Topic: How does this work ?
 Message: Posted by: DavidF (May 17, 2006 11:32AM)
Has anyone seen this web site ? http://digicc.com/fido/ Im sure there is a simple answer to this puzzle but I cannot work it out.I don't want the answer I would just like to know if anyone knows how it works
 Message: Posted by: landmark (May 17, 2006 03:08PM)
Cute website. It's a very old math effect. I don't think anyone will mind if I tell you that when you do that kind of subtraction, the digits must add up to a multiple of nine. Lots of other fun math stuff in this forum, check it out.

Jack Shalom
 Message: Posted by: Doctor Whoston (May 17, 2006 03:10PM)
I do!
DW
 Message: Posted by: Slim King (May 17, 2006 07:46PM)
Relax there Jack! That's my favorite secret...LOL
 Message: Posted by: TomasB (May 18, 2006 02:38AM)
That is definitely my favourite way to force a number to be a multiple of nine.

Here's a hint to why it works (n>m): a*10^n-a*10^m=a*10^m(10^(n-m)-1)=9*a*10^m*111...111 where the last number has n-m ones in it.

/Tomas
 Message: Posted by: landmark (May 18, 2006 02:58PM)
[quote]Here's a hint to why it works (n>m): a*10^n-a*10^m=a*10^m(10^(n-m)-1)=9*a*10^m*111...111 where the last number has n-m ones in it. [/quote]

That's easy for you to say . . . :)
 Message: Posted by: roi_tau (Jun 23, 2006 10:16AM)
What is it about the nine that so many tricks involve him.

will it work with 7 in a octali basis?

Have fun
ROi
 Message: Posted by: Nir Dahan (Jun 23, 2006 11:03AM)
[quote]
On 2006-06-23 11:16, roi_tau wrote:
What is it about the nine that so many tricks involve him.

will it work with 7 in a octali basis?

Have fun
ROi
[/quote]

yes

Gardner (who else) has written about it.
N