First contest problem I downloaded a contest and worked the first problem which is:

There exists a digit Y such that, for any digit X, the seven-digit number 1 2 3 X 5 Y 7 is not a multiple of 11.
  Compute Y.

My attempt:
$$\begin{align}
1+3+5+7&=2+X+Y \\
16&=2+X+Y \\
14&=X+Y \\
\implies\quad Y&=14-X \\
\end{align}$$
The values of Y for which the number can be divided by $11$ are: $5,6,7,8,9.$
How would I deduce that the only value of Y for which the number can't be divided by $11$ is only $4$?
 A: It looks like you're almost done. We can express the problem, from where you've left off, to find a $Y$ such that the equation
$$Y\equiv 14-X\pmod{11}$$
has no solution. Notice that we have to work mod 11 - the two sums you list don't have to be equal for a number to be divisible by $11$ - the sums may differ by a multiple of $11$. Doing this is relatively easy though - we can simply list the values the right hand side could obtain. These are, in particular:
$$14,\,13,\,12,\,11,\,10,\,9,\,8,\,7,\,6,\,5$$
where $X$ ranges from $0$ to $9$. Notably, none of these values are congruent to $4$ mod $11$, so if we set $Y=4$, we cannot match that value on the right side. Thus, $Y=4$ is the answer.
A: You say $Y =14-X$ if the number is divisible. Because X is a digit (in base ten), it ranges from $0$ to $9$; thus $Y$ ranges from $14-0=14$ to $14-9=5$. But note that $14$, $13$, $12$, $11$, and $10$ are not accepted values for a single digit. If the number has to be divisible by $11$, $Y$ must be $11$ less than the numbers above. Thus $Y$ can be 3, 2, 1, 0, -1 (this is still invalid), or from 5 to 9 as before.
To make the number not divisible, $Y$ must therefore be 4.
