I'm continuing on my journey learning about modular arithmetic and got confused with this question:

Find the last two digits of $9^{9^{9^{…{^9}}}}$ (nine 9s). The phi function is supposed to be used in this problem and so far this is what I've got:

$9^{9^{9^{…{^9}}}} ≡ x (\text{mod } 100)$ Where $0 ≤ x ≤ 100$

$9^{9^{9^{…{^9}}}} \text{ (nine 9s) }= 9^a$ In order to know $9^a (\text{mod } 100)$, we need to know $a (\text{mod } \phi(100))$ As $\phi(100)= 40$, we get $a = b (\text{mod } 40)$

$9^{9^{9^{…{^9}}}} \text{ (eight 9s) }= 9^b$ In order to know $9^b (\text{mod } 40)$, we need to know $b (\text{mod } \phi(40))$ As $\phi(40)= 16$, we get $b = c ( \text{mod }16)$

$9^{9^{9^{…{^9}}}}\text{ (seven 9s) }= 9^c $ In order to know $9^c (mod 16)$, we need to know $c (\text{mod } phi(16))$ as $\phi(16)= 8 $ we need to find $c (\text{mod } 8)$

As $9 = 1 (\text{mod } 8)$ $c = 1 (\text{mod } 8)$

I feel like I might have made a mistake somewhere along the way because I'm having a lot of trouble stitching it all back together in order to get a value for the last two digits. Could anyone please help me with this? Thank you!


Hint. Consider $a_n=9^{9^{2n+1}}$ and show by induction that $a_n\equiv 89\pmod{100}$ for $n\in\mathbb{N}$.

By the binomial theorem, we have that $$a_0=9^9=(10-1)^9\equiv 9\cdot 10^1 (-1)^8 +(-1)^9=90-1=89\pmod{100}.$$ Moreover, for $n\geq 0$, $$a_{n+1}=9^{81\cdot 9^{2n+1}}=a_n^{81}\equiv 89^{81}\equiv ?\pmod{100}$$ Can you take it from here? (you already noted that $\varphi(100)=40$).

  • $\begingroup$ Got the answer! Thank you very much! $\endgroup$ – Kami May 5 '19 at 14:47

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