bonanova
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Everything posted by bonanova
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A B C F T all Yes.
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D: Yes M: Yes U: Yes
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G: Yes N: 3/6
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It's a little earlier than that.
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J: Yes
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Hi Chronic, and welcome to the Den.
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Nice. And welcome to the Den.
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3/6
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I: Yes J: 3/6 and 4/6
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The old Grandfather clock in the corner has a 5-inch long hour hand and a 10-inch long minute hand. At midnight, and at twenty-two more times during the day, the hands line up. At those times, the tips of the hands are 5 inches apart. Midway between those times, they are 15 inches apart. During the first hour, when does that distance increase at the greatest rate?
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4/6 and 3/6.
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Hi cddude, and welcome to the Den. Can you find two distances that work?
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Check out whether that point could be on a diagonal and be consistent with a distance of 5 from another corner.
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You could do a few other words, discover the clue, then come back to Y_____.
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In Hill County there are two kinds of roads: hilly and level. The cabs in Andy's Hilltop Taxi fleet travel 72 mph downhill, 63 mph on level roads, and 56 mph going uphill. It took Andy 4 hours to pick up his fare in Green Valley and 4 hours and 40 minutes to return him to Hilltop. How far is it between the two cities?
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People can't believe this when I tell them, but I used to be nit-picky.
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I feel like there should be a better way of doing this though...I'm not really seeing any obvious patterns in the numbers... Yeah, I meant one at a time, so you only need enough for the biggest one [8580 blocks, 9 Sets]. Nice one. Don't recognize the format, tho. LISP? Only method I know is to set limits on a [4 and 9 are derivable limits] then solve for b and c for each a value. a=4: 8[b + c] = 16. No solution with 4 <= b <= c a=5: [b − 12][c − 12] = 120 - eight pairs of factors a=6: [b − 8][c − 8] = 48 - five pairs of factors a=7: [3b − 20][3c − 20] = 280 - four pairs of factors a=8: [b − 6][c − 6] = 24 - three pairs of factors a=9: [5b − 28][5c − 28] = 504. No factors with 9 <= b <= c Not elegant.
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Right on.
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Johnny received a Jumbo Set of 1000 building blocks for his birthday. He begins to assemble them into "bricks" of dimensions a x b x c where of course a, b and c are integers, and a <= b<= c. He notices that for each brick he builds, some of the blocks are "outside" blocks [visible] and some are "inside" blocks [hidden by the outside blocks.] He imagines that for some bricks the numbers of outside and inside blocks must be equal; call such a brick an Equal Brick. With his set of blocks, how many different Equal Bricks can Johnny make? Now for small bricks [a=1 or 2] there are no Inside Blocks. And for larger bricks, the Inside Blocks will eventually dominate. So there must be a finite number of Equal Bricks. If Johnny wants to make all possible Equal Bricks, how many Jumbo Sets of blocks will he need?
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That's correct, and fairly straightforward to prove.
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3/6. I added the Y word to the list of not really common words. Two of those left to get: F and Y.
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EDIT: Am going off for a bit now. Don't solve it too fast guys - leave some for me. 3/6.
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Yes.
