We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
190
1
avatar

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

 Aug 3, 2018
 #1
avatar+21991 
0

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

 Aug 3, 2018

25 Online Users

avatar