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${1,2,3,4,5,6,7,8,9,10}$ numbers are given and divided into 2 groups and let the groups be $A$ and $B$.Prove that $A$ or $B$ contains $a,b,a+b$. $A$ contains none of the elements in the $B$. $a$ is not equal to $b$

Is this problem can be solved? If not ${1,2,3....n}$ what is lowest value of n?

Any help would be appreciated. Well i tried dividing into groups and i didn't found any problem but i didn't check all possible groups.

Joshua
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  • Suppose $A$ has $n\geq5$ elements, and $B$ has $m$ elements. Say $A=\left{a_1,\cdots,a_n\right}$, with $a_1<\cdots<a_n$. Assume $a_1+a_2,a_1+a_3,\cdots,a_1+a_n,a_2+a_3,\cdots,a_{n-1}+a_n$ all belong to $B$. Then these are distinct elements. So $m\geq\frac{n(n-1)}2\geq 2n$. So $n+m\geq3n\geq15>10$, a contradiction. Hence some element $a_i+a_j\in A$. – awllower Mar 23 '20 at 07:11
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    I can't even find a way to divide the numbers $1,2,3,4,5$ into two groups without one group containing $a,b,a+b$. Unless there's some condition you didn't bother to tell us about, like $a\ne b$? – bof Mar 23 '20 at 08:30
  • $A$ contains none of the elements in $B$ – Joshua Mar 23 '20 at 08:40
  • In $1,2,3,4,5$ there is a contradiction$ (1,2,4) (3,5)$ – Joshua Mar 23 '20 at 08:50
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    But then A contains 1+1=2. – Jaap Scherphuis Mar 23 '20 at 08:51

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