# Building a BCD to 7-segment using 3x8 decoder

So.. I know how to build a BCD to 7-segment using 3x8 decoder but the question provides 4-digit binaries and asks to only input 3 of those.. my question is that won't two different digits be regarded as one.. If that's confusing, here's the question:

Consider a BCD digit $$\X_3X_2X_1X_0\$$ with $$\X_3\$$ being the MSB. Design a BCD to 7-segment converter using two 3x8 decoders and minimum number of gates, such that $$\X_3\$$, $$\X_2\$$ and $$\X_0\$$ are applied at the inputs of the decoders. The converter should be compatible with a common cathode display.

So.. in this case 0 is 0000 and 2 is 0010 but if we don't input $$\X_1\$$, we're only giving out 0000 in both cases.. so how would the decoder know which is 0 and which is 2? Do I use $$\X_1\$$ as enabler? And we're not supposed to use anything other than 3x8 decoder, a 7 segment and minimal gates. I'm really confused here. Please help!

By my logic, $$\X_{1}\$$ must be used as an enable on the decoders, directly to one and through a not gate to the other. This is because without using it as an enable, both decoders will have the same inputs and thus the same outputs, which is redundant.
I think the trick to this question is the realization that since $$\X_{3}\$$ is NOT the enable (A more common case), then every two decimal digits will produce a high output on a different decoder. Below are two images showing the decimal BCD codes that will produce a high value on each output.