Good sets and coloring of all positive integers.

freemind

#1

Good sets and coloring of all positive integers.

**Posted:** Sun Mar 30, 2008 4:05 pm

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The fate of equilibrium is to end the eternity...

bilarev

#2

http://www.mathlinks.ro/viewtopic.php?search_id=954790132&t=149160

**Posted:** Sun Mar 30, 2008 8:51 pm

SpongeBob · Poincare Conjecture Offline Joined: 01 Jul 2006 Posts: 188

#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 Smile

Bye

**Posted:** Tue Apr 01, 2008 1:51 am