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 $p-1$.