Prof. Templeton
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This is another old one. You have nine sacks that are numbered from 1 to 9 arranged in 5 bundles in a row like this... #7 - #2,#8 - #1,#9,#6 - #3,#4 - #5 So there is one sack, then two sacks together, then three sacks together, then two, then one again. Now we can see that the first single sack (7) when multiplied by the pair next to it (28) equals the three sacks in the center (196). The pair to the right (34) when multiplied by its neighbor (5) does not equal the three in the center (196). Your task is to rearrange the sacks so that when multipling from the left or right the correct product is the number in the middle. There are a few solutions that will satisfy this, but your job is to do it in the fewest moves.
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Correct. Quick work as well. Good job.
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Throughout this exercise only whole eggs were dispensed.
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I forgot to mention that throughout she never broke an egg.
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This is an old puzzle that I think was first published in the early 1900’s. An eccentric old woman, who ran a poultry farm outside of Baton Rouge, took her eggs to market one day. She always had a unique way of doing business and was a true lover of riddles. She was overheard talking with her friends about how she had dispensed the eggs. She sold half of her eggs to a customer and gave half an egg over. She then sold a third of the remaining eggs and gave a third of an egg over. To the next customer she sold a fourth of the remainder and gave a fourth of an egg over. The last customer was sold a fifth of the remaining eggs and was given an additional fifth of an egg. What was left after that she distributed evenly to thirteen of her friends. What was the least amount of eggs that she could have taken to market?
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I think the key to this question is to determine how many people do not share the same birthday. The answer can not be 20 if you look at it like this
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I agree with this explanation. Octopuppy also has a insightful way of looking at it.
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Congrats to Chuck, Bona, and Prime. I have 36/91 for Player One. And, of course, the last scenario is a zero sum game.
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Bonanova, Prime, and Octopuppy were sitting around a table discussing the expected probablity of a 6 to come up on a fair six-sided die. Bonanova then pulls such a die from his pocket and puts forth the following game. Starting with Bonanova, then Prime, then Octopuppy they will each take a turn rolling the die. When someone rolls a 6 they are declared the winner. If no six is rolled the first time around the table, then they continue in order until a six is rolled. What is the probablity that Bonanova will win this game? If they each put in a dollar to start, and they each put in another after every win and if after each win Bonanova will always be player number one, should Bonanova expect to make money? Should he expect to make money if he isn't always player number one?
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Good job everyone!!
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Why is it that, when I think the hints are more vague, it takes you less time to solve?
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My 6,13,1,10,7 is considered a small part, My 6,7,1,10,16 has a tail and icy heart, My 10,8,12,9,1,13 is something of a riddle, My 8,17,14,7,11,3 tends to run up the middle, My 2,8,5,14,12,1,7,3 is how celebs arrive, My 17,7,9,12,4 higher spirituality they strive, My 17,7,9,12,4 from the bell lacking in nutrition, My Whole has, of course, the best disposition.
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But what happens
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Just getting ready to post the whole answer and you guys have beat me to it. That's what I get for trying to get my work done first. A Y-san riddle should always be highest priority, I suppose.
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Itachi-san, As always, you've done a great job as host. Thank you, Tsunade. Each sucessive ABR has been better and more enjoyable than the prevoius one. And I really enjoyed ABR2. The narrative was engaging and well written. You stay true to the character's and I'm excited every time I see my characters name (or my character's aletr-ego, me) in the Night or Day posts. My Teammates, Excellent playing you guys. You were very involved and did a fantastic job with suggestions and scenarios which lead to good decision making (most of the time ). You guys helped make this the most fun ABR yet. *High Fives all around*555 Everybody, Good Game.
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Yes it is. I'm going to have to make these harder just for you. Maybe
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My 5,8,14,16,7,10,11,12 always precedes the story, My 15,3,8,9,7,10,17,1 is when a soldier needs more glory, My 1,2,15,6,9,3 is where usurpers set their eye, My 2,14,10,7,17,1,6,11,12 is like crystal from the sky, My 2,14,13,4,10,15,3 is when something is amiss, My whole is used when everything it consists What am I?
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Good job. I didn't want to make it too difficult for the first one. Kudos to PoudreAvenger who clearly had it figured out from the start. I'll work on some more for you.
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Yes.
