Question 62
Find positive integers $a_{1} \lt a_{2} \lt \ldots \lt a_{2010}$ such that
$$
a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} = (2011!)^{2010}.
$$
$$
a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} = (2011!)^{2010}.
$$