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riddles / hard / Re: power of 3 close to power of 2 
Jul 13^{th}, 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. 
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3 
riddles / hard / Identify your favourite childhood games 
Jun 29^{th}, 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 
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4 
riddles / easy / What are we? 
Jun 29^{th}, 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? 
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5 
riddles / medium / Find the words that end with the letters "ANT 
Jun 29^{th}, 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 goodnatured? 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 
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7 
riddles / hard / Re: fractal maze 
May 29^{th}, 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 
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8 
riddles / hard / Re: Magician having 100 cards 
Jan 24^{th}, 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 xd must be in the 1100 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, yd 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<y. If y1 is in the same pile as x, then you can pick consecutive numbers, a and a+1, from the pile with consecutive numbers, and a+y = (a+1)+(y1) gives an ambiguous sum. So y1 must be in the pile with consecutive numbers. If x1 is in the range 1100, it cannot be in the same pile as x otherwise x wouldn't be the smallest in that pile, nor in the same pile as y since y is smallest in that pile and, by assumption, greater than x. But if x1 is in the same pile as y1, then x+(y1) = (x1)+y gives an ambiguous sum. So x1 cannot be in any pile, so x=1. If there are more numbers in the same pile as x, call one of them x', and consider x'1. It can't be in the same pile because there are no consecutive pairs of numbers in that pile, and can't be in the same pile as y for the same reason y1 couldn't be in the same pile as x, but if it's in the same pile as y1, you again get an ambiguous sum, so that pile must only contain a single number, 1. Due to the symmetry of the situation, a very similar argument shows that y=100 and must be the only number in its pile. So if one pile contains consecutive numbers, it must be 299, and the other piles must be 1 and 100.  
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