| Manifesto | Bookstore | Contents | Amazon store | Term index | What changed? | Contact | Recommend |
Date: Thu, 24 Oct 96 16:01:48
From: Tobias Danielsson
Hi,
Nice pages.
Here's another variant of arithmetic problem I've seen somewhere, just for information.
Make the following statements true by adding operators and parenthesis (and square root and factorials), but no extra figures.
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
0 0 0 = 6
Of course there are alternative answers to most (or all) of these. Personally I think the 8-solution is neat. Could it be possible to do this with a different answer than 6? I suppose the 0's set some boundaries.
Regards,
Tobias
Copyright © 1996-2009 Alexander Bogomolny
| 34387996 |  |
