All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Alphabet Repeat Pattern Poser (Posted on 2010-10-18) Difficulty: 2 of 5
The string abbcccddddeeeee.... continuously repeats - with a repeating once, b repeating twice and so on, such that after the final z, the letters abbcccddddeeeee.... begin again.

What will be the 2010th letter in the above pattern?

As a bonus, what will be the 2010th letter in the above pattern, if instead - after the final z, the next a repeats 27 times, then b repeats 28 times and so forth?

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution a shortcut spoiler Comment 3 of 3 |

The cycle's length is   ( 1+26)*26/2=351

2010 mod 351= 255

lowest INTEGER n to satisfy n*(n+1)>2*255    IS 23

23th letter is w.


For bonus question -same reasoning:

lowest INTEGER n to satisfy n*(n+1)>2*2010    IS 63

63 mod 26=63-52=11

11th letter is k.



  Posted by Ady TZIDON on 2010-10-18 14:53:27
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information