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

Home > Just Math
Get one half by squaring (Posted on 2004-09-20) Difficulty: 4 of 5
Can you solve the following equation?

= 1/x + 1/y +...+ 1/z

All variables must be different, positive integers, and there must be a finite number of terms.

See The Solution Submitted by Federico Kereki    
Rating: 4.2500 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Cannot do with 6 | Comment 30 of 31 |
(In reply to Less than 9? by e.g.)

You can't do it with 6 numbers either.

Call them a,b,c,d,e,f where a<b<c<d<e<f

As seen before, a has to be 2.

We can also see that b has to be 3, because if we take 2,4,5,6,7,8 then we get a number less than 1/2.

By the same argument, c has to be 4, d has to be 5, and e has to be 6. So that leaves us only to determine f.

Plugging the above numbers in, we get the sum for a through e is 0.4913888..., which doesn't leave room for f to be any whole number.

By a similar argument, if we assume there are 7 numbers: a,b,c,d,e,f and g, we know that a=2, b=3, c=4, and d=5.

Edited on November 9, 2004, 10:50 am
  Posted by Avin on 2004-11-09 10:46:57

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 (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information