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I am working on a practice problem right now for my number theory class, where I have to find $101^{{101}^{99}} \pmod{7}$. My professor gave us the hint to reduce the first exponent, $101^{99}$, modulo 6. Which is what I did, and found that $101^{99} \pmod{6}\equiv -1$.

From here, I'm not sure where to go. I've been told to use Fermat's Little Theorem, but I feel like I am just not seeing how this is all tying together, or why reducing $101^{99} \pmod{6}$ is useful to me.

Any help is appreciated!

Bill Dubuque
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Kirsten
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  • Re your final query: Reducing $101^{99}$ modulo $6$ is useful because for any number $a$ coprime to $7$, $a^6 \equiv 1 \bmod 7$. Thus, if we can remove as many factors of $6$ as possible from the exponent $101^{99}$, we can use the remainder $b$ as a proxy for the original large exponent. All of the numerous occurrences of $101^6$ implicit in the large exponent simply yield factors of $1$ and can be ignored. – Keith Backman Oct 15 '23 at 17:18

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$$3^6\equiv 1 \mod 7 \text{ (Fermat's Little Theorem)}$$

$$101\equiv 5 \equiv -1 \mod 6$$

$$101^{99} \equiv (-1)^{99}\equiv -1\equiv 5 \mod 6$$

$$101^{101^{99}}\equiv 101^5 \equiv 3^5 \equiv 5 \mod 7$$

zxen
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Lion Heart
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    Yes, that's correct. We can also observe that $101^{{101}^{(2 \cdot m-1)}} \equiv 5 \pmod 7, \forall m \in \mathbb{Z}^+$ (since $99$ is odd and $99-2 \cdot k$ is also odd for $k=1,2,3,\dots,49$). – Marco Ripà Oct 15 '23 at 16:28
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 15 '23 at 17:27