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

Home > Numbers
Zero to 150 in 2003 (Posted on 2002-11-29) Difficulty: 4 of 5
Take the digits 2, 0, 0 and 3. Make equations equating to all the integers from 1 to 150 using these digits according to the following rules:-

a) The above digits are the only digits to be used and no other digits should appear anywhere in the equation (except on the side where the answer will be).

b) Use of any mathematical symbols are allowed.

c) The digits 2, 0, 0 and 3 should appear in the given order in the equation. e.g - 0 + 2 + 3 + 0 = 5 is not acceptable.

d) When using the mathematical symbols try using the most simplest forms as much as possible.

See The Solution Submitted by Raveen    
Rating: 4.0526 (19 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): cheating | Comment 19 of 65 |
(In reply to re: cheating by levik)

yes i gathered it was cheating. the reason for that
specific question is that via the identity

-e^(iπ) = 1

you could generate anything.

by the way, 0! = 1 is perfectly valid. it is a boundary condition in the recursive definition of !
  Posted by Cheradenine on 2002-12-01 22:48:58

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 (8)
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