Turing machine halting in at least 100 steps on all inputs L = {&lt; M >  $\mid$ L(M) = M halts on all inputs in atleast 100 steps }.

Answer to this question is given undecidable in my book without any explanation but I think it is decidable.



$\overline{L}$ =  {&lt; M > $\mid$ L(M) = M halts on some input in atmost 100 steps }

For the above language, we can feed strings of length ranging from 1-100 (finite amount of strings)  and run Turing machine for finite amount of steps (100) on each string and check if at least one of them halts in atmost 100 steps.

So, as the complement of L is recursive, L should also be recursive and decidable according to me.

Where am I going wrong ?
You have not correctly computed $\overline{L}$. It contains 


machines which halt in less than 100 steps for some input (which is a finite set of possibilities, as you say), and
machines which don't ever halt for some input.


The second part requires a solution to the halting problem, so $\overline{L}$ is not decidable (nor recursively enumerable).