 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Produce More (Posted on 2012-05-29) Among pairs of numbers whose sum is 16, 8*8=64 is the greatest possible product. However, if we allow for a sum of 17 then there are 3 distinct ways of achieving a higher product with two positive integers: 9*8, 10*7, and 11*6 are all greater than 64.

Among pairs of numbers whose sum is 2n, n*n=n2 is the greatest possible product. However, if we allow for a sum of 2n+1 then there are 2012 distinct ways of achieving a higher product with two positive integers.

Find the minimum value of n.

 No Solution Yet Submitted by Jer Rating: 5.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 5 of 5 | First we know that for a number of the form 2n+1, n being integer,and if the two parts of the product being integer,  the maximum value occurs at  n*(n+1). from this point we construct the inequality:

n*(n+1)*(n-a)/n*(n+a+1)/(n+1) - n^2 > 0

→ (n-a) (a+n+1)-n^2 > 0
→-a^2-a+n > 0

Solving for a at  -a^2-a+n = 1

we get for positive solutions:

a= 1/2 (Sqrt[4 n-3]-1)

Now, if we take the floor of the solution we have the point at the series stating at (n-a)*(n+1+a) where this expression stops being greater that n^2.So the unique integer pairs that satisfy this are equal to Floor[a] + 1(we have to include a=0),solving this for 2012 we have: n= 4046133.

Edited on June 13, 2012, 6:58 am
 Posted by John Dounis on 2012-06-13 06:30:27 Please log in:

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