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June 16th, 2018, 12:00 AM   #1
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Another Complex Problem

I have come across another tough problem. Please help me to solve it. My thinking is $\displaystyle (1004)^2$ should be the first maximum value and the final maximum value is $\displaystyle (2007)^2$. But i don't know how to prove it will start from $\displaystyle (1004)^2$.
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June 17th, 2018, 05:05 AM   #2
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Quote:
 Originally Posted by MathsLearner123 I have come across another tough problem. Please help me to solve it. My thinking is $\displaystyle (1004)^2$ should be the first maximum value and the final maximum value is $\displaystyle (2007)^2$. But i don't know how to prove it will start from $\displaystyle (1004)^2$.
Why not look at the problem with 1-7 instead of 1-2007 and see what you can do?

 June 17th, 2018, 06:45 AM #3 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 Thanks for the hint. Yes I took from 1 to 7 and observed this If I assume the following set {a1,a2,a3,a4,a5,a6,a7} => {7,6,5,4,3,2,1} I end up as 1*7, 2*6, 3*5, 4*4, 5*3, 6*2, 7*1 => 7, 12, 15, 16, 15, 12, 7. Hence 16 is the greatest value obtained, can i generalize as? $\displaystyle ((1+7)/2)*((1+7)/2) = 4^2 = 16$
June 17th, 2018, 07:17 AM   #4
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Quote:
 Originally Posted by MathsLearner123 Thanks for the hint. Yes I took from 1 to 7 and observed this If I assume the following set {a1,a2,a3,a4,a5,a6,a7} => {7,6,5,4,3,2,1} I end up as 1*7, 2*6, 3*5, 4*4, 5*3, 6*2, 7*1 => 7, 12, 15, 16, 15, 12, 7. Hence 16 is the greatest value obtained, can i generalize as? $\displaystyle ((1+7)/2)*((1+7)/2) = 4^2 = 16$
To generalise the result, try a proof by contradiction. Again, try first for 1-7. The question is what happens if you try to keep every product less than $4^2$.

 June 17th, 2018, 09:07 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,950 Thanks: 1842 For every product to be less than 1004², each of the 1004 integers from 1004 to 2007 must be paired with one of the 1003 integers from 1 to 2003, with no such number being used twice, which is impossible.

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