Modulus operation to find unknown

Question

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 ?

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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 ..

Answer 1

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 $x=2$, $y=3$.

Answer 2

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 $7\mid 538xy$...

Answer 3

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?