                            AWAPB> N= exponent, or power, in this case 13.
                            AWAPB> Jump: the number which with the end figures change.
                            AWAPB> 11^13 end on 3931, the problem ends on 3931.
                            AWAPB> My formula - on which I am a little proud- is

                            AWAPB> Jump = BN^(n-1) × power × 100. In this case:

                            AWAPB> 11^12×13×100= 721×13×100= ....7300.

                            AWAPB> Now we subtract 5831-3931= 1900. The question is yet
                            AWAPB> how many ...7300's go in ...1900. It is 3, so the answer ends
                            AWAPB> on 0311.

                            AWAPB> Regards,

                            AWAPB> Willem Bouman


                            Sincerely Yours, Oleg Stepanov.
                            http://stepanov.lk.net/
                          • A.W.A.P. Bouman
                            Dear Mr. Stepanov, Thanks again. Willem Bouman ... Van: Oleg Stepanov Aan: A.W.A.P. Bouman Verzonden: maandag 16 juni 2008 10:20 Onderwerp: Re: [Mental
                            Message 13 of 17 , Jun 16, 2008
                            • 0 Attachment
                              Dear Mr. Stepanov,

                              Thanks again.

                              Willem Bouman

                              ----- Oorspronkelijk bericht -----
                              Van: Oleg Stepanov
                              Aan: A.W.A.P. Bouman
                              Verzonden: maandag 16 juni 2008 10:20
                              Onderwerp: Re: [Mental Calculation] roots


                              Dear Mr. Bouman.
                              I do not know is it interesting. But here one more chapter from
                              that book:
                              http://stepanov.lk.net/mnemo/sm16.html

                              AWAPB> Dear Mr. Stepanov,

                              AWAPB> Thanks again for your interesting reaction and the link it contents.

                              AWAPB> My problem: it is more than 50 years ago that I left
                              AWAPB> the secundary school, I have to study the log questions and
                              AWAPB> mantissas thoroughly.

                              AWAPB> My solution: not so long ago I found a formula for
                              AWAPB> finding end figures of roots.
                              AWAPB> BN = basic number, in this case 11, I speak about the
                              AWAPB> second problem in your link, the 13 th rot of 75185.....85831.
                              AWAPB> N= exponent, or power, in this case 13.
                              AWAPB> Jump: the number which with the end figures change.
                              AWAPB> 11^13 end on 3931, the problem ends on 3931.
                              AWAPB> My formula - on which I am a little proud- is

                              AWAPB> Jump = BN^(n-1) × power × 100. In this case:

                              AWAPB> 11^12×13×100= 721×13×100= ....7300.

                              AWAPB> Now we subtract 5831-3931= 1900. The question is yet
                              AWAPB> how many ...7300's go in ...1900. It is 3, so the answer ends
                              AWAPB> on 0311.

                              AWAPB> Regards,

                              AWAPB> Willem Bouman

                              Sincerely Yours, Oleg Stepanov.
                              http://stepanov.lk.net/





                              [Non-text portions of this message have been removed]
                            • issam khneisser
                              Dear mathalthletes,   I had notice a great issue regarding 8 digit, my point of view   it had been agreed   8 digit x 8 digit mulltiplication even you can
                              Message 14 of 17 , Jun 22, 2008
                              • 0 Attachment
                                Dear mathalthletes,
                                 
                                I had notice a great issue regarding 8 digit, my point of view
                                 
                                it had been agreed
                                 
                                8 digit x 8 digit mulltiplication even you can do more than 8.
                                 
                                for square root, 8 digit answer
                                 
                                it may had been thought a root of 100 digit number
                                therefore they had choosen the 13 root, because it have 8 digit answer and
                                As ralf explanined once, there a little tricks regarding certain numbers in 13th root, so only look for the few first both end number and you can give a quick the answer, therefore
                                Alex and the big GERT during a show when more than 100 attempt had been displayed,
                                in one of their attempts they had had done that magic of less than 3 second answer for 13th root. which led to don t accept anymore one attempt shot for any mental calculation operation since 2004.
                                At that time it was for my inconvinience because i was preparing to break the 44.7 s gert metring record for square root. I remember while practicing once i had done like a square in 12.7 second in july 2004, for 20 % of number i can go for less than than 44.7 s and in others i go for between 80 to 90 s.
                                 
                                See you guys next weeks and have a great sunday, I hope that germany will be in final of euro 2008 so we will have a great gathering sunday night.
                                 
                                Final world, we need ALL to thank RALF for all his efforts for organizing the world championship for mental calcualtion, Big THANKS RALF and see you next week.
                                 
                                regards
                                Issam
                                 
                                 
                                 
                                 
                                 
                                 
                                 
                                 


                                --- On Sun, 6/15/08, Andy Robertshaw <robertshaw_andy@...> wrote:

                                From: Andy Robertshaw <robertshaw_andy@...>
                                Subject: Re: [Mental Calculation] roots
                                To: MentalCalculation@yahoogroups.com
                                Date: Sunday, June 15, 2008, 2:22 PM






                                Hi Willem,
                                I think one reason for this is because prime roots are more difficult than composite roots.
                                For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
                                So we've got the 7th, 11th, 13th and 17th roots to contend with.
                                I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
                                But certainly the 7th and 11th roots would be tricky enough!
                                Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
                                Andy

                                ----- Original Message ----
                                From: A.W.A.P. Bouman <awap.bouman@ casema.nl>
                                To: mental calculation <MentalCalculation@ yahoogroups. com>
                                Sent: Sunday, 15 June, 2008 10:40:11 AM
                                Subject: [Mental Calculation] roots

                                Dear Calculator friends,

                                What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

                                Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

                                And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

                                But why only 13th??

                                Regards,

                                Willem Bouman

                                [Non-text portions of this message have been removed]

                                ____________ _________ _________ _________ _________ _________ _
                                Sent from Yahoo! Mail.
                                A Smarter Email http://uk.docs yahoo.com/ nowyoucan. html

                                [Non-text portions of this message have been removed]


















                                [Non-text portions of this message have been removed]
                              Your message has been successfully submitted and would be delivered to recipients shortly.