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Thread: Last 5 digits

  1. #1
    Nov 2008

    Last 5 digits

    find the final 5 digits of,

    $\displaystyle 9^{9^{9^{^{.^{.^{.^{.^{9}}}}}}}}$ that is 1001 nines
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    A computing approach

    Define $\displaystyle f(x)=9^x$. So $\displaystyle f^2(x)=f(f(x))=9^{9^x}$ and so on. Phrased this way, we are asked to calculate $\displaystyle f^{1000}(9) (\bmod 100000)$. By nature of modular arithmetic, it does not matter if we calculate $\displaystyle f^{1000}(9)$ first, then reduce it modulo 100000, or if we calculate f(9), reduce it, then calculate f of the reduced answer, and so on. In other words, we can iterate via:

    $\displaystyle x_0=9, x_{n+1}\equiv 9^{x_n} (\bmod 100000)$

    Iterating 1000 times, reducing the answer each time. Setting this ball in motion, we get:

    $\displaystyle 9\to 20489 \to 77289 \to 65289 \to 45289 \to 45289 \to 45289 ...$

    Here we notice that $\displaystyle f^5(9)=f^6(9)$ and therefore $\displaystyle f^n(9)=f^5(9)=45289$ for all $\displaystyle n>5$. In other words, we found a fixed point in the function f and need go no further. Namely, $\displaystyle 9^{45289}\equiv 45289 (\bmod 100000)$. Thus we have found our answer.

    This seems like an odd coincidence, and begs the general question: Given a,b, for what value of x does $\displaystyle a^x\equiv x (\bmod b)$, and for what values a,b does no such x exist?
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