The given number is


    .    (1)



    +---------------------------------------------------------------+
    |    Notice that the number    gives the remainder 1,      |
    |    when is divided by 9, for ANY positive integer m.          |
    +---------------------------------------------------------------+



It implies that the first addend in  (1)  gives the remainder 4, when is divided by 9.

The second addend in  (1)  is divisible by 9 without the remainder (OBVIOUSLY).


    It makes it OBVIOUS that the number (1) is divisible by 9 without the remainder.



Next, the number (1) has the form  49000 . . .005, where 

    - the leading digit 4 is located in some O D D position (2n+1), counting from the most right "ones" position;

    - the digit 9 is placed in the next (E V E N) position (2n);

    - and the last, "ones" digit 5 is located in the most right position number ONE (counting from the right);

    - while all the other digits are zeros.


Applying the "divisibility by 11 rule", we see that the number (1) has alternate sum of digits  4 - 9 + 5 = 0.


        Since the alternate sum of digits equal 0 (i.e. is divisible by 11),
        it means that the number (1) itself is divisible by 11,
        according to the "divisibility by 11 rule.



Thus the number (1) is multiple of 9 and 11 --- HENCE, it is multiple of 99.