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Math Help - Congruent numbers

  1. #1
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    Post Congruent numbers

    what is the last 2 digits of 3 raised to 1994 and 7 raised to 1994??

    Plz help me this is urgent
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  2. #2
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    Hello, xXxSANJIxXx!

    I found two different methods for these problems . . .


    What are the last 2 digits of 3^{1994} ?

    Since 3^2)^{997} \:=\:9^{997} \:=\10 - 1)^{997}" alt="3^{1994} \:=\3^2)^{997} \:=\:9^{997} \:=\10 - 1)^{997}" />, consider the binomial expansion:

    . . 10^{997} - \binom{997}{996}10^{996} + \binom{997}{995}10^{995} - \hdots  +  \binom{997}{3}10^3 - \binom{997}{2}10^2 + \binom{997}{1}10 - 1
    . . . . \underbrace{\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad}
    . . . . . . . . . These do not affect the last two digits

    Therefore, the last two digits are: . \binom{997}{1}10 - 1\:=\:997\cdot10 - 1\:=\:99\boxed{69}



    What are the last 2 digits of 7^{1994} ?

    We note that: 7^4 = 2401
    . . Hence: (7^4)(7^4)(7^4)\cdots(7^4) will end in 01.

    Since 1994 \,=\,4\cdot498 + 2, we have: . 7^{1994} \:=\:7^{(4\cdot498 + 2)} \;= \;7^{4\cdot498}\cdot7^2

    . . . . . =\;\underbrace{(7^4)^{498}}\cdot\,49\quad \Longrightarrow\quad \text{a number ending in }49
    . . (a number ending in 01)

    Therefore, the last two digits of 7^{1994} are: \boxed{49}

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  3. #3
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    Quote Originally Posted by xXxSANJIxXx
    what is the last 2 digits of 3 raised to 1994 and 7 raised to 1994??

    Plz help me this is urgent
    I have a different approach. It involves the phi function.
    Since,
    \gcd (3,100)=1 you have,
    3^{\phi(100)}\equiv 1 (\mbox{mod } 10)
    To find, \phi(100) prime factorize 100,
    100=2^4\cdot 5^2
    Thus,
    \phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5} \right)=\frac{100\cdot 1\cdot 4}{2\cdot 5}=40
    Thus,
    3^{40}\equiv 1 (\mbox{mod }100)
    Raise both sides to the 50th power,
    3^{2000}\equiv 1 (\mbox{mod }100)
    Express as,
    3^{1994}\cdot 3^6 \equiv 1 (\mbox{mod }100)
    Let, x=3^{1994} then,
    3^6x\equiv 1(\mbox{mod }100)
    729x\equiv 1(\mbox{mod }100)
    Solve for x,
    (729 - 700)x\equiv 1 (\mbox{mod }100)
    29x\equiv 1(\mbox{mod }100)
    x\equiv 69
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  4. #4
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    Question THx

    Thank you guys... i kinda get it....
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