Vtyjkyuk

Find the smallest values of x,y that satisfy:

77x=13y+2

Please show how I can solve this using modular arithmetic, in simple steps(I am a novice to modular arithmetic).

Also, is there a way to solve this equation(except for simple trial and error) without using modular arithmetic?

Hint: You can write $$77xequiv 2mod 13$$ and you can solve it like $$xequiv frac277 mod 13$$ and you and add the modul to the numerator $$frac2+13+13+...+1377$$ until we get a whole number.

You are looking for an integer $x$ such that $77xequiv2pmod13$.

Also $77equiv-1pmod13$, I think you know how to go on from here.

Hope this helps.

I think for finding an answer we need trial and error but we know that one answer exists when $0le x<13$. One answer is $$x=11\y=65$$and the general answer would become $$x=11+13k\y=65+77k\kinBbb Z$$

For any pair of integers $ x,y$ satisfying the equation we can obtain the congruency $77x equiv 2 mod13$.

As you commented that you don't know how to solve such equations, i'm gonna try and explain them without assuming you know the basic theory behind it.

There is a standard way using divison algorithm successively to obtain at least one solution of a solvable linear congruency. I recommend referring to any elementary text on number theory to understand existence and construction of a solution formally. However, it is clear that if we consider,

$77x equiv 2 mod13$.

Knowing that $13*6=78$ we can rewrite the equation above as, $78x-x equiv 2+11-11 mod13 implies -xequiv -11 mod13$ which literally translates to is the set of all integers which upon division by 13 leave 11 as a remainder.

these can be represented as $ 13t+11 $ where$ t in BbbZ$ and are all the possible values of $x$.From the original equation we get

$y=frac77(13t+11)-213$

The smallest positive value being $x=11$ at $t=0$ and corresponding value of $y$ being $frac77*11-213=65$

Also note that for any integer $t<0$ we have $ y<0$ and so $(11,65)$ is the least positive integral solution to the equation.