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Math Help - 10 lucky digits

  1. #1
    Junior Member samsum's Avatar
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    10 lucky digits

    find the 10 digits of 3^2007.

    how many 4-digit numbers whose digits are all odd are multiples of 11?
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    Quote Originally Posted by samsum View Post
    find the 10 digits of 3^2007.
    I will use Congruences and Euler's Generalization of Fermat's Little Theorem..

    We note that,
    \gcd(3,100)=1

    And, \phi(100)=100\left(1-\frac{1}{2} \right)\left(1-\frac{1}{5}\right)=100(1/2)(4/5)=40

    Thus,
    3^{40}\equiv 1(\mbox{mod }100)

    Bring both sides to the power of 50 thus,
    3^{2000}\equiv 1(\mbox{mod }100)
    Note that,
    3^7\equiv 87(\mbox{mod }100)
    Thus, their multiplication yields,
    3^{2007}\equiv 87(\mbox{mod }100)

    Since it is congruent to 100 it shows the last two digits. The second to last digits it 8.
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    Quote Originally Posted by samsum View Post
    how many 4-digit numbers whose digits are all odd are multiples of 11?
    You have a 4 digit number,
    ABCD with A\not = 0
    und,
    0\leq A,B,C,D\leq 9

    The necessary and suffient condition for the divisibility of 11 to exist we need that the Alternating sum is a divisible by 11. Thus,
    D-C+B-A=0 \mbox{ or } 11 \mbox{ or } -11
    Thus,
    D+B=A+C
    D+B=A+C+11
    11+D+B=A+C
    Basic trial and error will produce all solutions.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by samsum View Post
    how many 4-digit numbers whose digits are all odd are multiples of 11?
    The smallest 4 digit number divisible by 11 is 91*11, and the largest
    4 digit number divisible by 11 is 909*11. Therefore there are 819 such
    numbers divisible by 11. Of these (819-1)/2 are even and the rest are odd
    so there are 410 odd numbers 4 digit divisible by 11.

    RonL
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