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Produce More (Posted on 2012-05-29) Difficulty: 2 of 5
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)

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Solution A very tricky problem (spoiler) | Comment 2 of 5 |
 Probably unintentional, but Jer never required n to be an integer, although 2n+1 is required to be a positive integer.

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If n is an integer, then the first of the 2012 distinct ways is (n+1)*n.  The 2012th way is (n+2012)*(n-2011).  This must be greater then n*n.  The next pair, (n*2013)*(n-2012) must be less than or equal to n*n.

In other words,

(n+2012)(n-2011) > n*n >= (n+2013)*(n-2012)

Multiplying.

n*n + n - 2011*2012 > n*n >= n*n + n - 2012*2013

Simplifying,

2011*2012 < n <= 2012*2013

Multiplying,

4046132 < n <= 4050156

So, if n is an integer. then its minimum value is is 4046133 and its maximum is 4050156.  

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BUT, if n is not an integer, then it must be .5 + an integer.  The first of the 2012 distinct ways is (n+.5)*(n+.5).  The 2012th way is (n+2011.5)*(n-2010.5).  This must be greater then n*n.  The next pair, (n*2012.5)*(n-2011.5) must be less than or equal to n*n.

In other words,

(n+2011.5)*(n-2010.5) > n*n >= (n+2012.5)*(n-2011.5)

Simplifying,    4044120.75 < n <= 4048143.75

So, if n is not an integer, then its minimum value is  4,044,121.5 and its maximum value is 4,048,143.5

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Combining the two results, we get:
Minimum n: 4,044,121.5 
Maximum n:  4,050,156  

I also agree with Broll that this is definitely not a difficulty 2 problem, at least not if interpreted literally.  And I am clearly feeling literal today.


Edited on May 29, 2012, 3:51 pm
  Posted by Steve Herman on 2012-05-29 13:45:57

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