check if large number $(9^{81}+6)$ is divisible by $11$

Question

I would like to know if there is a mathematical way to check whether number $9^{81}+6$ is divisible by $11$, without actually calculating the whole number.

Answer 1

By Fermat's little theorem: $$ 9^{81}+6 \equiv 9^{1}+6 \equiv 4\pmod{11}$$ hence the answer is negative.

Answer 2

$$9^1 \equiv -2 \bmod{11}$$ $$9^2 \equiv 4 \bmod{11}$$ $$9^6 \equiv 64 \equiv -2 \bmod{11}$$ $$9^{30} \equiv -32 \equiv 1 \bmod{11}$$ $$9^{81} \equiv 9^{30} 9^{30} 9^{6} 9^{6} 9^{6} 9^2 9^1 \equiv (1)(1)(-2)(-2)(-2)(4)(-2) \equiv 64 \equiv 9 \bmod{11}$$ $$9^{81} + 6 \equiv 15 \equiv 4 \bmod{11}.$$

EDIT: I accidentally wrote $-32 \equiv -1 \bmod{11}$ instead of $-32 \equiv 1 \bmod{11}$. The minus signs fortuitously cancelled!

Answer 3

Let's also throw the Euler's theorem variation in: $9$ and $11$ are coprime, hence

$$ 9^{81} + 6 \equiv 9 \cdot (9^{10})^8 + 6 \equiv 9 + 6 \equiv 4 \mod 11, $$

where we used that $\phi(11) = 11^1 - 11^0 = 10$, where $\phi$ is Euler's totient function.