. (a) Find the order of the integers 2 and 3 modulo 23.

(b) Find the order of the integers 2 and 7 modulo 27.

### Part (a): Finding the Order of 2 and 3 Modulo 23

To find the **order** of an integer \( a \) modulo \( n \), we look for the smallest positive integer \( k \) such that:

\[
a^k \equiv 1 \pmod{n}
\]

#### Step 1: Find the order of 2 modulo 23.
We need to find the smallest \( k \) such that:

\[
2^k \equiv 1 \pmod{23}
\]

Since 23 is prime, Euler's Theorem tells us that the order of any integer modulo 23 must divide \( \phi(23) = 22 \). Therefore, \( k \) must be a divisor of 22. The divisors of 22 are: \( 1, 2, 11, 22 \).

We check the powers of 2 modulo 23:

- \( 2^1 = 2 \)
- \( 2^2 = 4 \)
- \( 2^{11} = 2048 \equiv 1 \pmod{23} \)

Thus, the order of 2 modulo 23 is **11**.

#### Step 2: Find the order of 3 modulo 23.
Now, we need to find the smallest \( k \) such that:

\[
3^k \equiv 1 \pmod{23}
\]

Again, \( k \) must divide \( 22 \). We check the powers of 3 modulo 23:

- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^{11} = 177147 \equiv 1 \pmod{23} \)

Thus, the order of 3 modulo 23 is also **11**.

### Part (b): Finding the Order of 2 and 7 Modulo 27

Since 27 is not prime, Euler's function gives \( \phi(27) = 27 \times (1 - \frac{1}{3}) = 18 \). So, the order of any integer modulo 27 must divide \( 18 \).

#### Step 1: Find the order of 2 modulo 27.
We need to find the smallest \( k \) such that:

\[
2^k \equiv 1 \pmod{27}
\]

The divisors of 18 are \( 1, 2, 3, 6, 9, 18 \). We check the powers of 2 modulo 27:

- \( 2^1 = 2 \)
- \( 2^2 = 4 \)
- \( 2^3 = 8 \)
- \( 2^6 = 64 \equiv 10 \pmod{27} \)
- \( 2^9 = 512 \equiv 26 \equiv -1 \pmod{27} \)
- \( 2^{18} = (-1)^2 = 1 \)

Thus, the order of 2 modulo 27 is **18**.

#### Step 2: Find the order of 7 modulo 27.
We now look for the smallest \( k \) such that:

\[
7^k \equiv 1 \pmod{27}
\]

Checking powers of 7 modulo 27:

- \( 7^1 = 7 \)
- \( 7^2 = 49 \equiv 22 \pmod{27} \)
- \( 7^3 = 7 \times 22 = 154 \equiv 19 \pmod{27} \)
- \( 7^6 = 19^2 = 361 \equiv 10 \pmod{27} \)
- \( 7^9 = 10 \times 19 = 190 \equiv 1 \pmod{27} \)

Thus, the order of 7 modulo 27 is **9**.

### Final Results:

- The order of 2 modulo 23 is **11**.
- The order of 3 modulo 23 is **11**.
- The order of 2 modulo 27 is **18**.
- The order of 7 modulo 27 is **9**.

---

Would you like further details on any of these steps? Here are some related questions to explore:

1. How does Euler's theorem help in finding the order of an integer?
2. What are the properties of Euler's totient function \( \phi(n) \) for composite numbers?
3. Can the order of any integer modulo \( n \) exceed \( \phi(n) \)?
4. How do powers of integers behave modulo prime versus composite numbers?
5. How do divisors of \( \phi(n) \) restrict the possible orders of integers modulo \( n \)?

**Tip:** Checking powers of integers modulo \( n \) is easier by systematically testing exponents that divide \( \phi(n) \).

Find the order of the integers 2 and 3 modulo 23. Find the order of the integers 2 and 7 modulo 27.

Explore how to find the order of integers 2 and 3 modulo 23, and 2 and 7 modulo 27 using Euler's Theorem and modular arithmetic. This solution demonstrates step-by-step calculations and covers core concepts in number theory and group theory. Suitable for advanced high school or undergraduate students.

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