k-man
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Everything posted by k-man
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The objective of this puzzle is to solve this hexadecimal sudoku. Instead of 1-9 digits in 9x9 grid it uses 0-F digits in 16x16 grid.
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Nice puzzle, Bushindo.
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Clarification... If I bet on both a suit and a rank will the dealer nod if either of them matches the card or only if both match? For example, if I bet on K and D, will he nod 1) only if he has a King of Diamonds or 2) if he has any King or any Diamond?
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kevin beat me to the punch while I was editing spoilers
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Edit: added "w" back
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4 unknown coins - another weighing puzzle
k-man replied to k-man's question in New Logic/Math Puzzles
Adding the fifth coin under the same conditions increases the number of possibilities dramatically, so 5 weighings will not cover 5 coins. I'll do some calculations and if I find it interesting enough, I'll post a new topic for 5 coins. -
4 unknown coins - another weighing puzzle
k-man replied to k-man's question in New Logic/Math Puzzles
Very good, Captain. This is the same solution that I've found to be the best so far. Now, I've been trying to either find a way to do it in 4 weighings or prove that it cannot be done. -
4 unknown coins - another weighing puzzle
k-man replied to k-man's question in New Logic/Math Puzzles
Aye, Captain. You can interpret it that way. I put this condition there to exclude the possibility of one coin weighing as much as 2 others. -
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You have 4 coins that look identical, but you know nothing about their weights. They may all weigh the same or may have different weights or some of them may weigh the same and some be lighter or heavier - you get the idea. The difference (if any) is small relative to the weight of the coins Goal: You need to organize the coins based on their relative weight, i.e. sort them by weight grouping identical coins together. All you have is a balance scale that tells you which side of the scale is heavier or if the sides are in balance. Question 1: What's the minimum number of weighings required to achieve the goal? Question 2: Provide a method that accomplishes the goal in the minimum number of weighings. Have fun!!!
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Find the missing numbers in the square.
k-man replied to Rob_Gandy's question in New Logic/Math Puzzles
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Can you find other methods of solving? Oh, and happy Friday k-man. Thanks, you too Our posts are crossing in time. I added the spoiler to my last post with another way to solve it and here is another...
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Yeah, I missed the fact that the game continues after the first 2 draws if they're both unsuccessful. I'm being lazy right now - it's Friday Edit: added spoiler with the answer
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Now what is the probability of Coop winning any given game?
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Here is the more interesting (and much more difficult) version of the same puzzle. Same rules, but 15 people.
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Mulikar showed that for any number of points k, there is a method of arranging the points on the surface that creates (k-2)2 triangles. But nobody has proven that this method always produces the maximum possible number of triangles. What if there is a different method of placing points that can produce more triangles? What if this method produces the same number of triangles for k<x, but produces more triangles for k>x? That's what this new challenge is - to prove that (k-2)2 is the maximum for any k.
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I think you miscalculated. I&#39;m almost done with the solution (it&#39;s actually taking me longer to type than to figure it out) Starting out with 2 coins on each side of the scale breaks the possibilities down nicely - 21:26:26. I have to stop for today, so I&#39;m posting what I&#39;ve got so far&nbsp;and I will try to&nbsp;finish it tomorrow. If the scale is not balanced then let's say 1,2 is heavier than 3,4. We are down to 26 possibilities: NO. (0) - there must be fakes here 1H. #1 or #2 (2) 1L. #3 or #4 (2) LH. Either #3 or #4 is light and either #5 or #6 is heavy (4) HL. Either #1 or #2 is heavy and one of the remaining coins #3 through #6 is light (8) HH. [1,2],[1,5],[1,6],[2,5],[2,6] (5) LL. [3,4],[3,5],[3,6],[4,5],[4,6] (5)
