Berlin92 wrote:
Archit3110 wrote:
Bunuel wrote:
The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?
(A) 29
(B) 55
(C) 85
(D) 133
(E) 250
first term = 2005
2nd term = 133
third term = 55
4th term = 250
5th term= 133
cyclicity is of 3
2005th term from 1 = 2004 term = 4th term = 250 IMO E
I established the same sequence but then I counted wrong. I always do that, could you try to explain to me how we can establish the 2005th term? I don't understand your reasonening above
Berlin92once you have understood the cyclicity of the terms viz 3 in this case , then simply divide the cyclicity term with the desired nth no. i.e 2005 in this case , but since 1st term ie 2005 is not coming up again the 2004 term from 1st is to be determined
so now 2004/3 ; remainder 0 ; ie. 250 will be the value of 2005th term..