Democomputer Posted February 5, 2006 Share Posted February 5, 2006 Someone forwarded me a message with this website on it: http://www.mindreader.lx.ro There weren't a lot of people on the list, so I think that it may be pretty new. Anyway, it is a fake. How long will it take you to figure out how it works? Link to comment Share on other sites More sharing options...
Connor Posted February 5, 2006 Share Posted February 5, 2006 it has to be a multiple of 9, all the multiples of 9 are the same symbol *yawn* Link to comment Share on other sites More sharing options...
The Thing Posted February 6, 2006 Share Posted February 6, 2006 I'm predicting what it next shows in the crystal ball instead of it predicting what I'm thinking! T3h crystal ball hath been outwitted by t3h 1337 psych1c sk1llz0rz. No seriously, Connor's right. Multiples of 9. Mathematically, say you have a number 10x+y Add the digits: x+y Minus that from 10x+y, so 10x+y-x-y=9x. All multiples of 9, x depending on the tenth digit of your number. But that hardly matters when all of the multiples of 9 have the same symbol. Link to comment Share on other sites More sharing options...
SilentQ Posted February 6, 2006 Share Posted February 6, 2006 May be generalised to base [math]k[/math] saying that a number is equal to [math]\sum_{i=0}^nk^ix_i[/math]. [math]\sum_{i=0}^nk^ix_i - \sum x = \sum_{i=0}^n(k^ix_i-x_i)[/math] [math]= \sum_{i=0}^n(k^i-1)x_i[/math] [math]= \sum_{i=0}^n(x_i\sum_{j=0}^{i-1}(k-1)k^j)[/math] We may bring the constant factor of [math]k-1[/math] out to the front of both sums, giving [math](k-1)\sum_{i=0}^n(x_i\sum_{j=0}^{i-1}k^j)[/math], which is obviously then a multiple of [math]k-1[/math]. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now