SOLUTION: In the sequence 457, 473, 487, 506, ... each term is equal to the previous term added to the sum of its digits. A number NOT in the sequence is a) 1864 b) 1949 c) 3466 d) 45

First term, 457, has the sum of digits 4+5+7 = 16 = 1 (mod 3).

So, according to the divisibility rule by 3, the number 457 itself is 1 (mod 3).

    
    Therefore, the second term is 1+1 = 2 (mod 3).

    It means that the sum of the digits of the second term is 2 (mod 3).


It implies that the third term is 2+2 = 4 = 1 (mod 3)


It implies, in turn, that the fourth term is 2 (mod 3).


Thus we see that the given sequence of numbers is CYCLIC 1, 2, 1, 2 by modulo 3,

and it is true NOT ONLY for the first four terms, but IT IS TRUE for all other consecutive terms.


Now look at the given optional numbers.

Using the rule of divisibility by 3, you can easily check that the ONLY NUMBER of them 
which is DIVISIBLE by 3 is the number 4569.


So, we can conclude that this number, 4569, DOES NOT BELONG to the given sequence.

                 = 1 (mod 3)

     = 1 + 1     = 2 (mod 3)

     = 2 + 2 = 4 = 1 (mod 3)

     = 1 + 1     = 2 (nod 3)

While Ikleyn's solution is correct, I am not convinced of the validity of her
assumption of the generalization:

Edwin