If n is a positive integer, define n! =n×（n-1）×（n-2）×… X 3 × 2 × 1, let m = 1! +2！ +3！ +4！ +… +2003!+2004！ Then the sum of the last two digits of M is______ ．

If n is a positive integer, define n! =n×（n-1）×（n-2）×… X 3 × 2 × 1, let m = 1! +2！ +3！ +4！ +… +2003!+2004！ Then the sum of the last two digits of M is______ ．

The last two digits of ∵ 10! And above are 0, the last two digits of the sum of ∵ 10! To 2004! Are 00, ∵ M = 1! + 2! + 3! + 4! + +The sum of the last two digits of 2003! + 2004! Is the sum of the last two digits of 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!. the sum of the last two digits of ∵ 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 409113, and the sum of the last two digits of ∵ m is 1 + 3 = 4