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

Home > Just Math
Count the palindromes (Posted on 2015-11-11) Difficulty: 2 of 5
N is both a a palindrome and a sum of eleven consecutive positive integers.

How many possible values for N exist below 80,000?
Please provide the lowest and the highest samples.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Simplifying the question | Comment 5 of 7 |
(In reply to re: Simplifying the question by Ady TZIDON)

Oh, I don't solve perplexus problems with a computer program (except sometimes I do use Excel).  I was thinking about how to tackle it analytically.

Recognizing that all and only numbers divisible by 11 are the sum of eleven consecutive integers makes it trivial to analytically count the qualifying palindromes with an even number of digits (because they all qualify).  And it may help to count the palindromes with an odd number of digits, although I am not there yet. 

  Posted by Steve Herman on 2015-11-11 14:36:00
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (2)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information