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Post Posted: Mar 30, 2008, 7:05 am • # 1 


A non-empty set S of positive integers is said to be good if there is a coloring with 2008 colors of all positive integers so that no number in S is the sum of two different positive integers (not necessarily in S) of the same color. Find the largest value t can take so that the set S=\{a+1,a+2,a+3,\ldots,a+t\} is good, for any positive integer a.

P.S.

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Post Posted: Mar 30, 2008, 11:51 am • # 2 


http://www.artofproblemsolving.com/viewtopic.php?s ... 2&t=149160
 
 
Post Posted: Mar 31, 2008, 4:51 pm • # 3 


This really isn't some very hard problem, I'm surprised that nobody has posted some solution already...
Answer is t = 2\cdot 2007 = 4014.
Look at the set M = \{\frac {a}2, \frac {a }2 + 1, ..., \frac {a}2 + 2008\} for some even a. This set contains 2009 elements, so two of them have the same color. In other hand, if b,c\in M with b \not = c then a + 1 \leq b + c \leq a + 1 + 2\cdot 2007, so t must be less then 1 + 2\cdot 2007 because there is no good set S=\{a+1, a+2, ..., a+2\cdot 2007 + 1\} when a is even..
When t = 2\cdot 2007 we color the numbers like this: numbers 1,2,...,\lfloor \frac {a + 1}2 \rfloor have color 1, number \lfloor \frac {a + 1}2 \rfloor + 1 is in color 2, \lfloor \frac {a + 1}2 \rfloor + 2 have color 3, ..., \lfloor \frac {a + 1}2 \rfloor + 2006 have color 2007, and numbers from \lfloor \frac {a + 1}2 \rfloor + 2007 and grater have color 2008. You see that if you pick two different numbers with same color, their sum is either smaller then a + 1 or bigger then a + 2\cdot 2007, and we're done :)

Bye
 
 
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