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Thread: Big power

  1. #1
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    Big power

    I am trying to find a way to obtain the last 3 digits of k,
    where k = 4321^(10^4321 + 1).

    I can see that 10^4321 + 1 = log(k) / log(4321),
    or log(k) / log(4321) - 10^4321 = 1

    However, numbers are so huge here that I think the use of
    some trick or theory is required.

    Can anyone help. Thank you.
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  2. #2
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    Even though $\displaystyle 10^{4321}+1$ is a huge number, it still ends in 001.

    Therefore, if I am not mistaken, $\displaystyle 4321^{10^{4321}+1}$ will end in 321.
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  3. #3
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    Hello, Wilmer!

    Find the last 3 digits of $\displaystyle k$, where: .$\displaystyle k \:= \:4321^{10^{4321} + 1}$
    I used a very primtive approach . . .

    Since we are concerned with the last three digits only,
    . . we can raise $\displaystyle 321$ to various powers.

    I found that: .$\displaystyle 321^{25}$ ends in $\displaystyle 001.$

    That is, consecutive powers of 321 end in: .$\displaystyle 321,\,041,\,161,\,681,\,601,\,\cdots \,001$
    . . It goes through a 25-step cycle and starts over.


    The power $\displaystyle 10^{4321} + 1 \:=\:10^2\!\cdot\!10^{4319} + 1 \:=\:25\!\cdot\!4\!\cdot\!10^{4319} + 1$

    $\displaystyle \text{Then we have: }\;321^{25\cdot4\cdot10^{4319} + 1} \:=\;321^{25\cdot4\cdot10^{4319}}\cdot321^1 \:=\:\underbrace{(321^{25})}_{\text{This ends in 001}}\,\!\!\!\!\!\!\!^{4\cdot10^{4319}} \cdot321$

    So we have: .$\displaystyle (001)^{4\cdot10^{4319}}\cdot321 \quad\Rightarrow\quad (001)\cdot 321 \quad\Rightarrow\quad321$

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  4. #4
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    What we want is $\displaystyle 4321^(10^(4321)+1) mod 1000$

    $\displaystyle 4321^{10^{4321}+1} \equiv 321^{10^{4321}+1}$

    We need to use Euler's totient theorem $\displaystyle a^{\phi(n)}\equiv1$ mod n

    $\displaystyle \phi(1000)$ is the number of numbers smaller than 1000 and coprime with 1000.
    1000 = $\displaystyle 2^35^3$
    so we just need to count the number of multiples of 2 and 5 less than 1000. There are 1000/2 = 500 even numbers, 1000/5 = 200 multiples of 5 and 1000/10=100 multiples of 2 and 5.
    This gives us $\displaystyle \phi(1000)$ = 1000-500-200+100 = 400

    so we have
    $\displaystyle 321^{10^{4321}+1} \equiv 321^{10^{4321}+1 \pmod{400}}$
    but 400|$\displaystyle 10^4321$
    so $\displaystyle 321^{10^{4321}+1} \equiv 321^1$ =321
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  5. #5
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    Thank you very much everyone.
    I sure learned a lot.
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