Math Problem Statement

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

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

Solution

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

To find the order of an integer aa modulo nn, we look for the smallest positive integer kk such that:

ak1(modn)a^k \equiv 1 \pmod{n}

Step 1: Find the order of 2 modulo 23.

We need to find the smallest kk such that:

2k1(mod23)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 ϕ(23)=22\phi(23) = 22. Therefore, kk must be a divisor of 22. The divisors of 22 are: 1,2,11,221, 2, 11, 22.

We check the powers of 2 modulo 23:

  • 21=22^1 = 2
  • 22=42^2 = 4
  • 211=20481(mod23)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 kk such that:

3k1(mod23)3^k \equiv 1 \pmod{23}

Again, kk must divide 2222. We check the powers of 3 modulo 23:

  • 31=33^1 = 3
  • 32=93^2 = 9
  • 311=1771471(mod23)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 ϕ(27)=27×(113)=18\phi(27) = 27 \times (1 - \frac{1}{3}) = 18. So, the order of any integer modulo 27 must divide 1818.

Step 1: Find the order of 2 modulo 27.

We need to find the smallest kk such that:

2k1(mod27)2^k \equiv 1 \pmod{27}

The divisors of 18 are 1,2,3,6,9,181, 2, 3, 6, 9, 18. We check the powers of 2 modulo 27:

  • 21=22^1 = 2
  • 22=42^2 = 4
  • 23=82^3 = 8
  • 26=6410(mod27)2^6 = 64 \equiv 10 \pmod{27}
  • 29=512261(mod27)2^9 = 512 \equiv 26 \equiv -1 \pmod{27}
  • 218=(1)2=12^{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 kk such that:

7k1(mod27)7^k \equiv 1 \pmod{27}

Checking powers of 7 modulo 27:

  • 71=77^1 = 7
  • 72=4922(mod27)7^2 = 49 \equiv 22 \pmod{27}
  • 73=7×22=15419(mod27)7^3 = 7 \times 22 = 154 \equiv 19 \pmod{27}
  • 76=192=36110(mod27)7^6 = 19^2 = 361 \equiv 10 \pmod{27}
  • 79=10×19=1901(mod27)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 ϕ(n)\phi(n) for composite numbers?
  3. Can the order of any integer modulo nn exceed ϕ(n)\phi(n)?
  4. How do powers of integers behave modulo prime versus composite numbers?
  5. How do divisors of ϕ(n)\phi(n) restrict the possible orders of integers modulo nn?

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

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Group Theory
Number Theory

Formulas

a^k ≡ 1 (mod n)
Euler's Totient Function φ(n)

Theorems

Euler's Theorem
Order of an element in a group

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Grades 11-12)