Recent content by LarrrSDonald

In principle, zero is probably the correct guess if everyone plays perfectly - this would also yield a tie between everyone because everyone will be correct. In reallity though, I'd probably venture to guess slightly more then zero, perhaps even one. As eluded to, all it takes is one person...

My bid is: \omega^{\omega^\omega} Though my grasp of transfinite numbers is shakey at best, I think it's bigger then "infinity" (i.e. plain omega) and 3^infinity :-).

I've pondered the same "overanalytical" thing before when faced with similar questions. It certainly doesn't take much to realize where the problem means to aim with it, but it is a little dubious. Even with one wife, {6,6,1} could be a posibillity, in fact with the way my sisters and I were...

I assume it would be solvable by some combination of iffs (if and only if) or "if I asked you X, what would you say" type questions. Not having heard the question before, I don't understand the rules. "at" in 1. and 2. could mean "living at" or "currently at", 3. I'm not even sure what it might...

<joins in the applause> Very nice. I still feel that given continuous pieces of one color each, five still ought to be impossible as per the four color problem. Yes, I agree, the original problem is stated a bit haphazardly and doesn't prevent that and possibly other things. I figured that this...

I was thinking the answer was supposed to be the onomatopoetical "Hmm". But then perhaps I'm reading too much into the last part.

Sorry about not seeing the sticky at the time. I picked e9e9e9 by simply checking what the background rendered to, though with color profiles, websafeness and what not it may well be something else when it starts out (I just hit printscreen, tabbed over to photoshop, pasted in a screenshot and...

I was kind of thinking the same thing, i.e. if a single point is considered ok then the original 4 slice pie would have been fine. However, it might well be that non-single point contact can be made but reshaping the slices a bit and it cetainly ruins my previous argument.

None of the pre-set colors match the new skin, but if you use e9e9e9 it'll be invisible. I.e. <bracket>color="#e9e9e9"<bracket> Text <bracket>/color<bracket> (not sure how to make a bracket actually show up instead of being interpreted). Demo: Invisible text[/color]

Hmm, interesting. On an intuitive level, it certainly feels like four would be impossible, any attempt to "cross connect" the remaining pairs of the "three each" solution would encircle one of the others, something which obviously cannot happen with identical tiles. I'd suspect a proof would...

Same with the previous "70 cents... 3 quarters... 75 cents... 4 nickles... 20 cents... 4 pennies...", the three quaters, two dimes and a nickle would work as would numerious other combinations.

There are indeed half-dollar coins, though quater, dime, nickle and penny (25, 10, 5 and 1, respectively) are *way* the most common. The half-dollar is used pretty rarely, though they are definitely around. There have also been rounds of dollar coins for ages (gold/silver dollars, etc) though...

I can't figure out a way to get above $1.19 (such as 50+25+4*10+4*1). The only "split" that would take me above a dollar is the 25 which keeps it from happening since there's no 25 in 4*10 and there isn't enough pennies to make another 5. I may be wrong, but that'd be my guess.

Answer: 15 112 113 42 40 58 70 24 74 (leg leg hypotenuse) All three have area 840. Gotten by: I essentially used the old formula for generating primitive (i.e. non-reducable by division)pythagorian triangles: (m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2 ...where m>n and m and n are relativily prime...

You're right, lost a decimal point. Sorry 'bout that. [EDIT] If you're willing to consider 4th root, then perhaps you'd consider \sqrt[.4]{4} = 32 and thus 79 = \frac{(\sqrt[.4]{4} - .4)}{.4}