|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
24 Mar 2019, 23:13
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:18) correct
47%
(02:14)
wrong
based on 158
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 29
(B) 55
(C) 85
(D) 133
(E) 250
Archit3110
Major Poster
Updated on: 30 Mar 2019, 00:42
Originally posted by Archit3110 on 24 Mar 2019, 23:23.
Last edited by Archit3110 on 30 Mar 2019, 00:42, edited 1 time in total.
Last edited by Archit3110 on 30 Mar 2019, 00:42, edited 1 time in total.
Kudos
Bookmarks
Bunuel
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
28 Mar 2019, 07:02
Kudos
Bookmarks
Archit3110
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
