Question

what is solution of congruence ,3x=4 mode 7​

Answer 1

We're asked to find the solution of,

$\longrightarrow 3x\equiv4\pmod{7}$

Then let,

$\longrightarrow 3x=4+7a,\quad a\in\mathbb{Z}\quad\quad\dots(1)$

Here the least coefficient appears is 3 (ignoring constant term). Then take,

$\longrightarrow 3x\equiv4+7a\pmod{3}\quad\quad\dots(2)$

But we see that,

• $3\equiv0\pmod{3}$

• $4\equiv1\pmod{3}$

• $7\equiv1\pmod{3}$

Then (2) becomes,

$\longrightarrow 0\equiv1+a\pmod{3}$

$\longrightarrow a\equiv-1\pmod{3}$

Since $-1\equiv2\pmod{3},$

$\longrightarrow a\equiv2\pmod{3}$

Here the coefficient of $a$ is 1 itself. Then let,

$\longrightarrow a=3k+2,\quad k\in\mathbb{Z}$

Then (1) becomes,

$\longrightarrow 3x=4+7(3k+2)$

$\longrightarrow 3x=21k+18$

$\longrightarrow\underline{\underline{x=7k+6}}$

This is the solution of the equation.

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

We're asked to find the solution of,

$\longrightarrow 3x\equiv4\pmod{7}$

Then let,

$\longrightarrow 3x=4+7a,\quad a\in\mathbb{Z}\quad\quad\dots(1)$

Here the least coefficient appears is 3 (ignoring constant term). Then take,

$\longrightarrow 3x\equiv4+7a\pmod{3}\quad\quad\dots(2)$

But we see that,

$3\equiv0\pmod{3}$

$4\equiv1\pmod{3}$

$7\equiv1\pmod{3}$

Then (2) becomes,

$\longrightarrow 0\equiv1+a\pmod{3}$

$\longrightarrow a\equiv-1\pmod{3}$

Since $-1\equiv2\pmod{3},$

$\longrightarrow a\equiv2\pmod{3}$

Here the coefficient of $a$ is 1 itself. Then let,

$\longrightarrow a=3k+2,\quad k\in\mathbb{Z}$

Then (1) becomes,

$\longrightarrow 3x=4+7(3k+2)$

$\longrightarrow 3x=21k+18$

$\longrightarrow\underline{\underline{x=7k+6}}$

This is the solution of the equation.