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A FIRST and SECOND Problem (Posted on 2006-08-22) Difficulty: 2 of 5
There are n retailers and each of them have copies of two novels left, titled “FIRST” and “SECOND”. Each retailer has a different number of copies of "FIRST", which is always more than the number of copies of "SECOND" stocked by that retailer.

It is observed that the sum of squares of the number of copies of FIRST and the number of copies of SECOND for each of the retailers is always equal to:

For example, if the respective number of copies of FIRST and SECOND titles possessed by any given retailer i are xi and yi, then :
xi2+yi2 = (22+32)(32+42)(42+52); for all i = 1,2,....,n

How many retailers are there and how many copies of FIRST novel in total are available from all these (n) retailers?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts An idea, with no computer | Comment 3 of 5 |
Seeing that the sum was written as a product of sums of squares, I doodled a bit and got:

(ac+bd)^2+(ad-bc)^2= (a^2+b^2)(c^2+d^2)

so, applying this recursively,

(ace+bde+adf-bcf)^2+(acf+bdf-ade+bce)^2 = (a^2+b^2)(c^2+d^2)(e^2+f^2)

so I guess it should be just a matter of substituting 2, 3, 4, and 5, for a, b, c, d, e, and f.

Edited on August 22, 2006, 4:47 pm
  Posted by Old Original Oskar! on 2006-08-22 15:31:14

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