Question

: In exercise 8-13, solve the system of congruences

11.

$\begin{array}{l}\mathbf{x}\mathbf{\equiv }\mathbf{2}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{d}\mathbf{5}\mathbf{\right)}\\ \mathbf{x}\mathbf{\equiv }\mathbf{0}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{d}\mathbf{6}\mathbf{\right)}\\ \mathbf{x}\mathbf{\equiv }\mathbf{3}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{d}\mathbf{7}\mathbf{\right)}\end{array}$

Short answer

The system of congruence is obtained as $x\equiv -18\left(\mathrm{mod}210\right)$.

Step-by-step solution

Conceptual Introduction

Let $m_1,m_2\mathrm{....}{m}_f$ be pair wise relatively prime positive integers, (it means that $\left(m_i,m_j\right)=1$ whenever ) . Assume that the $a_1,a_2\mathrm{.....},a_r$, are any integers.

Then the system,

$\begin{array}{l}x\equiv a_1\left(mod{m}_1\right)\\ x\equiv a_2\left(mod{m}_2\right)\\ x\equiv a_3\left(mod{m}_3\right)\\ .\\ .\\ .\\ x\equiv a_r\left(mod{m}_r\right)\end{array}$

has a solution.

Evaluating the system of congruence for x≡2(mod 5)  and data-custom-editor="chemistry" x≡0(mod 6)

Assume that the numbers m and n are relatively prime integers. Then the following system of congruence equations has a solution.

$\mathrm{x}\equiv \mathrm{a}\left(\mathrm{modm}\right)$ …… (1)

$\mathrm{x}\equiv \mathrm{b}\left(\mathrm{mod}\mathrm{n}\right)$ …… (2)

Now from the above two equation solution can be obtained as,

$\mathrm{t}=\mathrm{bmu}+\mathrm{anv}$…… (3)

Here the numbers u and v are found such that the following equation is satisfied.

$\mathrm{mu}+\mathrm{nv}=1$…… (4)

Now the given congruence equations are:

$x\equiv 2\left(\mathrm{mod}5\right)$ …… (5)

And,

$\mathrm{x}\equiv 0\left(\mathrm{mod}6\right)$……. (6)

Compare the equations (5) and (6) with (1) and (2) to obtain the following values,

$m=5$, $n=6$, $a=2$ and $\mathrm{b}=0$

Now find out the values of and such that the equation (4) is satisfied.

$5\mathrm{u}+6\mathrm{v}=1$

Above equation will get satisfied with the values $u=1$ and $v=-1$

Now substitute 0 for b , 5 for m , -1 for u , 2 for a , 6 for n and 1 for v into the equation (3)

$\begin{array}{c}t=0\times 5\times \left(-1\right)+2\times 6\times 1\\ =12\end{array}$

So the required solution is given by,

$\begin{array}{c}x\equiv 12\left(\mathrm{mod}6\times 5\right)\\ \equiv 12\left(\mathrm{mod}30\right)\end{array}$ ...... (7)

Evaluating the system of congruence for  x≡2mod 7

Now the third system of congruence is given by,

$\mathrm{x}\equiv 3\left(\mathrm{mod}7\right)$ ...... (8)

Again compare the equations (7) and (8) with (1) and (2) to obtain the following values,

$\mathrm{m}=30$, $\mathrm{n}=7$, $a=12$ and $\mathrm{b}=3$

Substitute 30 for m and 7 for n into the above equation, and again find the values of u and v such that the equation (4) is satisfied.

$30u+7v=1$

From the heat and trial method, above equation holds true for $u=-3$and $v=13$

Now substitute 3 for b , 30 for m , -3 for u , 12 for a , 7 for n and 13 for v into the equation (3)

$\begin{array}{c}t=bmu+anv\\ =3\times 30\times \left(-3\right)+12\times 7\times 13\\ =822\end{array}$........ (9)

Now from the equation (7), (8) and (9), the required solution can be calculated,

$\begin{array}{c}x\equiv 822\left(\mathrm{mod}30\times 7\right)\\ \equiv 822\left(\mathrm{mod}210\right)\\ \equiv -18\left(\mathrm{mod}210\right)\end{array}$

Therefore the final solution is $x\equiv -18\left(\mathrm{mod}210\right)$.