Nice question.

But although the explanation is correct adding up the 1 bits is doing it the hard way. 2014 is 2047 - 33, so the bits with are 0 must be the bit that represents 1 and teh bit that represents 32 - that's much easier. Or doing it by elimination as suggested by Igor and Marcia - 2014 is oddeven so the last bit is zero; it's bigger that 1536, so the 2nd bit (from the significant end) is not zero. That eliminates 3 of the 4 options, and is a lot easier than adding up the powers of two indicated by 10 bits to see if they come to 2014 even if you start by testing the correct value: if you did it that way and came to the correct value last you would end up adding up 3 sets of 10 numbers - no need to ad up teh 4th set once you know the others are all wrong - really a lot of work compared with the not-so-hard ways of doing it.