superprismatic
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They are split within a word.
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Yes, Z=0. Of course, if you used Z=26, it wouldn't matter because the arithmetic is done mod 26.
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There are two piles, containing between 100 and 1000 items. A is allowed to choose any two items from the first pile, and B is allowed to choose any two from the second pile. If A has only two-thirds as much choice as B, how many items are in each pile? SUPERPRISMATIC'S LAST WALTER PENNEY PUZZLE: The Den has done a spectacular job of solving all but 1 (#40: Un-halving a Message) of Walter's puzzles. I'm sure that #45 will be solved. And I'm willing to bet that you people are so tenacious that #40 will also be solved within a few weeks. I hope you enjoyed his puzzles. I'll try my best to make up a few challenging ones each month or so.
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Edit: made my point clearer.
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A pair of dice have letters on the various faces in addition to the usual spots. One die has the letters A, L, and V on three faces with N, T, and W, respectively, on the opposite faces. The second die has the letters B, E, and F on three faces with D, S, and R, respectively on the opposite faces. The dice are rolled nine times, with successive throws being 12, 2, 5, 10, 10, 6, 7, 8, 10. In the process a certain message is spelled out. What is the message? SUPERPRISMATIC CLARIFICATION: On each throw, you get two letters. Here, you get to decide which of the two letters comes first with each roll. For example, if you find that A and D came up on the first roll and E and T came up with the second roll, the message could begin either of 4 ways: DAET, ADET, DATE, or ADTE.
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The only way a 3 pan thing can act the way people are answering is if the arm lengths are all the same and the angular distance between arms is 120 degrees. I was thinking of the more general case. Since I never looked at 3-pan puzzles here, I jumped to the wrong conclusions. Sorry.
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I didn't say a hypothetical isn't true. I just said that a 3 pan balance can balance when the weights aren't all the same.
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"for example consider a three way balance scale. you put objects in each of the three levers, and based on the results of the levers, you can tell which objects are heavier or lighter." This just isn't true. If the three moments add up to 0, the thing will balance. There are lots of ways for this to happen even when a, b, and c are all different.
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A certain 10-letter word having the second and fifth letters the same is treated as follows: the first two letters are multiplied together (using the values A=1, B=2, ..., Y=25, Z=0); then the next two letters are similarly multiplied together and the two products added (mod 26). The sum is reconverted into a letter using the same correspondence. The same operations are performed on letters 2 to 5 producing another letter. This is continued until the seventh letter is produced from the last four letters of the word. The letters so generated are, in order, FPSLMMZ. What is the word?
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there must be hundreds of valid words that do fit the pattern - how to find the ones that make sense when strung together?
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Bits 1 to 5 of a certain random binary stream are converted to a decimal number (from 0 to 31) and used to encipher the first letter of a message by mod-26 addition. Bits 2 to 6 of the stream are used to encipher the second letter, etc. The result is VSYOM ZIUEA GIIWA XTOBL ITNFX IUBL [/code]What is the message?
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The fraction 1/N (for a positive integer N) is found to have a decimal representation which is cyclic with period 26. The digits proceeding left to right from the decimal point, are assigned in order to the letters from A to Z. Note that by this scheme a digit may stand for more than one letter. The message: "A cab is a taxi" enciphered by means of these equivalents becomes: 0 100 54 0 3075. Find N.
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Actually, the arithmetic is done mod-26, so if the result is greater than or equal to 26, then 26 is subtracted. So, neither of your restriction rules apply. See Semper Rideo's post above. He gives a good example and only makes one minor mistake which I pointed out immediately.
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Sorry, I don't understand what you're saying.
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Well, the wheels are identical. They are just rotated relative to each other to show some word facing the same direction. You are told what can be seen on the wheels at some number of positions later. So, you can get several incomplete sequences, by position, around the scrambled sequence. They must be meshed together to get the partial sequence used for all wheels. If these wheels are set to show you IGYWVHPO, there will be a (possibly partial) word five positions ahead in the group of wheels.
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You are very nearly correct. Two of your subsets can be consolidated in the correct way to get the solution. Nice work!
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Yes, except that the penultimate number in the encoded message should be a 0 instead of a 26.
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The letters of the first half of a message are added (mod 26) to the corresponding letters of the second half, using the values A=1, B=2, etc. The result is: 3 10 16 9 22 24 11 23 6 5 24 19 16 7 19 8 22 3 13 19 4 14 3 22 12 18 13 4 14 25 14 [/code]Read the message.
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Nice going!
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Graphically?
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