should'nt we get 103 states without trap state.
101 --> used to identify 100 len string but we can't simply do transition on first two states on their next state as we also need to identify the first 3 letters.
say q0 --> q1 [on 1] and q1 --> q2 [on 0] and q2 --> q3 [on 1] then after normal on 0,1 to next state like q4 then q5... and so on to q101 as final state...but now what if 101th digit comes can we go back to q0 .. answer is no.. so we make another state q'1 and q101 --> q'1 [on 0,1 and this is 101th char] then q'1 --> q'2 [on 0,1 and this is 102nd char] and q'2 --> q3 [as after that is normal mod machine states..to q101]
so we need 101 + 2 more --> 103 states without trap state.