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MC weekend

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  • A.W.A.P. Bouman
    Dear Calculators, For the 8 May weekend I organised a MC weekend. My esteemed guests were Dr.Dr. Gert Mittring, Jan van Koningsveld and Dr. Andrew Robertshaw,
    Message 1 of 3 , May 15, 2009
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      Dear Calculators,

      For the 8 May weekend I organised a MC weekend. My esteemed guests were Dr.Dr. Gert Mittring, Jan van Koningsveld and Dr. Andrew Robertshaw, quality street so to say.

      We did a lot of exercizes, and inspired by the George Lane file I proposed a big division with a remainder. The challenge was to find the remainder without first finding the quotient.

      The dividend I choosed randomly, the divisor could be factorised. I had the silent hope that modulo calculation could help us, unknowing how.

      121068 = 2²×3³×19×59.

      139875219643 ÷ 121068.

      It was Andy who ran the gauntlet. He immediately calculated that the dividend = 3(2²), 13(3³), 8(19) and 37(59). After that he explained us the chinese remainder theorem, eventually to be found at internet.

      First step: to find a number which meets 3(2²) and 13(3³) . 3= 3(3³) so for 13 (27) we miss 10(27). Taking 2² × 4 we have the missing 10 but we are not 3(2²). Taking 13½×4 we are (0)27 and 2(2²) so then we have reached our first goal: 67 which meets the requirements.

      Now from starting point 67 we have to find a number 67(108) and 8(19).
      67=10(19), 108 = 13(19) and number ?=8(19) . So how many times we have to add 13(19) from 10(19) to get 8(19). In fact we miss 17(19). Adding 13-26-39 etc we arrive at 169 (13×13) then we have 17(19).
      So by adding 67 + (13×108) we arrive at 1471 which - indeed- meets the third requirement.

      Now the 37(59).
      We muliply 2²×3³×19 = 2052= 46(59) and have to add this so many times till we have a number which is 37(59).
      1471= 55(59).
      So 55(59) + ??×46(59) = 37(59). So we miss 41(59).
      Lady luck gave us 9×46= 1(59). So now we miss 9×41=15(59).

      Finally: when we take 15×2052 and add this at 1471 we have reached our final goal: 32251 which, a check learns this to us indeed meets all the requirements.

      I think this question is too big for a tournament, a smaller one would to my opinion be an enrichment in a tournament.
      What I fear - being no mathematician but a simple arithmetician - is if this type of questions could be answerd with a prime number as a divider.

      Nevertheless; it has put me on fire and I'll train every day 1 type of this question. The answer is always a smaller number than the product of the numbers, in this case 121068.

      Kindest regards,


      Willem Bouman

















      [Non-text portions of this message have been removed]
    • gururaj manchali
      Respected Mr Bouman I did not understand what is this chinese remainder theorm but I am very much interested to learn the same. Can you give 1 or 2 simple
      Message 2 of 3 , May 30, 2009
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        Respected Mr Bouman
        I did not understand what is this chinese remainder theorm but I am very much interested to learn the same. Can you give 1 or 2 simple examples like the one you have shown .It helps us a lot to know whether there will be a remainder in the answer when we are deviding by big numbers .Pl mail one or two examples describing the steps to be followed.Pl oblige.You can send the mail to gmanchali@... or manchaali@... I also have a method to findout but it is different from what you have shown.

        --- On Fri, 5/15/09, A.W.A.P. Bouman <awap.bouman@...> wrote:


        From: A.W.A.P. Bouman <awap.bouman@...>
        Subject: [Mental Calculation] MC weekend
        To: "mental calculation" <MentalCalculation@yahoogroups.com>
        Date: Friday, May 15, 2009, 4:10 PM








        Dear Calculators,

        For the 8 May weekend I organised a MC weekend. My esteemed guests were Dr.Dr. Gert Mittring, Jan van Koningsveld and Dr. Andrew Robertshaw, quality street so to say.

        We did a lot of exercizes, and inspired by the George Lane file I proposed a big division with a remainder. The challenge was to find the remainder without first finding the quotient.

        The dividend I choosed randomly, the divisor could be factorised. I had the silent hope that modulo calculation could help us, unknowing how.

        121068 = 2²×3³×19×59.

        139875219643 ÷ 121068.

        It was Andy who ran the gauntlet. He immediately calculated that the dividend = 3(2²), 13(3³), 8(19) and 37(59). After that he explained us the chinese remainder theorem, eventually to be found at internet.

        First step: to find a number which meets 3(2²) and 13(3³) . 3= 3(3³) so for 13 (27) we miss 10(27). Taking 2² × 4 we have the missing 10 but we are not 3(2²). Taking 13½×4 we are (0)27 and 2(2²) so then we have reached our first goal: 67 which meets the requirements.

        Now from starting point 67 we have to find a number 67(108) and 8(19).
        67=10(19), 108 = 13(19) and number ?=8(19) . So how many times we have to add 13(19) from 10(19) to get 8(19). In fact we miss 17(19). Adding 13-26-39 etc we arrive at 169 (13×13) then we have 17(19).
        So by adding 67 + (13×108) we arrive at 1471 which - indeed- meets the third requirement.

        Now the 37(59).
        We muliply 2²×3³×19 = 2052= 46(59) and have to add this so many times till we have a number which is 37(59).
        1471= 55(59).
        So 55(59) + ??×46(59) = 37(59). So we miss 41(59).
        Lady luck gave us 9×46= 1(59). So now we miss 9×41=15(59).

        Finally: when we take 15×2052 and add this at 1471 we have reached our final goal: 32251 which, a check learns this to us indeed meets all the requirements.

        I think this question is too big for a tournament, a smaller one would to my opinion be an enrichment in a tournament.
        What I fear - being no mathematician but a simple arithmetician - is if this type of questions could be answerd with a prime number as a divider.

        Nevertheless; it has put me on fire and I'll train every day 1 type of this question. The answer is always a smaller number than the product of the numbers, in this case 121068.

        Kindest regards,

        Willem Bouman

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



















        [Non-text portions of this message have been removed]
      • ralf_laue
        ... Before asking other people for help, I would recommend to do your homework: Option 1: Lookup Chinese Remainder Theorem at Wikipedia Option 2: Search for
        Message 3 of 3 , May 30, 2009
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          --- In MentalCalculation@yahoogroups.com, gururaj manchali <gmanchali@...> wrote:
          >
          > Respected Mr Bouman
          > I did not understand what is this chinese remainder theorm but I am very much interested to learn the same.

          Before asking other people for help, I would recommend to do your homework:

          Option 1: Lookup "Chinese Remainder Theorem" at Wikipedia
          Option 2: Search for "Chinese Remainder Theorem" at Google
          Option 3: Search the archive of this group for "Chinese Remainder Theorem" (where you will find my message #1815 with a good link)

          Ralf Laue
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