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Title: Cheerleaders in Formation Post by FiBsTeR on Mar 14th, 2009, 7:10pm A cool problem, I'm sure there's a few ways to do it: ===== A group of cheerleaders wish to arrange themselves on a field. They attempt to line up in a square such as . . . . . . . . . (etc) but find that five cheerleaders would be left over. They then line up in a rectangle with seven more columns than rows with none left over. What is the greatest number of cheerleaders that could be in the squad? |
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Title: Re: Cheerleaders in Formation Post by ronnodas on Mar 14th, 2009, 9:52pm I'm getting only one solution: [hide]294[/hide] |
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Title: Re: Cheerleaders in Formation Post by Grimbal on Mar 15th, 2009, 2:18am There would be another value matching the conditions: [hide] 30 = 25+5 = 3*10 [/hide] We want to solve n(n+7) = k2 + 5 2n(2n+14) = (2k)2 + 20 (2n+7)2 - 49 = (2k)2 + 20 (2n+7)2 - (2k)2 = 69 (2n+7+2k)(2n+7-2k) = 69 = 3*23 There are only 4 possible products: (2(n+k)+7, 2(n-k)+7) = (1, 69), (3, 23), (23, 3) or (69, 1) (n+k, n-k) = (-3, 31), (-2, 8), (8, -2) or (31, -3) (n, k) = (14,-17), (3, -5), (3, 5) or (14, 17) but n,k>0 (n, k) = (3, 5) or (14, 17) It shows no larger solution exists. PS: actually there are more products -1*-69, -3*-23, etc. but these all result in n<0. |
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Title: Re: Cheerleaders in Formation Post by FiBsTeR on Mar 15th, 2009, 7:55pm Cool! Here's my solution, though it doesn't allow you to find all possible numbers of cheerleaders very easily (I don't think): We want the maximal k so that: n(n+7)=k2+5 n2+7n-(k2+5)=0 By the quadratic formula: n=(1/2)(-7+\sqrt{(2k)2+69}) We took the positive root because there are a positive number of cheerleaders. Now the last time the discriminant is a perfect square is when adding 69 to a perfect square yields the next perfect square. (After this point, the "spacing" becomes too big. You know what I mean.) So: (2k)2+69=(2k+1)2 4k2+69=4k2+4k+1 4k=68 k=17 --> 172+5=294. |
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