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

Home > Just Math
Summing inverses II (Posted on 2012-03-16) Difficulty: 3 of 5
The sum of the reciprocal of the square root of all the positive integers up to n is denoted by F(n), that is:

F(n) = 1+1/√2 + 1/√3 +...+ 1/√n

Determine the maximum value of n such that the integer part of the base ten expansion of F(n) DOES NOT exceed 2012.

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

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution | Comment 2 of 12 |

Maximum such n is 1,013,505:

10   point 35
20   while T<=2012
30     PrevT=T
40     N=N+1:T=T+1/sqrt(N)
50   wend
60   print N-1,PrevT
70   print N,T

finds that F(1013505) ~= 2011.9998512 and F(1013506) ~= 2012.0008445.

Specifically, the output is:

Words for fractionals 35 (Decimals for display 168)
 1013505         2011.9998512067347756246902699409366862982247383561279030457342
16555503796507039814234377477881798284087714660672885032442019561678422057764197
776286526187964973375524958049
 1013506         2012.0008445213783854834399657352730035024527421198563401973979
52251518173653291207749387862141705733740828106568073177121727135534711192409921
224060824796931311757117336879


  Posted by Charlie on 2012-03-16 12:46:14
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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information