If a 10-digit number M30348462N is divisible by both 8 and 11, then what is the value of $M^2+N^2-18$ ?

7

10-digit number M30348462N is divisible by both 8 and 11

Number divisible by 8 if last 3 digits are divisible by 8: Thus 624 is divisible by 8, N=4

Number divisible by 11 if difference of numbers obtained by sum of alternate digits is divisible by 11: (M+0+4+4+2)-(3+3+8+6+N) = (M+0+4+4+2)-(3+3+8+6+4) = M+10-24 = 11x

M-14 = 11x

M is 3. 

M²+N²−18 = 3*3 + 4*4 -18 = 9+16-18 =7

The correct answer is option (3) : 7