better explanation
let xi ------> coin weight in bag i
hence xi can be 10 or 11
basic crux of problem is
odd +even=odd ---(0)
even +even =even--(1)
x1 + 2*x2 + 4*x3 +8*x4 + 16*x5 =323 --(2)
from (0), we get
x1 is odd since other terms are of form 2*x.
hence x1 can only be 11.
2*x2 + 4*x3 +8*x4 + 16*x5 =323 -11=312 --(3)
dividing (3) by 2
x2 + 2*x3 +4*x4 + 8*x5 =156 --(4)
here 156 is even and 2*x3 +4*x4 + 8*x5 is even,hence from (1), x2 must be even
i.e x2 = 10
so,
2*x3 +4*x4 + 8*x5 =156 -x2 =156-10=146 --(5)
dividing (5) by 2
x3 +2*x4 + 4*x5 =73 ---(6)
similarly continuing
x3 is odd hence x3=11
After taking (eq(6)-x3)/2
x4 + 2*x5 =31
x4 is odd hence x4 =11
implies x5=10
xi={11,10,11,11,10}
Answer is 1*3*4=12 . correct is (B)