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Does there exist a set $A\subset \{1,2,.......,100\}$ with 65 elements such that 65 can't be expressed as a sum of two elements of $A$?

I am not getting enough hints to start it

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    $\begingroup$ Hint: How many pairs of elements sum to 65? You would need to exclude at least one member of each of these pairs... $\endgroup$ – Johanna Nov 11 '14 at 15:02
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yes I guess . A= {${1,2,3,....32,65,66,67,....,97}$} works.

Working- If $x$ is there , just exclude $65-x$ and anything $\ge 65$ can be included in $A$

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There are 32 unordered pairs which sum up to 65 namely: $1+64, 2+63, ... 32 + 33$. So such a set would have to contain none of these pairs. But this is possible: Take $\{1,\cdots,100\}$ and take out one element of each pair, e.g. $1,\cdots ,32$. Then there is no possibility to sum up to 65, but the set has more than 65 elements

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