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Question

Question: Solution set of \[x\equiv 3(\bmod \,7),\,\,p\in Z\] is given by 1\. \[\\{3\\}\] 2\. \[\\{7p-3:p\...

Solution set of x3(mod7),pZx\equiv 3(\bmod \,7),\,\,p\in Z is given by
1. 3\\{3\\}
2. 7p3:pZ\\{7p-3:p\in Z\\}
3. 7p+3:pZ\\{7p+3:p\in Z\\}
4. None of these

Explanation

Solution

In this type of particular problem we must use the formula which we apply here for such types of problems, that is the formula which is given by xa(modn)x\equiv a(\bmod n)\, where pZp\in Z. So, here in this type of problem we will use these formulas to find the solution set of the value of x.

Complete step-by-step solution:
In this type of problem here it is given that x=3(mod7),pZ(1)x=3(\bmod \,7),\,\,p\in Z\,\,--(1)
So, here if you see then it is asked to find the solution set of x=3(mod7),pZx=3(\bmod \,7),\,\,p\in Z
To find the solution set of such a problem we need to use the formula
As we know already that above question given in the form of xa(modn)x\equiv a(\bmod n)
So, for that we have to use the formula for such forms
Formula is given by
xa=p×n(2)x-a=p\times n---(2)
Where n is the number which is present inside the mod\bmod that is n\left|n \right| and aa is the coefficient of n\left|n \right|
If you compare this formula with equation (1) then you can notice that we have the value of a and n
We get the solution set in the term of p.
a=3a=3
n=7n=7
Substitute the value of a=3a=3 and n=7n=7 on equation (2) we get:
x3=p×7x-3=p\times 7
By simplifying this above equation we get:
x3=7px-3=7p
By simplifying further we get:
x=7p+3x=7p+3
So, above value of x=7p+3x=7p+3 is represented on solution set which is represented as \\{\\}
So, the solution set of x3(mod7)x\equiv 3(\bmod \,7) is: 7p+3:pZ\\{7p+3:p\in Z\\}
Hence, the correct option is “option 3”.

Note: In this type of problem we have always applied the formula correctly for such types of problems.
Don’t make silly mistakes while simplifying the steps. Solve these problems step by step to avoid confusion. So, the above solution is preferred for such types of problems.