Problem (Number Theory):
There exists a digit $Y$ such that, for any digit $X$, the seven-digit number $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is not a multiple of 11. Compute $Y$.

Solution:
Consider the ordered pairs of digits $(X, Y)$ for which $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is a multiple of 11 . Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a multiple of 11 . Because $1+3+5+7=16$, the sum of the remaining digits, namely $2+X+Y$, must equal 5 or 16 . Thus $X+Y$ must be either 3 or 14 , making $X=3-Y$ (if $Y=0,1,2$, or 3 ) or $14-Y$ (if $Y=5,6,7,8$, or 9 ). Thus a solution $(X, Y)$ exists unless $Y=4$.

Final answer: 4