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

Home > Numbers
Ring in the New - again and again and again.. (Posted on 2019-01-01) Difficulty: 3 of 5

Let a be the smallest prime greater than 1000.

Let b=sqrt((2a+1)^2+4)+2a+1

Divide b by 2, and express the result as a continued fraction.

Happy New Year.

No Solution Yet Submitted by broll    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Answer | Comment 1 of 2
a=1009
b=sqrt((2(1009)+1)^2+4)+2(1009)+1=sqrt((2018+1)^2+4)+2018+1=sqrt(2019^2+4)+2019=sqrt(4076361+4)+2019=sqrt(4076365)+2019
b/2=(sqrt(4076365)+2019)/2=[2019, 2019, 2019, 2019, 2019, ...]


  Posted by Math Man on 2019-01-01 14:11:19
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information