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Find the last two digits of \(9^{8^7}\)

 Oct 21, 2021
 #1
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By Euler's Theorem, $9^{8^7} \equiv 27 \pmod{100}$.

 Oct 22, 2021
 #2
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(9^8^7) mod 10^10 ==1626926081 - these are the last 10 digits.

 Oct 22, 2021
 #3
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Thank you so much!!

 Oct 22, 2021

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