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Thread: 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 $\displaystyle 3^{1994}$ ?

    Since $\displaystyle 3^{1994} \:=\3^2)^{997} \:=\:9^{997} \:=\10 - 1)^{997}$, consider the binomial expansion:

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

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



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

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

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

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

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

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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,
    $\displaystyle \gcd (3,100)=1$ you have,
    $\displaystyle 3^{\phi(100)}\equiv 1 (\mbox{mod } 10)$
    To find, $\displaystyle \phi(100)$ prime factorize 100,
    $\displaystyle 100=2^4\cdot 5^2$
    Thus,
    $\displaystyle \phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5} \right)=\frac{100\cdot 1\cdot 4}{2\cdot 5}=40$
    Thus,
    $\displaystyle 3^{40}\equiv 1 (\mbox{mod }100)$
    Raise both sides to the 50th power,
    $\displaystyle 3^{2000}\equiv 1 (\mbox{mod }100)$
    Express as,
    $\displaystyle 3^{1994}\cdot 3^6 \equiv 1 (\mbox{mod }100)$
    Let, $\displaystyle x=3^{1994}$ then,
    $\displaystyle 3^6x\equiv 1(\mbox{mod }100)$
    $\displaystyle 729x\equiv 1(\mbox{mod }100)$
    Solve for $\displaystyle x$,
    $\displaystyle (729 - 700)x\equiv 1 (\mbox{mod }100)$
    $\displaystyle 29x\equiv 1(\mbox{mod }100)$
    $\displaystyle x\equiv 69$
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  4. #4
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    Question THx

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