superprismatic
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Everything posted by superprismatic
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Well, I got it down to 39 moves, but I can't see how to shorten it.
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That was a fun puzzle. Thanks, Wolfgang!
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What is the length of a side of an equilateral triangle which has a point inside of it that is at distances 3, 4, and 5 from its vertices?
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Yes, I went through all 13! permutations.
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No, that's not correct. I haven't worked through your answer, but you must have made a mistake somewhere. Your method seems fine but I think the 30/7 is wrong.
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Very nice puzzle. Thanks!
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Begin with an isosceles triangle with sides 13, 13, and 10. Put the 10-long side on the bottom and create an infinite stack of circles inside this triangle as follows: Place the largest inscribed circle in the triangle (it will be tangent to the three sides of the triangle). Next, place the largest circle you can fit in the triangle which sits atop the first circle. Keep doing this to make an infinite stack of circles with each circle tangent to sides of the triangle and to the circles above or below it. What is the sum of the circumferences of all the circles in the stack?
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So, you stop flipping cards when you've flipped the ace, which is your last point (unless it's the first card flipped)? I suppose that's what the expression "shot down by the bullet" means.
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Why is it that the binomial coefficient nCk is an odd number if and only if, thinking of the numbers in their binary representations, k = n AND k ?
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Ya gotta find one with integer coefficients.
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Is there a polynomial P(x) with integer coefficients such that P(1)=3 and P(3)=2?
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That is known as Fermat's Little Theorem. You can find a simple proof on line here.
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You guys made short work of that one!
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Is [(2n-1)(2n-2)(2n-22)(2n-23)···(2n-2n-1)] ÷ n! an integer? Why or why not?
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That is a very clear proof of the correct function. Nice!
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Well, Bushindo is saying that, taking enough steps, there must be a rotation which is used at least 64 times (since there are a finite number of them -- 64 to be exact). Whichever rotation that is, call it m, (I-Am)64 = 0.
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Is (100!)÷(50!×250) an integer? Why or why not?
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Sorry, I made a clarification on the OP. What I'm looking for is the number of N×N matrices satisfying the conditions as a function of N.
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How many N×N matrices are there which satisfy the following? 1. All entries are non-negative integers 2. Every row and every column sums to 3 3. There are no more than two non-zero entries in any row and any column. Find the number of N×N matrices satisfying the conditions as a function of N.
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Nice way of getting 2/3! It works well with this problem.
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Nice! You have it nailed. Now, you can decide with probability p exactly with coin flips. Simply write p as a binary number and flip until the number the flips give is not equal to the binary p. Flip number less than p, call the decision made; greater than p, call it not made; equal to p so far, flip again. You could be in for a lot of flips, however. It's interesting to see what the winning flip probability and associated win probability are as the number of players approach infinity.
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Two really talented mathematicians are chosen to play a little game. Neither mathematician is told the identity or whereabouts of the other. So, we can assume that there is no possibility that these two can communicate regarding this game until it is completed. The game itself is simple: Each is given a fair coin. Each must then decide whether he will make an "official" coin flip with this coin. If at least one of the two mathematicians makes an "official" coin flip AND every "official" coin flip comes up heads, each of them will receive $1,000,000. If neither of them makes an "official" coin flip OR an "official" coin flip comes up tails, they lose and get nothing. What strategy should these mathematicians employ to maximize the probability that they will win? What is the winning probability associated with the best strategy? Please note: I use the term "official" coin flip because I didn't want to preclude allowing the mathematicians the ability to use the coin in some way to decide whether to make an "official" coin flip or not.
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Nice Puzzle. Thanks!
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Yes, thanks for pointing that out. Perhaps Bonanova can put some initial conditions on the lamps which will make this problem work.
