Marvin Mccormick

2020-10-28

Solve the following systems of congruences.

$x\equiv 2(mod\text{}5)$

Sadie Eaton

Skilled2020-10-29Added 104 answers

Formula used:

1) Theorem: System of congruences:

Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences

$x\equiv a(mod\text{}m)$

$x\equiv b(mod\text{}n)$

Furthermore, any two solutions x and y are congruent modulo mn.

2) Theorem: Addition and Multiplication Properties:

If$a\equiv b(mod\text{}n)$ and x is any integer, then $a+x-=b+x(mod\text{}n)\text{}and\text{}ax\equiv bx(mod\text{}n)$ .

Explanation:

Consider the system of congruences$x\equiv (mod\text{}5)$

$x\equiv 3(mod\text{}8)$

Since 5 and 8 are relatively prime, then$(5,8)=1$ .

Then, by using a theorem, there exists an integer x that satisfies the system of congruences.

From the first congruence$x=2+5k$ for some integer k and substitute this expression for x into the second congruence.

$2+5k\equiv 3(mod\text{}8)$

By using addition property,

$2+5k+(\u20142)\equiv 3+(\u20142)(mod\text{}8)$

$=5k=1(mod\text{}8)$

By using multiplication property,

$\Rightarrow 2+5k+(-2)\equiv 3+(-2)(mod\text{}8)$

$\Rightarrow 5k\equiv 1(mod\text{}8)$

By using multiplication property,

$5\ast 5k-\equiv 5\ast 1(mod\text{}8)$

$\Rightarrow 25k\equiv 5(mod\text{}8)$

Since$25\equiv 1(mod\text{}8)$ ,

$\Rightarrow k\equiv 5(mod\text{}8)$

Thus,$x=2+5(5)=27$ satisfies the system and $x\equiv 27(mod\text{}5\ast 8)\text{}or\text{}x\equiv 27(mod\text{}40)$ gives all solutions to the given system of congruences.

1) Theorem: System of congruences:

Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences

Furthermore, any two solutions x and y are congruent modulo mn.

2) Theorem: Addition and Multiplication Properties:

If

Explanation:

Consider the system of congruences

Since 5 and 8 are relatively prime, then

Then, by using a theorem, there exists an integer x that satisfies the system of congruences.

From the first congruence

By using addition property,

By using multiplication property,

By using multiplication property,

Since

Thus,