The remainder when 6^(6^6) is divided by 11 is?

Steps to Solve

1. Understand the problem using modular arithmetic

We want to find the remainder of $6^{(6^6)}$ when divided by 11. This can be expressed using modular notation as $6^{(6^6)} \mod 11$.

2. Apply Fermat's Little Theorem

According to Fermat's Little Theorem, if $p$ is a prime number and $a$ is an integer not divisible by $p$, then

$$ a^{p-1} \equiv 1 \mod p $$

Here, $p = 11$ and $a = 6$. Therefore:

$$ 6^{10} \equiv 1 \mod 11 $$

3. Reduce the exponent $6^6$ modulo 10

Since $6^{10} \equiv 1 \mod 11$, it suffices to calculate $6^6 \mod 10$ to find the effective exponent for $6^{(6^6)}$.

First, calculate $6 \mod 10$:

$$ 6^1 \equiv 6 \mod 10 $$

$$ 6^2 \equiv 6 \times 6 \equiv 36 \equiv 6 \mod 10 $$

It follows that $6^n \equiv 6 \mod 10$ for any positive integer $n$. Thus:

$$ 6^6 \equiv 6 \mod 10 $$

4. Calculate $6^{6} \mod 11$

Now we need to calculate:

$$ 6^6 \mod 11 $$

First, we calculate:

$$ 6^2 \equiv 36 \mod 11 \equiv 3 $$

Then compute $6^4$:

$$ 6^4 = (6^2)^2 \equiv 3^2 \equiv 9 \mod 11 $$

Finally, calculate $6^6$:

$$ 6^6 = 6^4 \times 6^2 \equiv 9 \times 3 \equiv 27 \mod 11 \equiv 5 $$