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The original question is: $3^{3^3} \mod 10$
I got to $3^{27} \mod 10$
Then did the following:

$3^1 \mod 10 = 3$
$3^2 \mod 10 = 3^2 \mod10 = 9 \mod 10 = 9$
$3^4 \mod10 = 9^2 \mod10 = 81\mod10 = 1$ <-- This is where I get stuck. Could someone explain how to correctly solve this modulo?

Thanks!

k.Vijay
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Monil
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2 Answers2

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$$3^{4n+3}=(80+1)^n\cdot3^3\equiv1\cdot27\pmod{10}\equiv?$$

lab bhattacharjee
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  • Thanks for your answer, but I am still confused. Where do you get the 4n + 3? Could you further explain in detail how you would solve this? – Monil May 07 '17 at 18:07
  • 27=4*6+3, so your exponent is of the form 4n+3. – TMM May 07 '17 at 19:30
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$$ 3^{27} = 3^{26} \cdot 3 = 9^{13} \cdot 3 \equiv (-1)^{13}\cdot 3 = -3 \equiv 7 \bmod 10$$

lhf
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