To solve the problem using Fermat's Little Theorem, let's go through the steps:

Step 1: Recall Fermat's Little Theorem

Fermat's Little Theorem states that if $p$ is a prime number and $a$ is an integer such that $a$ is not divisible by $p$, then: $a^{p-1} \equiv 1 \pmod{p}$ Here, $p = 401$ (which is prime) and $a = 111$. Since 111 is not divisible by 401, the theorem applies.

Step 2: Apply the Theorem

According to Fermat's Little Theorem: $111^{400} \equiv 1 \pmod{401}$

Step 3: Reduce the Exponent

Now, we need to find $111^{401} \mod 401$. Notice that: $111^{401} = 111^{400} \cdot 111$ Using the congruence from Fermat’s Little Theorem: $111^{400} \equiv 1 \pmod{401}$ Thus: $111^{401} \equiv 1 \cdot 111 \equiv 111 \pmod{401}$

Conclusion

The least residue of $111^{401}$ modulo 401 is: $\boxed{111}$

Would you like a deeper explanation of any part of the solution or have any further questions?

Further Questions

1. How does Fermat's Little Theorem work when $a$ is divisible by $p$?
2. Can Fermat’s Little Theorem be applied when $p$ is not a prime number?
3. What happens if the base and the modulus are the same?
4. How can we generalize Fermat’s Little Theorem for cases where the modulus is not prime?
5. How would the solution change if $a$ were greater than $p$?

Tip

Always check if the base is coprime with the modulus when applying Fermat’s Little Theorem. This ensures the theorem’s conditions are met.