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Topic: Easy: Inverse sum of squares (Read 2937 times) 

Qwerty
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Posts: 1


Easy: Inverse sum of squares
« on: Mar 21^{st}, 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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Posts: 49


Re: Easy: Inverse sum of squares
« Reply #1 on: Mar 21^{st}, 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
Uberpuzzler
Posts: 866


Re: Easy: Inverse sum of squares
« Reply #2 on: Mar 27^{th}, 2003, 5:07pm » 
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on Mar 21^{st}, 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 27^{th}, 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
Uberpuzzler
Posts: 866


Re: Easy: Inverse sum of squares
« Reply #4 on: Apr 16^{th}, 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 16^{th}, 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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