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The solution of 8x+1 == 5 mod 12 is x == a mod m for some positive integers m >= 2 and a

Guest Aug 3, 2018
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The solution of 8x+1 == 5 mod 12 is x == a mod m for some positive integers m >= 2 and a

 

\(\begin{array}{lrcl} &8x+1 &\equiv& 5 \pmod{12} \\ \text{or } & 8x+1 &=& 5 + 12n,\ \quad n\in Z \\\\ & 8x+1 &=& 5 + 12n \quad & | \quad -1 \\ & 8x &=& 4 + 12n \quad & | \quad :4 \\ & 2x &=& 1 + 3n \\ &&& \boxed{\mathbf{2x-3n} \mathbf{=} \mathbf{1} } \\ \end{array} \)

 

\(\begin{array}{|rclrcl|} \hline 2x-3n &=& 1 \quad | \quad x \text{ has the smallest coefficient} \\\\ 2x &=& 1+3n \\\\ \mathbf{x} & \mathbf{=}& \mathbf{\dfrac{1+3n}{2}} \\\\ x &=& \dfrac{1+2n+n}{2} \\\\ x &=& \dfrac{2n+(1+n)}{2} \\\\ x &=& n+\underbrace{\dfrac{ 1+n }{2}}_{=b} \qquad a \in Z \\\\ && b = \dfrac{ 1+n }{2} \\\\ && 2b = 1+n \quad | \quad n \text{ has the smallest coefficient} \\\\ && n = -1+2b \\ && \quad \quad | \quad \text{Now on the right side of the equation there is no}\\ && \quad \quad | \quad \text{break and none of the variables any more is included.} \\ && \quad \quad | \quad \text{By inserting in reverse order, now in all equations,} \\ && \quad \quad | \quad \text{in which one variable has been isolated,} \\ && \quad \quad | \quad \text{eliminates the other variables.} \\\\ x &=& \dfrac{1+3n}{2} \quad | \quad n = -1+2b \\\\ x &=& \dfrac{1+3(-1+2b)}{2} \\\\ x &=& \dfrac{1-3+6b }{2} \\\\ x &=& \dfrac{-2+6b}{2} \\\\ \mathbf{x }&\mathbf{=}& \mathbf{-1+3b} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x & = & \underbrace{ -1 }_{=a} + \underbrace{3b}_{=m} \\\\ && \text{The positive integer $ m\ge 2$ , is $3$ or multiples of $3$}. \\ && \mathbf{m= 3,6,9,12,\ldots} \text{ and } \mathbf{a=-1} \\ \hline \end{array} \)

 

\(\begin{array}{|r|c|c|c|c|c|} \hline b & m=3b & m\ge 2 & x = -1+m & 8x+1 \equiv 5 \pmod {12} & x \equiv -1 \pmod{m} \\ \hline 1 & 3 & \checkmark & 2 & 17 \equiv 5 \pmod {12}\ \checkmark & 2 \equiv -1 \pmod {3}\ \checkmark \\ \hline 2 & 6 & \checkmark & 5 & 41 \equiv 5 \pmod {12}\ \checkmark & 5 \equiv -1 \pmod {3}\ \checkmark \\ \hline 3 & 9 & \checkmark & 8 & 65 \equiv 5 \pmod {12}\ \checkmark & 8 \equiv -1 \pmod {3}\ \checkmark \\ \hline \ldots & \ldots &\ldots & \\ \hline \end{array}\)

 

laugh

heureka  Aug 3, 2018

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