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I know that I have to utilize (mod 100) for the problem.

Getting use of Euler's theorem as $(139, 100) = 1$ yields

$$139^{{139}^{100}} \equiv 139^{{139}^{100} \pmod{\Phi(100)}} \pmod{100}$$

$$ \equiv 139^{{139}^{100} \pmod{40}} \pmod{100}.$$

How to proceed from this point?

curiousMind_85
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  • Use the same idea to determine $139^{100}\equiv 19^{100} \mod 40$ – Pomponazzo Dec 06 '21 at 14:41
  • And, since $\phi(40)=16$, $$19^{100}\equiv19^4\cdot(19^{16})^6\equiv19^4\cdot1\bmod{40}$$ – Rushabh Mehta Dec 06 '21 at 14:50
  • $19^{100} = (20-1)^{100}$. If you expand by binomial theorem, every term is divisible by $40$ except the last one, which equals $1$. – B. Goddard Dec 06 '21 at 14:52
  • I'd personally consider it mod $8$ and $5$ and use Chinese remainder theorem. – Jakobian Dec 06 '21 at 14:52
  • Or use $19^2\equiv1$. – J.G. Dec 06 '21 at 14:53
  • $\bmod 10!:\ k:=139^{100}\equiv (-1)^{100}\equiv 1,,$ so $,\color{#c00}{k= 1+10j},$ so $!\begin{align}\bmod 100!:\ 139^k\equiv (-1!+!40)^k\equiv&, (-1)^k + (-1)^{k-1} 40\color{#c00} k\ \ {\rm by}\ \ 40^2\equiv 0\ \ &\ \ \rm Binomial\ Thm\ \equiv &\ {-}1 + 40(\color{#c00}{1+10j})\equiv -1+40 \end{align}$ – Bill Dubuque Dec 06 '21 at 15:24
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    $\begin{align} &\bmod 8!:\ 139\equiv\ 3,\ \ \ &\ \ \ \ \ \ \ 3^2,\equiv\ 1\ & \bmod 5!:\ 139\equiv -1\ \ &\ \ (-1)^2\equiv 1\end{align}$ so $,139^2\equiv 1\bmod 5,8,$ so also $!\bmod 40,$ by CCRT – Bill Dubuque Dec 06 '21 at 15:36

2 Answers2

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You want to evaluate $139^{100}\bmod{40}$. Note that $139\equiv19\bmod{40}$.

Finally, since $19=20-1$, $19^2=20^2-2\cdot20+1$,

$19^2\equiv1\bmod{40}$.

Thus, $$139^{19^{100}}\equiv139^1\equiv39\bmod{100}$$

Rushabh Mehta
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Lemma : $N^{20k} \equiv 1 \bmod 100$ for any natural number N that is coprime to 100 and any natural number k.

This is because $\lambda(100) = 20$. (See Carmichael Function) So we only have to find $139^{100} \bmod 20$

$$139 \equiv -1 \bmod 20 \implies 139^{100} \equiv -1^{100} \equiv 1 \bmod 20$$ Therefore $$139^{139^{100}} \equiv 139^{20k+1} \equiv 139 \equiv 39 \bmod 100$$

sato
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