, 2015

Consider the value of $U$ in the diagram shown. [[IMAGE0]] Since a 3 already occurs in the second row, then $U$ cannot equal 3 (each of the numbers $1,2,3,$ occur only once in each row). Since a 1 already occurs in the third column, then $U$ cannot equal 1 (each of the numbers $1,2,3,$ occur only once in each column). Since $U$ cannot equal 3 or 1, then $U=2$. Therefore, a 2 and a 3 already occur in the second row and so $X=1$. At this point, a 2 and a 1 already occur in the third column and so $Y=3$. The value of $X+Y=1+3=4$.