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Math Help - Japanese Samurai and his soldiers after a battle

  1. #1
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    Japanese Samurai and his soldiers after a battle

    A Japanese Samurai counted his soldiers remaining after a battle, by lining them in rows of different lengths, counting the number left over each time, and calculating the total from these remainders. A Samurai had 1200 soldiers at the start of a battle. After the battle, there were 3 left over when they lined up 5 at a time, 3 left over when they lined up 6 at a time, 1 left over when they lined up 7 at a time, and none left over when they lined up 11 at a time. How many soldiers were lost in the battle?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by dkmathguy View Post
    A Japanese Samurai counted his soldiers remaining after a battle, by lining them in rows of different lengths, counting the number left over each time, and calculating the total from these remainders. A Samurai had 1200 soldiers at the start of a battle. After the battle, there were 3 left over when they lined up 5 at a time, 3 left over when they lined up 6 at a time, 1 left over when they lined up 7 at a time, and none left over when they lined up 11 at a time. How many soldiers were lost in the battle?
    x\equiv 3\ (mod\ 5)

    x=5a+3

    -------------------------------------------------------------

    x\equiv 3\ (mod\ 6)

    5a+3\equiv 3\ (mod\ 6)

    5a\equiv 0\ (mod\ 6)

    a\equiv 0\ (mod\ 6)

    a=6b

    x=5a+3=5(6b)+3=30b+3

    -------------------------------------------------------------

    x\equiv 1\ (mod\ 7)

    30b+3\equiv 1\ (mod\ 7)

    30b\equiv -2\ (mod\ 7)

    15b\equiv -1\ (mod\ 7)

    b\equiv -1\ (mod\ 7)

    b=7c-1

    x=30b+3=30(7c-1)+3=210c-27

    -------------------------------------------------------------

    x\equiv 0\ (mod\ 11)

    210c-27\equiv 0\ (mod\ 11)

    210c\equiv 27\ (mod\ 11)

    c\equiv 5\ (mod\ 11)

    c=11d+5

    x=210c-27=210(11d+5)-27=2310d+1023

    -------------------------------------------------------------

    Clearly, d=0

    So, x=1023

    No. of soldiers lost in the battle =1200-1023=177
    Last edited by alexmahone; April 30th 2011 at 12:37 AM.
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  3. #3
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    alexmahone, you are correct. Thanks, I never saw you could do negative too, lol. I did the hard way by making the -1 to a 6.

    15b = -1 (mod 7)

    b = -1 (mod 7)


    15b = 6 (mod 7)

    b = 6 (mod 7)
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  4. #4
    Super Member Quacky's Avatar
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    That is one awesome way of counting.
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  5. #5
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    Can be done by simple reasoning:

    a = alive: we want a / 11 = integer; so maximum a = 1199 ; then 1188, 1177, .....

    we want a / 6 to have remainder of 3, so a is odd ; so a = 1199, 1177, 1155, 1133, ....

    we want a / 5 to have remainder of 3, so a must end with 3 (being odd), so a = 1133, 1023, 913, ...

    we want a / 7 to have remainder of 1:
    1133 / 7 = 161 remainder 6
    1023 / 7 = 146 remainder 1 : YA!

    So deads = 1200 - 1023 = 177

    Much quicker (and easier) is find minimum that accomodates 5,6,11,
    which of course is 33; this number must go up by 330 (to keep accomodating 5,6,11),
    so: 33, 363, 693, 1023; only 1023 has remainder of 1 if divided by 7.
    Last edited by Wilmer; May 1st 2011 at 10:54 PM.
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