Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?

You are wrong considering $x,y\in\left[0,6\right]$.
$30+x+y$ is divisible by $9$. That's correct, but it may be $30+x+y=36$ or $30+x+y=45$. It cannot be $30+x+y=54$, since the greatest value of $\left(x+y\right)$ is $9+9=18$.
From these two options you have $x+y=6$ or $x+y=15$. Remember this.
From the division by $11$, you achieved that $12+x-y$ must be divisible by $11$. That's also correct.
Note that you cannot get $12+x-y=22$, since the maximum value $\left(x-y\right)$ can reach is $9-0=9$, so you have to use negative values of $\left(x-y\right)$ like $12+x-y=11$, i.e, $x-y=-1$. It means $x$ and $y$ are two consecutive numbers where $y=x+1$; it also means that one of them is odd and the other is even, because of this, there is no way $x+y=6$, it must be $x+y=15$.
Now, you have a linear system
$$\begin{cases}x+y=15\\x-y=-1\end{cases}$$
Solve this and you will have your answer.
