To calculate \( 5^{-1} \mod 23 \) using Fermat's Little Theorem, we proceed as follows:

### Fermat's Little Theorem
Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is any integer such that \( \gcd(a, p) = 1 \), then:
\[
a^{p-1} \equiv 1 \mod p
\]
This implies that:
\[
a^{p-2} \equiv a^{-1} \mod p
\]
Thus, the modular inverse of \( a \mod p \) is \( a^{p-2} \mod p \).

### Applying Fermat's Theorem
Here, \( a = 5 \) and \( p = 23 \) (a prime number). To compute \( 5^{-1} \mod 23 \), we need to calculate:
\[
5^{23-2} \mod 23 = 5^{21} \mod 23
\]

Now, let's compute \( 5^{21} \mod 23 \).

We can use **modular exponentiation** to simplify the process:

\[
5^2 = 25 \equiv 2 \mod 23
\]
\[
5^4 = (5^2)^2 = 2^2 = 4 \mod 23
\]
\[
5^8 = (5^4)^2 = 4^2 = 16 \mod 23
\]
\[
5^{16} = (5^8)^2 = 16^2 = 256 \equiv 3 \mod 23
\]

Now, combine powers:
\[
5^{21} = 5^{16} \times 5^4 \times 5 = 3 \times 4 \times 5 = 60 \equiv 14 \mod 23
\]

Thus, \( 5^{-1} \mod 23 = 14 \).

### Final Answer
\[
5^{-1} \mod 23 = 14
\]

Would you like further details on modular exponentiation or Fermat's theorem?

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Here are 5 related questions:
1. How can Fermat's Little Theorem be used to calculate larger modular inverses?
2. What are other ways to calculate the modular inverse aside from Fermat's theorem?
3. Can the extended Euclidean algorithm be used to find the inverse of 5 modulo 23?
4. How do we compute large powers mod \( p \) efficiently using exponentiation by squaring?
5. What is the relationship between Fermat's Little Theorem and Euler's Theorem?

**Tip**: To compute modular inverses efficiently for larger numbers, consider using the extended Euclidean algorithm, which is faster for practical use than Fermat's theorem.

Learn how to find the modular inverse of 5 modulo 23 using Fermat's Little Theorem. This problem explores modular arithmetic, number theory, and modular exponentiation, providing a step-by-step explanation suitable for students in Grades 11-12.

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