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So, I know that $495 = 5times 9times 11$. 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 $11$?

$30+x+y$ doesn't have to equal 9; it has to equal a multiple of 9. So it could be $36$ or $45$ or $54ldots$. And $x$ could be $7$, if $y$ is 8, for example. But they can't be anything at all; if $x$ is 7 then $y$ must be $8$; no other value of $y$ works. And if $x$ is 3 then $y$ must be $3$ also. So make a table that shows for each $x$, what can $y$ be, if $30+x+y$ is to be a multiple of 9.

Then do the same for the $12+x-y$ must be divisible by $11$: if $x$ is 1, then $y$ must be $2$, and so on.

Then see which pairs are in both tables.

$30+x+y$ must be a multiple of 9 which leads to:$$30+x+y=9a$$$$x+y=9a-30tag{1}$$
$12+x-y$ must be a multiple of 11 which leads to:$$12+x-y=11b$$$$1+x-y=11ctext{ where c=b-1}tag{2}$$
the biggest difference between two digits is 9 therefore the maximum value of $x-y$ is 9 which leads to (from (2)):$$11cle1+9le10$$therefore $c=0$ which leads to:$$y=x+1tag{3}$$substitute (3) into (1) to get:$$2x+1=9a-30$$$$2x=9a-31$$Hopefully you can solve from here.

You are wrong considering $x,yinleft[0,6right]$.

$30+x+y$ 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 $left(x+yright)$ 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 $left(x-yright)$ can reach is $9-0=9$, so you have to use negative values of $left(x-yright)$ 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
$$begin{cases}x+y=15\x-y=-1end{cases}$$

Solve this and you will have your answer.