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Find the last $2$ digits of $$\large7^{7^{7^{10217}}}$$

So far I have: $17\cdot 601=10217$ and $7=7 \pmod{10}$. Any help greatly appreciated.

Ángel Mario Gallegos
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Rick Owens
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  • What's the question again? The title and the body are somewhat related, but ... Also, do you mean the iterated power? Won't the answers to the Related questions on the sidebar not settle this as well? – Jyrki Lahtonen Jan 28 '16 at 22:05
  • Welcome to Math.SE, but to get useful answers you also need do your part and make a clear question as well as search before asking. – Jyrki Lahtonen Jan 28 '16 at 22:06
  • Also, read this, or this or any of the other scores of questions where the technique is explained. – Jyrki Lahtonen Jan 28 '16 at 22:12
  • See http://math.stackexchange.com/questions/1631143/find-last-two-digits-using-modular-arithmetic – user236182 Jan 28 '16 at 22:45

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Note that $100=4\cdot 25$. We can find $7^{7^{7^{10217}}}$ mod $4$, $25$, then combine this to get the value mod $100$ using the Chinese Remainder Theorem.

$7^4\equiv \left(7^2\right)^2\equiv (-1)^2\equiv 1\pmod{25}$

Therefore $$7^{7^{7^{10217}}}\equiv 7^{7^{7^{10217}}\pmod 4}\equiv 7^{(-1)^{7^{10217}}\pmod 4}\pmod{25}$$

$$\equiv 7^{-1\pmod{4}}\equiv 7^{3}\equiv \left(7^2\right)\cdot 7\equiv (-1)\cdot 7\equiv 18\pmod{25}$$

Also

$$7^{7^{7^{10217}}}\equiv (-1)^{7^{7^{10217}}}\equiv -1\equiv 3\pmod{4}$$

Using the Chinese Remainder theorem, $7^{7^{7^{10217}}}\equiv 43\pmod{100}$.

user236182
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