To reduce \( 2^{2016} \mod 7 \), we can use **Fermat's Little Theorem**. The theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[
a^{p-1} \equiv 1 \mod p
\]

Here, \( p = 7 \) and \( a = 2 \), so according to Fermat's Little Theorem:

\[
2^{6} \equiv 1 \mod 7
\]

Given this, \( 2^{2016} \) can be rewritten as:

\[
2^{2016} = (2^6)^{336}
\]

Since \( 2^6 \equiv 1 \mod 7 \), it follows that:

\[
(2^6)^{336} \equiv 1^{336} \equiv 1 \mod 7
\]

Therefore:

\[
2^{2016} \equiv 1 \mod 7
\]

So, \( 2^{2016} \mod 7 = 1 \).

Would you like further details on Fermat's Little Theorem or any other related topic?

Here are five related questions you might find interesting:
1. How can Fermat's Little Theorem be applied to different bases and moduli?
2. What is the significance of the order of an element in modular arithmetic?
3. How can you calculate powers of large numbers mod small primes more efficiently?
4. What other theorems are related to Fermat's Little Theorem in number theory?
5. How does Fermat's Little Theorem relate to the RSA encryption algorithm?

**Tip:** When dealing with large exponents in modular arithmetic, look for patterns or use theorems like Fermat's Little Theorem to simplify the calculations.

Learn how to reduce 2^2016 mod 7 using Fermat's Little Theorem with a detailed step-by-step solution. This problem covers modular arithmetic, exponentiation, and number theory, focusing on Fermat's Little Theorem for prime numbers. Suitable for advanced high school students and anyone interested in number theory.

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