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

Home > Just Math
Point to 2012 (Posted on 2012-04-04) Difficulty: 3 of 5
Determine the minimum value of a positive integer N such that the four digits immediately following the decimal point in the base ten expansion of √N is 2012.

*** For an extra challenge, solve this puzzle without using a computer program.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Non computer solution (pretty much) | Comment 3 of 7 |
We seek N such that for some integer a
(a+.2012)^2 < N < (a+.2013)^2

For simplicity lets look for N ≈ (a+.20125)^2
N ≈ (a + 161/800)^2 = a^2 + 161a/400 + 25921/600000
Since a^2 is already an integer we need
161a/400 + 25921/640000 ≈ b for some integer b
Solving for a gives
a ≈ 400b/161 - 161/1600

So what is sought is a value of b that makes a close to an integer. 
A quick chart shows
b = 23, a=57.042 which is close to 57
57.20125^2 = 3271.983
√3272 = 57.201398 close but not quite

b = 25, a=62.011 which is even closer to 62
62.20125^2 = 3868.99955
√3869 = 62.20128616 bingo!

The solution is N = 3869

  Posted by Jer on 2012-04-04 13:38:30
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (12)
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