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 1 riddles / hard / Re: power of 3 close to power of 2 Jul 13th, 2022, 2:09pm Started by jollytall | Last post by jollytall I also got something like this. I did not find any reason why the difference (as a ratio) MUST converge to zero. If the irrational number would have any sequence of digits for sure, then e.g. lots of zeros after each other would make a rational number suddenly a very good approximation. Even if not all decimal sequences occur there might still be rational numbers that are suddenly give a "very good" approximation. Still I do not see that this chance exists as we (or at least I) know so little about the value (digits, actually in any base) of log2(3).   Going back to the first part of the question. Any idea on the absolute value (not the ratio)? Would there be a number occur infinite times, that would immediately mean a convergence to 1. Even if the difference has some limits as a function of 2^n but less than linear, it would also make the ratio close to 1. Reply Quote Notify of replies

 2 riddles / easy / Re: How much would an underwear cost? Jun 30th, 2022, 11:59am Started by livinggod29 | Last post by rmsgrey \$45, but lingerie would be \$5 cheaper Reply Quote Notify of replies

 3 riddles / hard / Identify your favourite childhood games Jun 29th, 2022, 10:57pm Started by livinggod29 | Last post by livinggod29 Identify your favourite childhood games 1. Kevin runs to avoid being touched by Sahlan and Vhanz in hot pursuit 2. Declining number of seats ensure the inevitability of the last person standing.... or sitting 3. Count, then search 4. Throw and chart the course to retrieve it one foot 5. Demolish a pile and reconstruct before you are hit 6. Your ability to stand still will be a life saver 7. No need to invite a former American Idol judge but just do what he tells 8. Tiny round glass balls, you are crazy if you have lost them 9. This is the right opportunity to jump over the other riddlers 10. Get from Point A to Point B.... in a coarse bag 11. Keep off the ground at any cost 12. Back when you thought law enforcement would pursue the bad guys on foot 13. You snoop something.... 14. Riddlers on their knees. Opponents running amok. Zsacksaw chases, tags Sanjay to continue the chase 15. Pull with all your might, bring victory to your side Reply Quote Notify of replies

 4 riddles / easy / What are we? Jun 29th, 2022, 10:55pm Started by livinggod29 | Last post by livinggod29 What are we?   1. I am always excited 2. I tell people where things are at 3. You can play a game on me 4. I'm very rich 5. I give percentages 6. I am always looking up 7. And......... 8. I look like a light in the sky 9. I keep things arranged that are to my right 10. I keep things arranged that are to my left   Share with your friends and family and see if they can figure out what are we? Reply Quote Notify of replies

 5 riddles / medium / Find the words that end with the letters "ANT Jun 29th, 2022, 10:47pm Started by livinggod29 | Last post by livinggod29 Find the words that end with the letters "ANT" in the clues given below. eg. What kind of an ant works with figures? An accountANT.   1. Lives in the jungle? 2. Is far away? 3. Is extraordinarily large? 4. Works for a master? 5. Is good-natured? 6. Is unchanging? 7. Is luxurious? 8. Is one who takes part? 9. Is a very small child? 10. Is sleeping? 11. Is very bright? 12. Is empty? 13. Is immediate? 14. Is plentiful? 15. Has moved to a different country? 16. Is meaningful? 17. Is something that grows? 18. Has influence over others? 19. Is unsure and indecisive? 20. Lives in a certain place?   Source : Find the words that end with letters ANT Reply Quote Notify of replies

 6 riddles / what happened / Re: Miracle of Murcia Jun 3rd, 2022, 11:25am Started by alien2 | Last post by Grimbal It is possible with a pilot license.   With the adequate license it is allowed to fly a helicopter by night. Reply Quote Notify of replies

 7 riddles / hard / Re: fractal maze May 29th, 2022, 3:04pm Started by towr | Last post by Grimbal OK, well, I wrote a program.  Here is the result. It should be minimal, not necessarily unique.   path: - C5 (5 A29 (29 13) A13 D7 (7 17) D17 H12 (12 15) H15 G1 (1 16) G16 16) C16 + length: 11   Contacts are numbered clockwise from the top left. Between the brackets is what happens inside of a circuit.   Anyway, all answers are here: https://www.mathpuzzle.com/fractalmaze.txt Reply Quote Notify of replies

8   riddles / hard / Re: Magician having 100 cards  Jan 24th, 2022, 3:25am
Started by navdeep1771 | Last post by rmsgrey
I've spotted a hole in my "proof" - still figuring out how to patch it, if it's possible to patch.

edit: And I've figured out the patch:

The hole is that x+d could be in the same pile as a and b without creating an ambiguous sum - it's only if it's in the third pile that it creates problems

The patch:
 hidden: The smallest gap between two numbers in the same pile must be no more than 3 - if you take four consecutive numbers, the largest gap between any two of them is 3, and by the pigeonhole principle, when they're divided among three piles, at least two of them must be in the same pile.   If that smallest gap is bigger than 1, then take a, b as before as the two numbers either side of one such gap, d the difference between them, and let x be some number between a and b. x must be in a second pile, otherwise it would be in the gap between a and b At least one of x+d and x-d must be in the 1-100 range - without loss of generality, assume x+d is. It cannot be in the third pile - suppose x+d is in the third pile, then a+(x+d) = (a+d)+x = b+x gives you the same sum from two different pairs of piles. If it's in the first pile, then the gap between b and x+d would be smaller than the gap between a and b, contradicting our definition of a and b. So x+d must be in the same pile as x, but then x and x+d can be used in place of a and b, with b in place of x, to repeat the same argument to show that b+d (if it's not over 100) must be in the first pile. Repeating the argument, alternating piles, lets you fill in the first pile with a+kd and the second with x+kd for any integer k that gives a number between 1 and 100.   The third pile must contain some number, call it y, and y must either be in one of the gaps of the first pile (in which case the exact same argument works) or it must be at one of the ends of the range. If it's at the bottom, y+d will be less than d away from the lowest number in the first pile; at the top, y-d will be close to the highest - either way, it can't be in the first pile without creating a smaller gap, nor the second pile without creating an ambiguous sum.   The rest of the proof runs as above - d must be 3, and the cards sorted modulo 3.   The only remaining case is where two consecutive numbers are in the same pile. If a second pile also contains consecutive numbers, then the third pile must contain a number, call it y, where y has at least one neighbour in a different pile - without loss of generality, assume y+1 is in a different pile. Since both other piles have consecutive numbers in, you can pick x and x+1 from the pile that doesn't contain y nor y+1 so that x+(y+1) and (x+1)+y give the same value from two different pairs of piles.   So at most one pile can contain consecutive numbers.   If there is a pile with consecutive numbers, then the other two piles contain at least one number each, let the smallest number in each be x and y respectively, with x