2x + 4 = 0 (mod 5)
Understand the Problem
The question is asking us to solve a modular arithmetic equation involving the variable x. The equation states that 2x + 4 is congruent to 0 modulo 5, and we need to find the value(s) of x that satisfy this condition.
Answer
$$ x \equiv 3 \mod{5} $$
Answer for screen readers
The solution for $x$ is:
$$ x \equiv 3 \mod{5} $$
Steps to Solve
- Set up the modular equation We start with the given equation:
$$ 2x + 4 \equiv 0 \mod{5} $$
This means we want to find the values of $x$ such that when we substitute $x$ into the equation, the result is congruent to $0$ modulo $5$.
- Isolate the variable Next, we can isolate the term that contains $x$ on one side of the congruence:
$$ 2x \equiv -4 \mod{5} $$
Since we are working modulo $5$, we need to find an equivalent expression for $-4$:
$$ -4 \equiv 1 \mod{5} $$
Thus, we can rewrite the equation:
$$ 2x \equiv 1 \mod{5} $$
- Multiply by the modular inverse To solve for $x$, we need to find the modular inverse of $2$ modulo $5$. The modular inverse $y$ must satisfy:
$$ 2y \equiv 1 \mod{5} $$
By checking values, we find that $y = 3$ works because:
$$ 2 \cdot 3 = 6 \equiv 1 \mod{5} $$
- Multiply both sides by the inverse We now multiply both sides of the congruence $2x \equiv 1 \mod{5}$ by the inverse $3$:
$$ 3 \cdot 2x \equiv 3 \cdot 1 \mod{5} $$
Simplifying gives:
$$ 6x \equiv 3 \mod{5} $$
- Reduce modulo 5 Reduce $6x$ and $3$ modulo $5$:
$$ x \equiv 3 \mod{5} $$
This means the solution for $x$ is all integers that can be expressed as $3 + 5k$, where $k$ is any integer.
The solution for $x$ is:
$$ x \equiv 3 \mod{5} $$
More Information
The solution $x \equiv 3 \mod{5}$ indicates that values of $x$ can be expressed in the form of $3$, $8$, $13$, etc., depending on the integer $k$. This area of mathematics is useful in topics such as cryptography and computer science.
Tips
- Forgetting to reduce: Students often forget to reduce their results correctly modulo the given base.
- Incorrect identification of the inverse: It's crucial to correctly identify the modular inverse; mistakes here lead to the wrong solution.
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