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About araver

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  1. Double up!

    I haven't been able to come up with a better than n=11 alternative. I'm pretty sure (branching my results at n=15) that there's nothing between 12 and 14.
  2. dividing infinity by 17

    Assuming exponentiation.
  3. The mean of four numbers

    Agreeing with Rob. Or kinda disagreeing with both
  4. Double up!

    n=11 in my post above.
  5. Double up!

    I keep forgetting to click "Sort by Date" and when it sorts by rating I get confused. My question was different, let me rephrase: Assume there was no round before (A,B,C) =(4, 2, 5) with B and C active gets at the end of round 1 to (4, 4, 3). 1) Do we stop here even if the active players are not the ones equal? Bonanova's response does imply so i.e. Game ends at the end of the round if ANY 2 players are equal (different wording than OP though). So we count a minimum of 1 round before game with (4,2,5) ends. 2) If welet it go to the nextround since the two active players (B and C) are not equal: 2A) assumingA and B could be active AND if we don't check the condition before we subtract, then one of them gets 0, and that may cause infinite rounds. 2B) If we do check the condition A=B since they are the active, then we call the game ending in Round 2. What I meant is that different wording makes either N=1 or N=2. In any case, I'm assuming and counting using rule "Game ends at the end of the round if ANY 2 players are equal". And another submission from me:
  6. Double up!

  7. Double up!

    Question on wording of the goal:The game ends when the two active players in any round have exactly the same stake. If (A,B,C) go into a round, B and C are active and B = A at the end of the round (e.g. B = A/2, B<C), does that count as stopping the game OR the gamegoes to the next round where only if C sits out the goal is met?
  8. Sorry, butI was wrong about my "trapezoid" solution. At most two trapezoids would be similar, forcing the third to be similar would make all congruent. I've circled around and am pretty sure now there's no other way to split a triangle into 3 similar yet non-congruenttrianglesunless it's follows Bonanova's solution. And I think the samehypothesis holdsfor showing no 3 simlar quadriterals can be obtained either (limit case is 3 congruent trapezoids for an equilateral triangle). With more than 4 sides ... that's even more unlikely, as they won't have the necessary number of sides to begin with(i.e. partition a triangle into 3 pentagons /hexagons).
  9. I don't have spoilers on mobile interface, sorry. Does the problem state ANY triangle? I can do this for right triangles (which are not isosceles) with 3 similar non congruent triangles and I can do it for equilateral triangles with 3 trapezoids similar non-congruent. But not for any triangle. EDIT: just saw Bonanova's that is my solution as well, but it does not work for A=B.
  10. Rest in Peace, Games Forum

    I still pop into here from time to time, but I rarely have time to host a game. However, if you see this, message/PM me back - I'm willing to host a quick one (6-8) in August / September 2015 if you're interested.
  11. Fun with real numbers

    Answer for Part II