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    23 Apr '05 03:27
    Find a number divisible by 41 whose digits are all nines. Without doing the division find the remainder when
    90 000 800 007 000 060 000 500 004 000 030 000 200 001 is divided by 41.
  2. back in business
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    25 Apr '05 05:38
    Originally posted by elopawn
    Find a number divisible by 41 whose digits are all nines. Without doing the division find the remainder when
    90 000 800 007 000 060 000 500 004 000 030 000 200 001 is divided by 41.
    oh no..i have only ten fingers
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    25 Apr '05 07:032 edits
    😵😵😵😵

    i think i got it
  4. Joined
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    25 Apr '05 12:161 edit
    Originally posted by Jusuh
    oh no..i have only ten fingers
    Need a clue?

    Think about dividing 100, 1000, 10000, 100000 by 41 and the remainders of them. Eg they are 18, 16, 37 and 1.

    How does this relate to the question?
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    25 Apr '05 17:57
    Originally posted by elopawn
    Find a number divisible by 41 whose digits are all nines. .
    FLT


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    25 Apr '05 21:35
    oh, i got a headache.🙄
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    26 Apr '05 05:24
    9999999999
    243902439
  8. Joined
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    26 Apr '05 07:55
    Originally posted by THUDandBLUNDER
    FLT


    What's FLT stand for?

    Anyway I'll post the answer if no one gets it in a day.
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    26 Apr '05 09:522 edits
    Originally posted by elopawn
    What's FLT stand for?
    Fermat's Little Theorem:

    http://mathworld.wolfram.com/FermatsLittleTheorem.html
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    26 Apr '05 10:311 edit
    Fermat's Little Throrem? The problem does not relate to it.
    I'll post the answer now if you want it, it actually after you see it is rather simple.
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    26 Apr '05 10:401 edit
    Ok so we know after dividing bu 100, 1000, 10000, 100000 by 41 we get-- 18, 16, 37, 1

    Thus 99999 is divisible by 41.

    Let x=90 000 800 007 000 060 000 500 004 000 030 000 200 001

    =9* 10power40 + 8* 10power35 +7* 10power30 + 6* 10power25 + 5* 10power20 + 4* 10power15 + 3* 10power10 + 2* 10power5 + 1

    From the above, 10power5 leaves a remainder of 1 when divided by 41.

    Thus the remainders of each term of the sum are 9,8,7,6,5,4,3,2,1 !!

    The sum of the remainders is 45 and so the remainder when x is divided by 41 is 4 !!

  12. Joined
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    26 Apr '05 11:462 edits
    Originally posted by phgao
    Fermat's Little Throrem? The problem does not relate to it.
    a^(p-1) - 1 = 0 (mod p) where a is a positive integer and p is a prime.

    In your problem, a = 10 and p = 41

    Therefore the number 10^40 - 1 is divisble by 41

    As you didn't ask for the smallest such number, this answers your first question rather more quickly than your trial-and-error method.
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    26 Apr '05 13:19
    Originally posted by THUDandBLUNDER
    a^(p-1) - 1 = 0 (mod p) where a is a positive integer and p is a prime.

    In your problem, a = 10 and p = 41

    Therefore the number 10^40 - 1 is divisble by 41

    As you didn't ask for the smallest such number, this answers your first question rather more quickly than your trial-and-error method.
    Oh thanks for that insight T&B! I appreciate it.
  14. Joined
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    26 Apr '05 15:42
    Originally posted by phgao
    Oh thanks for that insight T&B! I appreciate it.
    Sorry, I thought I was replying to the original poster (elopawn).
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