Modulus operation to find unknown If the $5$ digit number $538xy$ is divisible by $3,7$ and $11,$ find $x$ and $y$ .
How to solve this problem with the help of modulus operator ?

I was checking the divisibility for 11, 3:
$5-3+8-x+y = a ⋅ 11$ and $5+3+8+x+y = b⋅3$ and I am getting more unknowns ..
 A: This is a solution with divisibility properties only:
$$11\mid 538xy \implies 11\mid y-x+8-3+5 = y-x+10$$
so $$11\mid y-x-1\implies y-x-1=0$$
Also $$3\mid 538xy \implies 3\mid y+x+8+3+5 = y+x+16$$ so $$3\mid  y+x+1 = 2(x+1)\implies x\in\{2,5,8\}$$
Now try for each pair $(x,y)\in \{(2,3),(5,6),(8,9)\}$ if $7\mid 538xy$...
A: From modulus 11,
$$
53800 + 10x + y \equiv 5 - 3 + 8 - x + y \equiv -1-x+y \equiv 0 \pmod {11}\\
\implies y\equiv 1+x \pmod {11}
$$
but $y$ and $x$ are digits, so $0\le y\le 9$ and $1\le 1+x \le 10$, so it must hold that $y=1+x$.
From modulus 3,
$$
53800 + 10x + y \equiv 5 + 3 + 8 + x + (x+1) \equiv 2+2x \equiv 0 \pmod {3}\\
\implies x\equiv -1 \pmod {3}
$$
so $x=2,5,8$.
Eventually, from modulus 7,
$$
53800 + 10x + y \equiv 5 + 3x + (x+1) = 6+4x \equiv 0 \pmod {7}\\
\implies x\equiv 2 \pmod {7}
$$
so $x=2,9$.
The only choice is thus $x=2$, $y=3$.
A: As $53800\equiv208\pmod{231},$
$$10x+y=231z-208$$
Now $99\ge10x+y\ge0$
$$99\ge231z-208\ge0$$
$$\implies 2>307/231\ge z\ge208/231>0$$
$$\implies z=1$$
Consequently $$10x+y=23\iff y=23-10x$$
Can you take it from here?
