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   Easy: Inverse sum of squares
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   Author  Topic: Easy: Inverse sum of squares  (Read 3010 times)
Qwerty
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Easy: Inverse sum of squares  
« on: Mar 21st, 2003, 8:24am »
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By adding inverses of perfect squares (e.g. 1/4, 1/9, 1/16, etc...) to each other, create a summation that equals exactly 1/2.
 
Constraints:
You may not repeat any value.
You may not use the inverse of any values greater than 100^2.
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prince
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Re: Easy: Inverse sum of squares  
« Reply #1 on: Mar 21st, 2003, 1:17pm »
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Using your max allowable value as a hint, I came up with the following:
1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/225 + 1/400 + 1/900 + 1/3600 + 1/5625 + 1/10000
 
Is there more than 1 solution?
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SWF
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Re: Easy: Inverse sum of squares  
« Reply #2 on: Mar 27th, 2003, 5:07pm »
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on Mar 21st, 2003, 1:17pm, prince wrote:
Is there more than 1 solution?

 
Yes, there are many solutions.  Below are seven groups of integers whose sum of reciprocal squares is 0.5:
  • 2 3 4 5 7 12 15 20 28 35
  • 2 3 4 5 7 12 13 28 35 39 52
  • 2 3 4 5 7 10 20 28 30 35 60
  • 2 3 4 5 7 9 28 35 36 45 60
  • 2 3 4 5 6 15 21 28 60 70 84
  • 2 3 4 5 6 18 21 24 30 60 87
  • 2 3 4 5 6 12 36 45 60 90
  • 2 3 4 5 6 15 20 36 45 60 90

Note that there are 8 groups above, but only 7 are correct. Basic arithmetic problem:  which one is bogus?
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cho
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Re: Easy: Inverse sum of squares  
« Reply #3 on: Mar 27th, 2003, 6:10pm »
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The bogus one kind of stands out if you notice what the numbers in any group have in common.
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SWF
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Re: Easy: Inverse sum of squares  
« Reply #4 on: Apr 16th, 2003, 5:46pm »
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What do you mean, cho?  The common thing among 7 of them is that the sum of their reciprocals is 1/2.  Do you mean something else?
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cho
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Re: Easy: Inverse sum of squares  
« Reply #5 on: Apr 16th, 2003, 6:18pm »
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What they have in common is factors. In the valid answers there is no factor that occurs only in one number. The bogus answer has 2 numbers with unmatched factors.
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