Question
Find the least positive value of \(x\) such that:
$$89 \equiv (x + 3) \pmod{4}$$Solution
The given congruence is:
$$89 \equiv (x + 3) \pmod{4}$$By the definition of modular congruence, this means that the difference \(89 - (x + 3)\) is a multiple of 4. Let's represent this with an integer \(n\):
$$89 - (x + 3) = 4n$$Simplifying the left side of the equation:
$$89 - x - 3 = 4n$$ $$86 - x = 4n$$This tells us that \(86 - x\) must be a multiple of 4. To find the least positive value for \(x\), we need to find the largest multiple of 4 that is less than or equal to 86.
Let's check the multiples of 4 near 86:
- \(4 \times 20 = 80\)
- \(4 \times 21 = 84\)
- \(4 \times 22 = 88\)
The multiple we need is 84. Now, we set \(86 - x\) equal to 84 to solve for \(x\):
$$86 - x = 84$$ $$x = 86 - 84$$ $$x = 2$$Since 2 is the least positive value for x that satisfies the condition, it is our answer.
∴ The least positive value of x is 2.
