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ganeshie8

  • one year ago

show that 12345678987654321 ≡ 0 (mod 12345679)

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  1. UsukiDoll
    • one year ago
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    so we're in modulo 12345679 ... and we need a zero remainder.. oh wow. x.x

  2. UsukiDoll
    • one year ago
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    like how many cycles of 12345679 do we have to go through to reach 12345678987654321

  3. ganeshie8
    • one year ago
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    Yes im sure it has a pretty neat solution :)

  4. imqwerty
    • one year ago
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    i've done this problem before :D

  5. ParthKohli
    • one year ago
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    Facebook pages have taught me that \(111111111^2 = 12345678987654321\)

  6. UsukiDoll
    • one year ago
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    -_-! This is from the same book that I've used last year. I'm not sure if I made my professor solve this one lol

  7. ganeshie8
    • one year ago
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    Haha what has that anything to do with the present problem

  8. ParthKohli
    • one year ago
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    The missing 8 in 12345679 is mildly annoying.

  9. UsukiDoll
    • one year ago
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    that's number 7b. page 52 x)

  10. UsukiDoll
    • one year ago
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    personally , I like the previous problems on page 51 XD

  11. ParthKohli
    • one year ago
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    \[12345679 \times 10^9 = 12345679000000000\]subtract 12345679 from this number. Woohoo.

  12. ParthKohli
    • one year ago
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    I know that it's ugly, but what's a better way to show that something divides something other than actually finding their ratio? :P

  13. anonymous
    • one year ago
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    12345678987654321=12345679987654321-1000000000 =12345679000000000+987654321-1000000000 the first term divided by 12345679 has remainder =0 And -1000000000+987654321=-12345679 divided by 12345679 remain 0 hence 12345678987654321\(\equiv\) 0 (mod 12345679)

  14. zzr0ck3r
    • one year ago
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    12345678987654321=999999999*12345679

  15. zzr0ck3r
    • one year ago
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    qed

  16. ikram002p
    • one year ago
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    was thinking of that the moment i saw it @zzr0ck3r :P

  17. ganeshie8
    • one year ago
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    that factorization is pretty @ParthKohli / @OOOPS method is really clever!

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