Recent content by vamfun

One could argue that the opening statement is included in the logic set. So there would now be 11 statements. the opening is true, 1 is F and 2 would then be F... But my friend..you must now solve the puzzle under thiis assumption or are you asing us to do the work:)

This could be worked with Baysian conditional prob. Define events: N=no disease P(N)=1/10000 Y=have disease P(Y)=1-P(N) Y'=test positive P(Y')=P(Y'/N)*P(N)+P(Y'/Y)*P(Y) N'=test negitive WRONG TEST RESULT P(Y'/N) =1/100= P(N'/Y) note: not always equal but assumed here. GOOD TEST...

Another answer with logic This is my solution and logic: ***1f,2t,3t,4t,5f,6t,7t,8f,9f,10f with x=420 *** Logic: Find x = minimum admissible number. If 1 is T then 2 is a paradox so 1F. 1F implies 9f and 10f. 2T if 1F 6T else logic paradox. Implies 7T or 8T 9F implies num of divisors...

Contradiction? It seems to me that statement 10 is true now since there are no three consecutive trues.

Hi all, I wanted to post this comment re the second teasor. Is there a way to have it appear in a relavent place rather than here? "I agree with this answer... But it must allow 0 hairs. I would reason a little differently. To get colony started 1st man must have 0 hair to meet...

Thanks... not sure how to do it but here goes.

Ok, I am with you now. If a coin is either light or heavy then an additional weighing is needed against the reference coins for most of the cases (excluding n=10,11 and 12) I will update the table when I get a chance. Sorry for the confusion. PS I had worked this problem before when...

You're right! Correction: The above assumed that one coin was heavy. This is acutally unknown , so an additional weighing is necessary against known good coins for many of the cases (excluding n=10,11 and 12 at least) Sorry

I agree that it is a groups of three problem as described by denverdoc. Extending mathmatically we can state the problem solution for any number of coins: n= number of coins w= number of weighings required find minimum w such that 3 exp (w) >= n...

A solution T=truth teller L= lier T'=mistaken truth teller It is assumed that T and L both see T' as a T (they don't know of mistake) possible adjacent seating for mistaken truth teller T' are LT'L and TT'T all other seats must be LTT LTT LTT LTT LTT etc so a solution is...