Prove that there exist a set of $2010$ natural numbers such that product of any $1006 $ numbers is divisible by product of remaining $1004$ numbers.
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Tags: inequalities, number theory unsolved, number theory
02.10.2010 02:19
Pick up these numbers : $2^{a_1} , 2^{a_2} , ...., 2^{a_{2010}}$ Where $a_i =10^{99999} (1+\frac{1}{10^{100i}})$ for $1 \leq i \leq 2009$ Yeah , they are too big
02.10.2010 09:12
\[a_i=\frac{N}{p_i}, N=\prod_{i=1}^{2010}p_i, \ i=1,2,...,2010\], were $p_i$ - distinct prines. Maximal $a_i$ is about $e^{2010}.$
14.10.2010 02:22
Subject: set of integers mahanmath wrote: Pick up these numbers : $2^{a_1} , 2^{a_2} , ...., 2^{a_{2010}}$ Where $a_i =10^{99999} (1+\frac{1}{10^{100i}})$ for $1 \leq i \leq 2009$ Yeah , they are too big can anyone plz explain how this works in detail?
14.10.2010 15:54
$2^{b_1} . 2^{b_1} ... .2^{b_{1004}} \mid 2^{b_{1005}} . 2^{b_{1006}} ... .2^{b_{2010}} $ Iff $b_1 + b_2 + ... + b_{1004} \leq b_{1005} + b_{1006} + ... + b_{2010} $ And now you can see this inequality holds for the $a_i$ (in my post) To prove this just close $10^{99999}$ from both sides and you need to prove this trivial inequality : $\frac{1}{10^{100}} +\frac{1}{10^{200}} + ... + \frac{1}{10^{201000}} \leq 2$ [Note that I just prove inequality for $a_1 , ... , a_{1004} $ , the other cases are similar .]