Question

find the least positive incongruent of the solution of 13x=9(mod 25)​

Answer 1

Answer:

REMAINDER IS -6..

Here is the key observation which enables us to solve linear congruences. By definition of congruence, ax ≡ b (mod m) iff ax − b is divisible by m. Hence, ax ≡ b (mod m) iff ax − b = my, for some integer y. Rearranging the equation to the equivalent form ax − my = b we arrive at the following result..

Hope it helps

Answer 2

Given to solve,

$\longrightarrow13x\equiv9\pmod{25}$

Let,

$\longrightarrow x\equiv9u\pmod{25}$

$\longrightarrow x=9u+25k,\quad k\in\mathbb{Z}$

Then,

$\longrightarrow13(9u+25k)\equiv9\pmod{25}$

$\longrightarrow13\times9u+13\times25k\equiv9\pmod{25}$

As $13\times25k\equiv0\pmod{25},$

$\longrightarrow13\times9u\equiv9\pmod{25}$

$\longrightarrow13u\equiv1\pmod{25}$

$\longrightarrow13u=1-25m,\quad m\in\mathbb{Z}$

$\longrightarrow25m+13u=1\quad\quad\dots(i)$

Now we make use of Euclid's Division Lemma.

$\longrightarrow25=1\times13+12\quad\quad\dots(1)$

$\longrightarrow 13=1\times12+1\quad\quad\dots(2)$

Then, from (2),

$\longrightarrow 1=13-12$

From (1),

$\longrightarrow 1=13-(25-13)$

$\longrightarrow 1=13-25+13$

$\longrightarrow 1=25(-1)+13(2)\quad\quad\dots(ii)$

Comparing (i) and (ii) we get,

$\longrightarrow u=2$

Then,

$\longrightarrow x\equiv9u\pmod{25}$

$\longrightarrow x\equiv9\times2\pmod{25}$

$\longrightarrow x\equiv18\pmod{25}$

So the solution is,

$\longrightarrow x=25q+18,\quad q\in\mathbb{Z}$

Since $0<18<25,$ the least positive incongruent is,

$\longrightarrow\underline{\underline{x=18}}$