SOLUTION: Alice replaces each of the 2008 numbers 6, 7, 8, …, 2012, 2013 with the sum of its digits. Brian replaces each of Alice’s numbers with the sum of its digits, and Colin replac

A)  After Alice completed her operations, she obtained 2008 numbers A(n) ranged from 1 to 28 inclusively 

    (notice that 1 is the sum of digits of the number 10, while 28 is the maximum of the sums of the digits to each 

     of the numbers from 6 to 2013; namely, 28 is the sum of digits of the number 1999).


    Again, when Alice completed her operations, she obtained 2008 numbers A(n) each of which is between 1 and 28 inclusively. 



B)  After Brian completed his operations, he obtained 2008 numbers B(n) ranged from 1 to 10 inclusively 

    (notice that 10 is the sum of digits of the number 28 from n.A) above).


    Again, when Brian completed his operations, he obtained 2008 numbers B(n) each of which is between 1 and 10 inclusively. 



C)  After Colin completed his operations, he obtained 2008 numbers C(n) ranged from 1 to 9 inclusively 

    (notice that 9 is the sum of digits of the number 9).


    Again, when Colin completed his operations, he obtained 2008 numbers C(n) each of which is between 1 and 9 inclusively. 



          It gives me an idea to consider the numbers at each step by the modulo of 9.

          The second idea is that taking the sum of the digits of any integer number gives the number which has the same remainder 
                                                                                
          when divided by 9  as the original number n has.  //  It is well known "rule of divisibility by 9" in the extended form.

          Combination of these two ideas provides the solution as follows.



At step A), each number n    modulus 9 is equal to A(n) modulus 9:           n (mod 9) = A(n) (mod 9).

At step B), each number A(n) modulus 9 is equal to B(A(n)) modulus 9:        n (mod 9) = A(n) (mod 9) = B(A(n)) (mod 9).

At step C), each number B(A(n)) modulus 9 is equal to C(B(A(n))) modulus 9:  n (mod 9) = A(n) (mod 9) = B(A(n)) (mod 9) = C(B(A(n)) (mod 9).


Thus finally  n (mod 9) = C(n) (mod 9),


or, in other words, after step C), the obtained number C(n) has the same remainder when divided by 9 as the original number n has.


It means that the final sequence of numbers C(n) is sequential row of remainders n (mod 9), starting from 6 and ending by 2013 (mod 9) = 6:


Thus this row is  6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, . . . . , 1, 2, 3, 4, 5, 6.


So cycles of the length 9 from 6 to 5 are repeating cycles, and the last number "6" is the  first  AND THE LAST number  after the last loop.


ANSWER.  The number which Colin obtains most frequently is "6".