To reduce $2^{2016} \mod 11$, we can again use Fermat's Little Theorem. This theorem states that for a prime number $p$ and any integer $a$ such that $p$ does not divide $a$:

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

Here, $p = 11$ and $a = 2$. So, according to Fermat's Little Theorem:

$2^{10} \equiv 1 \mod 11$

To find $2^{2016} \mod 11$, we first express 2016 in terms of multiples of 10:

$2016 = 10 \times 201 + 6$

Thus,

$2^{2016} = 2^{10 \times 201 + 6} = (2^{10})^{201} \times 2^6$

Since $2^{10} \equiv 1 \mod 11$, it follows that:

$(2^{10})^{201} \equiv 1^{201} \equiv 1 \mod 11$

Therefore:

$2^{2016} \equiv 2^6 \mod 11$

Now, calculate $2^6$:

$2^6 = 64$

Finally, reduce $64 \mod 11$:

$64 \div 11 = 5 \text{ with a remainder of } 9$

So:

$64 \equiv 9 \mod 11$

Thus,

$2^{2016} \equiv 9 \mod 11$

Would you like more details on the steps or any related concepts?

Here are five related questions you might find interesting:

1. How can modular exponentiation be used in cryptography?
2. What are some methods to efficiently compute large powers modulo a number?
3. How does the Chinese Remainder Theorem relate to modular arithmetic?
4. What is the significance of primitive roots in modular arithmetic?
5. How can you determine the smallest positive integer $x$ such that $a^x \equiv 1 \mod p$?

Tip: When working with large exponents in modular arithmetic, reducing the exponent first using the modulus can simplify calculations significantly.