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Topic: Cheerleaders in Formation (Read 331 times) |
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FiBsTeR
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Cheerleaders in Formation
« on: Mar 14th, 2009, 7:10pm » |
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A cool problem, I'm sure there's a few ways to do it:
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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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Ronno
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Re: Cheerleaders in Formation
« Reply #1 on: Mar 14th, 2009, 9:52pm » |
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I'm getting only one solution: 294
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Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
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Grimbal
wu::riddles Moderator
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Re: Cheerleaders in Formation
« Reply #2 on: Mar 15th, 2009, 2:18am » |
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There would be another value matching the conditions: 30 = 25+5 = 3*10
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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| « Last Edit: Mar 15th, 2009, 2:50am by Grimbal » |
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FiBsTeR
Senior Riddler
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Re: Cheerleaders in Formation
« Reply #3 on: Mar 15th, 2009, 7:55pm » |
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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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