2. Let the six-digit number $N=\overline{x 1527 y}$ be a multiple of 4, and $N$ leaves a remainder of 5 when divided by 11. Then, $x+y$ equals ( ).
(A) 8
(B) 9
(C) 10
(D) 11
2. B.
Since $N$ is a multiple of 4, $\therefore \overline{7 y}=70 x y$ is a multiple of 4,
$\therefore y=2$ or $y=6$.
If $y=2$, then $\overline{15278}$ is a multiple of 11,
$(x+5+7)-(1+2+3)=x+1$ is a multiple of 11,
no solution.
If $y=6$, then $\overline{x 15271}$ is a multiple of 11, i.e., $(x+5+7)-(1+2+1) \approx x+8$ is a multiple of 11, then $x=3$. Thus, $x+y=9$.