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 $11$?
 A: $30+x+y$ doesn't have to equal 9; it has to equal a multiple of 9.  So it could be $36$ or $45$ or $54\ldots$.  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.
A: $30+x+y$ must be a multiple of 9 which leads to:$$30+x+y=9a$$$$x+y=9a-30\tag{1}$$
$12+x-y$ must be a multiple of 11 which leads to:$$12+x-y=11b$$$$1+x-y=11c\text{ where c=b-1}\tag{2}$$
the biggest difference between two digits is 9 therefore the maximum value of $x-y$ is 9 which leads to (from (2)):$$11c\le1+9\le10$$therefore $c=0$ which leads to:$$y=x+1\tag{3}$$substitute (3) into (1) to get:$$2x+1=9a-30$$$$2x=9a-31$$Hopefully you can solve from here.
A: 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.
A: I do not follow "So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are..." and I think that is the source of trouble.
If $9$ divides $30+x+y$, then either $30+x+y=36$ or $30+x+y=45$.
If $11$ divides $12+x-y$, then for sure $12+x-y=11$.
Add one of the first two equations to the second. Either $42+2x=47$ or $42+2x=56$. The first can't be, since $47$ is odd. So it's the second situation, and $x=7$. Then $y$ must be $8$.
A: Notice $495 = 5\times 99$. It will be easier to look at everything modulo $99$.
$$0 \equiv \overline{273x49y5}
\equiv 27 + \overline{3x} + 49 + \overline{y5} 
\equiv \overline{yx} + 111
\equiv \overline{yx} + 12 \pmod{99}
$$
This leads to $\overline{yx} = 87 \implies (x,y) = (7,8)$.
