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 1 riddles / what am i / Re: Mystery of a Sphere May 21st, 2024, 9:49am Started by alien2 | Last post by alien2 SPOILER: It's a soccer ball. Reply Quote Notify of replies

 2 riddles / what am i / Re: The Message May 21st, 2024, 9:48am Started by alien2 | Last post by alien2 SPOILER: If you connect the dots (the offered numbers) on a dial pad of a fixed telephone a message to You will appear. Reply Quote Notify of replies

 3 riddles / easy / Re: What are we? Apr 25th, 2024, 4:22am Started by livinggod29 | Last post by rmsgrey Not everywhere uses the same standard.   Here, the second and third would be " and £ Reply Quote Notify of replies

 4 riddles / easy / Re: Geometric puzzle, two triangles Apr 15th, 2024, 8:24pm Started by rmsgrey | Last post by rloginunix That, clearly, means that that trajectory of the crossing point, shown in red, belongs to a (smaller) circular arc, as a I tried to capture that fact in the animation below.   The center of the said circle is located trivially: just construct the perpendiculars to the sides of the given equilateral triangles, etc. Reply Quote Notify of replies

 5 riddles / what happened / Re: Miracle of Murcia Mar 29th, 2024, 6:12am Started by alien2 | Last post by alien2 I modified the riddle a bit.   Spoiler: Religion plays a big role here, and so does pun.     Perhaps, the riddle is a bit silly or too sappy. Reply Quote Notify of replies

 6 riddles / easy / Re: Symbol Mar 29th, 2024, 6:03am Started by alien2 | Last post by alien2 Clue (possible spoiler): Too bad that there is no toilet inside the room... Or, perhaps, that fact just might spark a lifesaving idea! Reply Quote Notify of replies

7   riddles / easy / Re: Superduck; or, Superman  Feb 11th, 2024, 10:54am
Started by alien2 | Last post by alien2
on Feb 11th, 2024, 10:39am, Grimbal wrote:
 1. A duck is not a chick. https://www.urbandictionary.com/define.php?term=bootyhunter

But maybe the silly hunter doesn't know that. Or she is Scottish and her surname is Duck.
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 8 riddles / easy / Re: The Spider-Man of Paris Jan 17th, 2024, 7:54am Started by alien2 | Last post by rmsgrey For the third variation, your expected return from taking 6 paintings is always the same, so, if you're just concerned with the monetary return, the only question is how risk-averse or risk-seeking you are. If you're risk-averse, taking 3 of each type guarantees a 3-painting payout; if you're risk-seeking, taking 6 and 0 gives you either all or nothing.   You can get more interesting by delving into utility curves - if, rather than 2 paintings being twice as valuable to you as 1, 3 paintings being triple, and so on, if you instead have some other relative values for different numbers of paintings, then that changes your expected value for various strategies.   You have four options, each of which gives you a 50% chance of each of two possible outcomes - taking all the paintings from one set gives you 0 or 6; taking five from one and one from the other gives you 1 or 5; taking four and two gives you 2 or 4; and taking three of each gives you 3 or 3, which is just a guaranteed 3. Whichever pair gives you the highest total utility across the two possibilities would be your personal best pick. Reply Quote Notify of replies

 9 riddles / easy / Re: Tempus fugit, riddles and beyond Dec 25th, 2023, 10:32am Started by alien2 | Last post by alien2 Merry Christmas!     And for 3, yes, you can play with watches. Reply Quote Notify of replies

 10 riddles / hard / Re: power of 3 close to power of 2 Nov 13th, 2023, 10:10am Started by jollytall | Last post by jollytall If I remember well, I proved few month ago that indeed this aL is the best sequence, but that was never validated by anyone.   Nonetheless, I am more interested in the smallest absolute difference (i.e. not ratio) between 3^n and 2^m, where 2^m > 3^n.   I came up with a related question to prove:   2^m - 3^n > 2^(m-n) if 2^m > 3^n > 3.   With other words, if we write 3^n in base 2 then the first (i.e. highest value) n bits have at least one 0 among them.   Yet with other words, we proved earlier that 3^n/2^m > 1-e can be solved for any e, but even that solution is not that close, i.e. 3^n/2^m < 1-2^-n Reply Quote Notify of replies