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Question: The solution set of \[8x \equiv 6\left( {\bmod 14} \right),{\text{ }}x \in z\] are; \(\left( 1 \ri...

The solution set of 8x6(mod14), xz8x \equiv 6\left( {\bmod 14} \right),{\text{ }}x \in z are;
(1)[8][6]\left( 1 \right)\left[ 8 \right] \cup \left[ 6 \right]
(2)[8][14]\left( 2 \right)\left[ 8 \right] \cup \left[ {14} \right]
(3)[6][13]\left( 3 \right)\left[ 6 \right] \cup \left[ {13} \right]
(4)[8][6][13]\left( 4 \right)\left[ 8 \right] \cup \left[ 6 \right] \cup \left[ {13} \right]

Explanation

Solution

This question is based on linear congruence. Linear congruence is represented by the form axb(modn)ax \equiv b\left( {\bmod n} \right) ; where xx is an unknown integer whose power is 11 , therefore the expression is called linear congruence in one variable. There are certain steps to solve this type of linear congruent problems which are given below.

Complete step-by-step solution:
Step I : Find GCD(a,n){\text{GCD}}\left( {{\text{a,n}}} \right) , let us name it as d{\text{d}} .
Step II : Find ratio bd\dfrac{b}{d} ; if we get a whole number on division then there exists a solution for the given problem.
Step III : Find d mod n{\text{d mod n}} ; this gives us how many solutions are possible for the given problem.
Step IV : Divide both the sides by dd .
Step V : Multiply both sides by the multiplicative inverse of aa i.e. a1{a^{ - 1}} .
Step VI : The general solution equation is given by , xk=x0+k(nd){x_k} = {x_0} + k\left( {\dfrac{n}{d}} \right) ; where k = \left\\{ {0,1,2,3,.......\left( {d - 1} \right)} \right\\}
Now, using these rules let us solve the given problem;
We know the general form of linear congruence is axb(modn)ax \equiv b\left( {\bmod n} \right)
8x6(mod14) ......(1)\Rightarrow 8x \equiv 6\left( {\bmod 14} \right){\text{ }}......\left( 1 \right) (Given)
Comparing our question and the standard form to get the respective values, we get;
a=8, b=6 and m=14\Rightarrow a = 8,{\text{ }}b = 6{\text{ and }}m = 14 ;
Step I : Finding GCD(8,14){\text{GCD}}\left( {8,14} \right) ;
8=2×2×2\Rightarrow 8 = 2 \times 2 \times 2
14=2×7\Rightarrow 14 = 2 \times 7
Therefore, GCD(8,14)=2{\text{GCD}}\left( {8,14} \right) = 2 ; means d=2d = 2 .
Step II : Dividing bd i.e. 62=3\dfrac{b}{d}{\text{ i}}{\text{.e}}{\text{. }}\dfrac{6}{2} = 3 ; since 33 is a whole number therefore the given linear congruence problem is solvable. If bb is not completely divisible by dd , then we can stop here only and say that the given problem is not solvable.
Step III : Finding d mod nd{\text{ mod }}n i.e. 2 mod 14 = 22{\text{ mod 14 = 2}} ; therefore there will be two solutions for the given problem.
Step IV : Divide both the sides by dd which is 22 , we get;
8x6(mod14)\Rightarrow 8x \equiv 6\left( {\bmod 14} \right)
4x3(mod14)......(2)\Rightarrow 4x \equiv 3\left( {\bmod 14} \right)......\left( 2 \right)
Step V: We have to find the multiplicative inverse of a i.e. 41a{\text{ i}}{\text{.e}}{\text{. }}{{\text{4}}^{ - 1}}, according to the above equation;
(4×a)mod7=1\Rightarrow \left( {4 \times a} \right)\bmod 7 = 1
Now, we have to find such a value of aa; for which the above condition is satisfied and that value will be the multiplicative inverse of 44 .
Therefore, we will get a=2i.e. 41=2 \Rightarrow a = 2{\text{i}}{\text{.e}}{\text{. }}{{\text{4}}^{ - 1}} = 2
41×4×x41×3(mod7)\Rightarrow {4^{ - 1}} \times 4 \times x \equiv {4^{ - 1}} \times 3\left( {\bmod 7} \right)
On simplification, the above expression reduces to;
x41×3(mod7)\Rightarrow x \equiv {4^{ - 1}} \times 3\left( {\bmod 7} \right)
Now, put the value of 41=2{4^{ - 1}} = 2 ;
x6(mod7)\Rightarrow x \equiv 6\left( {\bmod 7} \right)
Therefore, x0=6{x_0} = 6 .
Step VI : The general solution is given by, xk=x0+k(nd){x_k} = {x_0} + k\left( {\dfrac{n}{d}} \right)
Put k=1k = 1 , x0=6{x_0} = 6 and n=14n = 14 according to equation (1)\left( 1 \right) , we get;
x1=6+(142)\Rightarrow {x_1} = 6 + \left( {\dfrac{{14}}{2}} \right)
x1=13\Rightarrow {x_1} = 13
Therefore the two solutions for the given linear congruence relation are x0=6 and x1=13{x_0} = 6{\text{ and }}{x_1} = 13 .
Hence the correct answer for this question is option (3)\left( 3 \right) i.e. [6][13]\left[ 6 \right] \cup \left[ {13} \right].

Note: The linear congruence relation axb(modn)ax \equiv b\left( {\bmod n} \right) means that integers a and ba{\text{ and }}b are congruent modulo nn if they produce the same remainder on division by nn . Ultimately we have to find an integer xx that satisfies this condition and those values of xx are called congruent solutions for the given linear congruence . The number of congruent solutions depends on the G.C.D. ( greatest common divisor ) as explained above. There can be different congruent solutions belonging to different congruent classes (different values or patterns), those solutions are called incongruent solutions.