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.

Learn how to solve the problem of reducing 2 raised to the power of 2016 modulo 11 using Fermat's Little Theorem. Detailed step-by-step solution and explanation provided. Suitable for advanced high school students and beyond.

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