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        riddles >> easy >> Easy: Inverse sum of squares
        
(Message started by: Qwerty on Mar 21^st, 2003, 8:24am)

Title: Easy: Inverse sum of squares
Post by Qwerty on Mar 21^st, 2003, 8:24am

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.

Title: Re: Easy: Inverse sum of squares
Post by prince on Mar 21^st, 2003, 1:17pm

Using your max allowable value as a hint, I came up with the following:
[hide] 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/225 + 1/400 + 1/900 + 1/3600 + 1/5625 + 1/10000 [/hide]

Is there more than 1 solution?

Title: Re: Easy: Inverse sum of squares
Post by SWF on Mar 27^th, 2003, 5:07pm

on 03/21/03 at 13:17:56, 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?

Title: Re: Easy: Inverse sum of squares
Post by cho on Mar 27^th, 2003, 6:10pm

The bogus one kind of stands out if you notice what the numbers in any group have in common.

Title: Re: Easy: Inverse sum of squares
Post by SWF on Apr 16^th, 2003, 5:46pm

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?

Title: Re: Easy: Inverse sum of squares
Post by cho on Apr 16^th, 2003, 6:18pm

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.