The problem involves solving the equation 6x + 3 = 1 in the context of modular arithmetic, specifically within ℤ8. The original poster notes that the textbook provides two solutions: x = 1 and x = 5, and seeks clarification on how to derive x = 5.
The discussion is ongoing, with participants exploring different interpretations of the problem and the implications of the modular arithmetic involved. Some guidance has been provided regarding the relationship between the factors of the equation and the divisibility conditions.
There is an emphasis on the constraints of working within ℤ8 and the specific properties of the numbers involved, particularly regarding the lack of a unit for 6 in this modular system.
Since ##6 \in \mathbb{Z}_8## is no unit, we cannot divide by ##6##. Thus from ##6x=6## we get ##6(x-1)=0## and we need to find all numbers, for which ##6y=0## in ##\mathbb{Z}_8##. Now which multiples of ##6## are divisible by eight?Math9999 said:Homework Statement
Solve 6x+3=1 in ℤ8.
Homework Equations
None.
The Attempt at a Solution
6x+3=9
6x=6
x=1
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The answers x=1, or x=5 in the textbook. But how do I get x=5?