Problem 192

Define a super-special number to be a number whose decimal expansion only consists of $0$'s and $1$'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\text{1 7 =0.142857142857}$. This can be done in $8$ numbers if we take
$$\text{0.111111 , 0.011111 , 0.010111 , 0.010111 , 0.000111 , 0.000101 , 0.000101 , 0.000100}$$
Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$.) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $\text{0.1428}$, there must be $8$ super-special numbers, so the answer is $\boxed{\textbf{(B)}\ 8}$.