Math Problem Statement

Find the multiplicative inverse of 17 modulo 23. (Demonstrate how you’ve found it. Guessing is not accepted.)

Solution

To find the multiplicative inverse of 17 modulo 23, we need to find an integer xx such that:

17x1(mod23)17x \equiv 1 \pmod{23}

This equation means that 17x117x - 1 is divisible by 23, or equivalently:

17x=1+23k17x = 1 + 23k

for some integer kk. The standard way to find this xx is by using the Extended Euclidean Algorithm.

Step 1: Apply the Euclidean Algorithm

We need to find the greatest common divisor (GCD) of 17 and 23 and express it in the form:

GCD(17,23)=17x+23y\text{GCD}(17, 23) = 17x + 23y

Since 17 and 23 are coprime (i.e., their GCD is 1), we will use the Extended Euclidean Algorithm to find xx.

First, apply the Euclidean algorithm to find the GCD: 23=17×1+623 = 17 \times 1 + 6 17=6×2+517 = 6 \times 2 + 5 6=5×1+16 = 5 \times 1 + 1 5=5×1+05 = 5 \times 1 + 0

So, the GCD is 1 (as expected since 17 and 23 are coprime).

Step 2: Use the Extended Euclidean Algorithm

Now, we backtrack to express 1 as a linear combination of 17 and 23.

Starting from the second to last equation: 1=65×11 = 6 - 5 \times 1

Substitute 55 from the equation before: 1=6(176×2)×11 = 6 - (17 - 6 \times 2) \times 1 1=617+6×21 = 6 - 17 + 6 \times 2 1=6×3171 = 6 \times 3 - 17

Now substitute 66 from the equation 6=2317×16 = 23 - 17 \times 1: 1=(2317×1)×3171 = (23 - 17 \times 1) \times 3 - 17 1=23×317×3171 = 23 \times 3 - 17 \times 3 - 17 1=23×317×41 = 23 \times 3 - 17 \times 4

Thus, we have: 1=23×317×41 = 23 \times 3 - 17 \times 4

Step 3: Interpret the Result

From this equation, we see that: 17×41(mod23)-17 \times 4 \equiv 1 \pmod{23}

Multiplying both sides by 1-1, we get: 17×4122(mod23)17 \times 4 \equiv -1 \equiv 22 \pmod{23}

Thus, the multiplicative inverse of 17 modulo 23 is 19\boxed{19}, since 221(mod23)22 \equiv -1 \pmod{23} and 234=1923 - 4 = 19.

Verification

We can verify by direct multiplication: 17×19=32317 \times 19 = 323 Now calculate 323mod23323 \mod 23: 323÷23=14 remainder 1323 \div 23 = 14 \text{ remainder } 1 So, 3231(mod23)323 \equiv 1 \pmod{23}

Thus, the multiplicative inverse of 17 modulo 23 is indeed 19\boxed{19}.

Do you want any more details or have any questions?


Related Questions:

  1. What is the multiplicative inverse of 5 modulo 13?
  2. How does the Extended Euclidean Algorithm work in general?
  3. Can the multiplicative inverse of a number modulo nn be found for non-prime nn?
  4. What are the properties of numbers with multiplicative inverses in a given modulus?
  5. How does modular arithmetic apply in cryptography, such as in RSA?
  6. What is the significance of the modular inverse in solving linear congruences?
  7. Can the Chinese Remainder Theorem be used to find multiplicative inverses?
  8. What happens if a number does not have a multiplicative inverse modulo nn?

Tip: When verifying your results in modular arithmetic, it can be helpful to multiply the original number by the calculated inverse and ensure the result is congruent to 1 modulo the base.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Extended Euclidean Algorithm

Formulas

Extended Euclidean Algorithm for finding modular inverses

Theorems

Existence of modular inverses

Suitable Grade Level

Advanced High School