Vtyjkyuk

Is there a formula or solution for this :
"$n$ is a number that is a multiple of $x$ and when divided by $b$ has remainder of $y$".

Is there a fast way to finding this number?

The short answer is "No". Here is the long answer.

$n equiv 0 pmod x implies n = alpha x textfor some integer $alpha$$

$n equiv y pmod b implies alpha x equiv y pmod b$

Let $g = gcd(x, b)$. Then there is no solution unless $g mid y$.

If $g mid y$...

Let $X = dfrac xg, quad Y = dfrac yg, quad B = dfrac bg.$

Then $alpha X equiv Y pmod B$ and $gcd(X,B) = 1$.

So there exists integers $xi, beta$ such that $xi X + beta B = 1$.

Then $xi Y equiv alpha xi X equiv alpha - alpha beta B equiv alpha pmod B$.

That is $alpha equiv xi Y pmod B$

Example.

Find a number $n$ that is a multiple of $20$ and leaves a remainder of $24$ when divided by $34$.

We have $x = 20, quad y=24, quad b=34$.

$n equiv 0 pmod20 implies n = 20 alpha textfor some integer $alpha$$.

$n equiv 24 pmod34 implies 20 alpha equiv 24 pmod34$

Let $g = gcd(20, 34) = 2$. Then there is no solution unless $2 mid 24$. Which it does.

Let $X = dfrac202 = 10, quad Y = dfrac242=12, quad B = dfrac342=17.$

Then $10 alpha equiv 12 pmod17$ and $gcd(10,17) = 1$.

So there exists integers $xi, beta$ such that $10 xi + 17 beta = 1$.

beginarrayr
& 17 & 1 & 0 & 17 = (1)17 + (0)10\
-1 & 10 & 0 & 1 & 10 = (0)17 + (1)10\
hline
-1 & 7 & 1 & -1 & 7 = (1)17 + (-1)10\
-2 & 3 & -1 & 2 & 3 = (-1)17 + (2)10 \
hline
& 1 & 3 & -5 & 1 = (3)17 + (-5)10
endarray

We find $xi = 3, quad beta = -5$

So $alpha equiv -5 cdot 12 = -60 pmod17 = 8 pmod17$

and $n = 20alpha = 160$

CHECK:

$160$ is a multiple of $20$

$160 = 4 cdot 34 + 24$