Math Problem Statement

4x = 10 ( mod 14 )

Solution

The given equation is:

4x10 (mod 14)4x \equiv 10 \ (\text{mod} \ 14)

We need to solve this congruence equation for xx. Let's break it down step by step:

Step 1: Simplify the equation

Since this is a modulo equation, we're working in the system of integers modulo 14. The equation reads that 4x4x is congruent to 10 modulo 14, which means: 4x10 is divisible by 14.4x - 10 \text{ is divisible by } 14.

Step 2: Check if a solution exists

We want to simplify 4x10 (mod 14)4x \equiv 10 \ (\text{mod} \ 14). First, note that 4 and 14 have a greatest common divisor (GCD) of 2. Therefore, the equation can only have a solution if the right-hand side, 10, is divisible by this GCD (2), which it is.

Step 3: Divide by the GCD (which is 2)

We can now divide the entire equation by 2:

2x5 (mod 7)2x \equiv 5 \ (\text{mod} \ 7)

Step 4: Solve the new congruence

Now, we need to solve:

2x5 (mod 7)2x \equiv 5 \ (\text{mod} \ 7)

We need to find the multiplicative inverse of 2 modulo 7. To find this, we test values of yy such that:

2y1 (mod 7)2y \equiv 1 \ (\text{mod} \ 7)

Through trial or using the Extended Euclidean Algorithm, we find that y=4y = 4, because:

2×4=81 (mod 7)2 \times 4 = 8 \equiv 1 \ (\text{mod} \ 7)

Step 5: Multiply both sides by the inverse of 2 (which is 4)

Now, multiply both sides of 2x5 (mod 7)2x \equiv 5 \ (\text{mod} \ 7) by 4:

x5×4 (mod 7)x \equiv 5 \times 4 \ (\text{mod} \ 7)

x20 (mod 7)x \equiv 20 \ (\text{mod} \ 7)

Since 206 (mod 7)20 \equiv 6 \ (\text{mod} \ 7), we have:

x6 (mod 7)x \equiv 6 \ (\text{mod} \ 7)

Step 6: General solution

Now that we know x6 (mod 7)x \equiv 6 \ (\text{mod} \ 7), we can write the general solution. Since we divided by 2 earlier, the original modulus was 14, so the general solution is:

x=6+7kfor some integer k.x = 6 + 7k \quad \text{for some integer } k.

Thus, the solution to the equation 4x10 (mod 14)4x \equiv 10 \ (\text{mod} \ 14) is:

x=6+7kfor any integer k.x = 6 + 7k \quad \text{for any integer } k.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How do we find the multiplicative inverse in modular arithmetic?
  2. What happens if the GCD is not a divisor of the right-hand side in a modular equation?
  3. Can you explain how the Extended Euclidean Algorithm helps in finding the inverse?
  4. How do you determine if a modular equation has no solution?
  5. What are some real-world applications of modular arithmetic?

Tip: In modular arithmetic, when solving axb (mod m)ax \equiv b \ (\text{mod} \ m), always check the GCD of aa and mm to see if a solution exists!

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Congruence Equations
Greatest Common Divisor (GCD)
Multiplicative Inverse

Formulas

ax ≡ b (mod m)
Multiplicative Inverse: a^(-1) (mod m) such that a * a^(-1) ≡ 1 (mod m)

Theorems

Extended Euclidean Algorithm
Solving Linear Congruences

Suitable Grade Level

Grades 9-12