IDM Mailing Archive

2001-04-05 21:03Brian MacDonald* Think of any number that has at least two digits. (The number cannot be all zeroes) * Ad

From:: Brian MacDonald <brianm@kuci.org>
To:: <idm@hyperreal.org>
Date:: Thu, 5 Apr 2001 14:03:18 -0700 (PDT)
Subject:: [idm] Puzzle for the chamber

* Think of any number that has at least two digits. (The number cannot be all zeroes) * Add up all the digits in that number * Subtract this total from the original number * Scratch out one of the digits from this result, as long as it is NOT a zero. * Give me the list of the remaining digits. (If there are no remaining digits, you must tell me "there are no remaining digits") ..and.. <HEAVY REVERB> I WILL TELL YOU WHAT DIGIT YOU SCRATCHED OUT!!! </HEAVY REVERB> What is my secret? And why does it work? ======================================================================= Brian MacDonald <brianm@kuci.org> "Capital punishment turns the state into a murderer. But imprisonment turns the state into a gay dungeon-master." -- Emo Philips ======================================================================= --------------------------------------------------------------------- To unsubscribe, e-mail: idm-unsubscribe@hyperreal.org For additional commands, e-mail: idm-help@hyperreal.org

2001-04-06 16:37Matthew KorfhageI'm on digest, so this was probably already responded to, but: Oh come on, end the geek pa

From:: Matthew Korfhage <pomomofo2000@hotmail.com>
To:: <idm@hyperreal.org>, <brianm@kuci.org>
Date:: Fri, 06 Apr 2001 09:37:57 -0700
Subject:: [idm] Puzzle for the chamber

I'm on digest, so this was probably already responded to, but: Oh come on, end the geek parade. Didn't we all figure this sort of thing out watching Square One when we were little? Simple algebra: (10x + y) - (x + y) = 9x All you have to do, when given the number, is add whatever number would make the sum of the digits divisible by 9. Here, prove this instead: x^n + y^n = z^n has no solutions for n >=3 M. Brian MacDonald <brianm@kuci.org> wrote:

quoted 20 lines * Think of any number that has at least two digits. (The number cannot be

>* Think of any number that has at least two digits. (The number cannot be >all zeroes) > >* Add up all the digits in that number > >* Subtract this total from the original number > >* Scratch out one of the digits from this result, as long as it is NOT a >zero. > >* Give me the list of the remaining digits. (If there are no remaining >digits, you must tell me "there are no remaining digits") > >..and.. > ><HEAVY REVERB> >I WILL TELL YOU WHAT DIGIT YOU SCRATCHED OUT!!! ></HEAVY REVERB> > >What is my secret? And why does it work?

_________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com --------------------------------------------------------------------- To unsubscribe, e-mail: idm-unsubscribe@hyperreal.org For additional commands, e-mail: idm-help@hyperreal.org

2001-04-06 17:17laermOn Fri, 6 Apr 2001, Matthew Korfhage wrote: > Here, prove this instead: > > x^n + y^n = z^

From:: laerm <laerm@soulfood.org>
To:: <idm@hyperreal.org>
Date:: Fri, 6 Apr 2001 13:17:27 -0400 (EDT)
Subject:: Re: [idm] Puzzle for the chamber
Reply to:: [idm] Puzzle for the chamber

On Fri, 6 Apr 2001, Matthew Korfhage wrote:

quoted 3 lines Here, prove this instead:

> Here, prove this instead: > > x^n + y^n = z^n has no solutions for n >=3

http://www.best.com/~cgd/home/flt/flt08.htm micah stupak /"\ laerm@soulfood.org ascii ribbon campaign \ / international bright young kook against HTML email X / \ --------------------------------------------------------------------- To unsubscribe, e-mail: idm-unsubscribe@hyperreal.org For additional commands, e-mail: idm-help@hyperreal.org

Re: [idm] Puzzle for the chamber