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 Discussion: Traffic Jam Applet tool Topic: New Freind Related Item: http://mathforum.org/mathtools/tool/10/

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 Subject: RE: New Freind Author: Steve Date: Apr 7 2004
I am unfamiliar with the problem, but I am assuming that knights always tell the
truth and knaves always lie??  If that's the case, then W and Y must be knights,
and the remainder must be knaves.  I just experimented with each possible number
of knights, going from 0 up to 6.

On Apr 07, 2004, Kamali wrote:

Can anyone answer this easy deduction?!

You meet a group of people who speak to you as follows:
U says: none of us is a knight
V says: at least three of us are knights
W says: at most three of us are knights
X says: exactly five of us are knights
Y says: exactly two of use are knights
Z says: exactly one of us is a knight
Which are knights and which are knaves?

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