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araver added an answer to a question 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.

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araver added an answer to a question dividing infinity by 17
Assuming exponentiation.

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araver added an answer to a question The mean of four numbers
Agreeing with Rob. Or kinda disagreeing with both

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araver added an answer to a question Double up!
n=11 in my post above.

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araver added an answer to a question 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 we let it go to the next round since the two active players (B and C) are not equal:
2A) assuming A 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:

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araver added an answer to a question Double up!

araver added an answer to a question 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 game goes to the next round where only if C sits out the goal is met?

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araver added an answer to a question Partitioning a Triangle into similar pieces
Sorry, but I 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 noncongruent triangles unless it's follows Bonanova's solution.
And I think the same hypothesis holds for 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).

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araver added an answer to a question Partitioning a Triangle into similar pieces
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 noncongruent. But not for any triangle.
EDIT: just saw Bonanova's that is my solution as well, but it does not work for A=B.

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araver added an answer to a question Circumscribing puzzle

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araver added a post in a topic 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 (68) in August / September 2015 if you're interested.

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araver added an answer to a question Fun with real numbers
Answer for Part II

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araver added an answer to a question if a,b,c, and d are natural numbers revisited.
Proof for part 2

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araver added an answer to a question Fun with real numbers
For part I.

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araver added a question in New Logic/Math Puzzles
CylindersLong time no see. I have a puzzle that's eating me for more than a week and I think you'll like working on it Credits in "spoilers".
Assume you have a cylinder around which you've wrapped a rectangular grid of cells (R rows x C columns). Each cell contains an integer greater or equal to 1. Each integer is equal to the number of neighboring cells (neighbor = sharing an edge, not just a corner) having the same integer. E.g. a cell containing 1 can have exactly one neighbor cell with 1 inside, the other neighbors can have any other different integers.
Two decorations are the same if by rotating (around the cylinder's axis of symmetry) one of the cylinders (whole rectangular grid rotation  rotating rows/columns separately is not possible) and putting them side to side you can view them as identical from any viewpoint. Also, the top and bottom of the cylinder are considered as being different so if the rectangular grid is not symmetrical around the cylinder's halffull/halfempty axis, a cylinder decoration identical to the original turned upside down would count as not the same as the original.
How many unique decorations can you make with such a RxC cylinder?
1. For R,C <=6.
2. A closedform expression or algorithm for any R,C.
Note: I am stuck at 2. I have an algorithm, but I'm rather more interested if there is a closedform expression i.e. using basic arithmetic operations (addition, subtraction, multiplication, and division), exponentiation to a real exponent, logarithms, and trigonometric functions.
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