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1 pointn factorial (represented as n!) is the product of all positive integers up to n. So the number you are looking for is 1,000,000! ("one million factorial")

1 pointWith a tip of the hat to @BMAD for his interesting puzzle. For your amusement, here's an interesting spin on this genre: One night you encounter a twohour traffic delay due to an accident (the tow truck had difficulty clearing the road.) So, for a time interval 13 of two hours you were constrained to travel at 0 mph. You called home and said, sorry dear, but I'll be two hours late getting home. The next night, for some unimaginable reason, you were also constrained to travel part of the way at 0 mph, this time for a distance of one inch. What do you say now when you call home?

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1 pointCrocodile Sophism  Back to the Paradoxes A slim crocodile living in Nile took a child. Mother begged to give him back. The crocodile could not only talk, he was also a great sophist, and so he stated: "If you guess (Edited: predict the fate = guess correctly), what I will do with him, I will return him. However, if you don't guess his fate I'll eat him." What statement shall the mother make to save her child (what about a vicious circle ...)?

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1 pointWe are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too. If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true? We assume that the TV host knows the pebbles distribution in the boxes  in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another. 1. If we swapped the boxes, we would have higher chance of picking up precious pebble next. 2. If we swapped the boxes, we would have same chance of picking up precious pebble next. 3. If we swapped the boxes, we would have lower chance of picking up precious pebble next. 4. In the beggining of the game, there've been exactly 2 boxes with 2 precious pebbles each. 5. In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble. 6. In the beginning of the game, the precious and not precious pebbles have been equal in count. 7. In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles. 8. The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game. 9. None of the above must be true.

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1 pointQuestion about #13, which sounds to me like a conflict. should it read “ nothing is south of the one black rook, which is on the edge”? or should it read “ nothing is south of the white rook that is on the edge” or even “one of the white rooks is on the edge, and has nothing south of it”? for that matter, should it name a different piece than white rook?

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1 pointIt looks like you're taking the first letter and essentially rotating it through the different positions while the other letters remain in the same order. I noticed that all the possible permutations begin with two letters that were next to each other in the original sequence (or in the first and last spot). For example, I don't think you can get a sequence starting with AC. So this does not seem to hold up.

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1 pointAn oldie but a goody (IMO): ****************************** Fivewalled open house Shut my door and make it four. I know exactly who you are, Where you sleep, friends you keep. A loyal soldier out on tour I take my leaves overseas. So pin the gold upon my breast, Give me orders crossing borders.

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1 pointThough seldom seen, I'm world renowned Ever in flight above the ground. Teeth chewing, gnashing in time gnawing on knots, shunning the find. My hands at the ready, arms at the stay Heart like the sun, burning away. Ears like a rabbit's, pinned to the side Listening close, far and wide.

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1 pointI’m the acrobat at your command, Give me a turn, or put me on the stand. I’m the sentinel, doorman you need, Give me a go, a goahead from me. Hit me, poke me, nail me in your wake, Without my flips, weak knees would quake. I live in one state and travel to another, Sunny side up or fumble in the other.

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1 pointThe answer should be 25. My "proof" is a brute force programming solution. I have a psuedocode c++ bruteforce solution. I can give the full version if requested. It takes a few seconds to brute force all possible paths. This might count as a proof, depending on if you trust the computer to be reliable. Basically what it does is it generates all possible paths, and then prunes as it goes along so the program dosen't take basically forever, or 6^27 moves. Now a proof will involve some theorems in graph theory, which I don't yet know all that well. Assume the XYZ plane void recursive_dumb_solution( A 3D cube,Position of X, Y, and Z of current spot,direction it went twice ago, direction it went to get here){ if traveled along a direction twice in a row, return to the above function. else if( x<0  x > 2  y < 0  y > 2  z < 0  z > 2) AKA if it exited the cube, return. else if(cube[x][y][z] == 1) AKA if visited spot already visited, then return. else{ //mark the current position of the cube as visited. cube[x][y][z] = 1; /////Moves in every direction possible. recursive_dumb_solution(cube, x+1, y, z, length+1,'x',prev_dir); recursive_dumb_solution(cube, x1, y, z, length+1,'x',prev_dir); recursive_dumb_solution(cube, x, y+1, z, length+1,'y',prev_dir); recursive_dumb_solution(cube, x, y1, z, length+1,'y',prev_dir); recursive_dumb_solution(cube, x, y, z+1, length+1,'z',prev_dir); recursive_dumb_solution(cube, x, y, z1, length+1,'z',prev_dir); cube[x][y][z] = 0; return; } }

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1 pointA joke from the internet yesterday: Trump really delivered when he said he would run the government just like his businesses. It's already shutdown! Don't worry, he still has three years to bankrupt it.

1 pointAssuming N people, the last standing (S) is given by: S=2*(N2^m)+1, where m=floor[ln(N)].

1 pointRight on, PG, and glad you like it. This one I think I'll keep in a Sundae Best collection.

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1 pointsHi guys, I love riddles so much that I wanted to insert it in game me and my friend are making. The game is 2D RPG roguelike survival game happening in a randomly generated universe. Here is the fun part, one of the random encounters will be the character in a game that will approach the player and ask him a riddle. If a player answers correctly he will be awarded else Riddler will attack the player. (About 10% of NPC encounters in the game will be by riddler). I think that's the cool way to promote riddles and make kids (and grownups) use the brain. Also please write me some riddles and I will insert it in game. Thank you, Josip