Don't forget that as the two Ds and one M appear in three entries then their cells must either be
(a) unches in three separate lights, or
(b) two unches, with the third being in one of the two lights concerned.
Therefore in the case of m with 5 digits we need to identify a light with 5 cells whose first cell is either unchecked or intersects with a light whose first cell is unchecked. Given that the intersection of 7a (9 cells) and 4d (5 cells) cannot contain an L, then there is only one 5-celled light that is available for m.