Using Fermat’s theorem, find 3201 mod 11

To solve \( 3201 \mod 11 \) using Fermat's Little Theorem, we proceed as follows:

### Step-by-Step Solution

Fermat's Little Theorem states:
> If \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \).

In this case:
- \( p = 11 \) (a prime number),
- \( a = 3 \), which is not divisible by \( 11 \).

According to the theorem:
\[
3^{10} \equiv 1 \mod 11
\]

Since we want \( 3^{201} \mod 11 \), we need to simplify \( 201 \) in terms of powers of \( 10 \) (because \( 3^{10} \equiv 1 \mod 11 \)).

1. **Divide 201 by 10 to find the remainder:**
   \[
   201 = 20 \cdot 10 + 1
   \]
   So, \( 3^{201} = (3^{10})^{20} \cdot 3^1 \).

2. **Apply Fermat's Little Theorem:**
   Since \( 3^{10} \equiv 1 \mod 11 \), we have:
   \[
   (3^{10})^{20} \equiv 1^{20} \equiv 1 \mod 11
   \]

3. **Simplify the remaining term \( 3^1 \):**
   \[
   3^{201} \equiv 1 \cdot 3 \equiv 3 \mod 11
   \]

### Conclusion
\[
3^{201} \mod 11 = 3
\]

The answer is:
\[
3201 \mod 11 = 3
\]

Would you like a more detailed explanation or have any questions?

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Here are 5 related questions to deepen your understanding:
1. How would you use Fermat's Little Theorem to find \( 5^{120} \mod 13 \)?
2. Can Fermat's theorem be used if the base is divisible by the prime \( p \)?
3. How does Fermat's theorem relate to the Euler’s theorem?
4. What is the result of \( 2^{201} \mod 7 \) using Fermat’s theorem?
5. Can we apply Fermat's Little Theorem when \( p \) is not a prime?

**Tip:** When using Fermat's Little Theorem, first check if the exponent can be simplified using powers of \( p-1 \).

Explore how to use Fermat's Little Theorem to find 3^201 mod 11, a detailed solution ideal for high school students. This problem involves key concepts of modular arithmetic and exponentiation, making it suitable for advanced mathematics students in Grades 11-12.

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