Determine the minimum value of n for which, from every subset of $\{1,2, ..., 999\}$ with $n$ elements, it is possible to select four distinct integers $a, b, c, d$ such that $a + 2b + 3c = d$
This problem is from a 1998 Brazilian Math Olympiad, reference link here.
I have tried figuring out some sort of restricion on $d$ as it is the largest value between $a, b, c, d$ but I could not find a way of solving yet. Is there any useful property of this sum that can be used?
Thanks in advance.
I'll address the problem for subsets of $\{1,2,\ldots,N-1\}$ where the original problem is for $N=1000$.
First, I'll provide a case for which there is no solution: ie a set $U$ of size $n$ for which there are no $\{a,b,c,d\}\subset U$ with $a+2b+3c=d$.
Let $U=\{N-n,N-n+1,\ldots,N-1\}$. That makes $a+2b+3c\ge 6(N-n)+4$. If $6n< 5N+5$, then $a+2b+3c>N-1$ which ensures no solution for $d$ exists. Thus, if $n<5(N+1)/6$, there is a set $U$ of size $n$ for which no solution exists.
Next, let's assume we have a set $U$ of size $n$. Let $c<b<k$ be the three smallest elements of $U$, and let $U'=U\setminus\{b,c\}$. For $m=2b+3c$, we are looking for $a,d\in U'$ so that $a+m=d$. If there is no such solution, then the union of $U'$ and $U'+m$ must be $2n-4$ distinct numbers ranging from $k$ to $N+m-1$, which requires $2n-4\le N+m-k$, otherwise the pigeonhole principle ensures a collision between $U'$ and $U'+m$. As $c<b<k$, we have $m=2b+3c\le 5k-8$, so $N+m-k\le N+4k-8$.
If $2n-4>N+4k-8$, we get $2n-4>N+m-k$ which is sufficient to ensure there are $a,d\in U'$ such that $a+m=d$. We know that $k\le N-n+2$, so $N+4k-8\le 5N-4n$. Thus, if $2n-4>5N-4n$, then we have $$ 2n-4>5N-4n\ge N+4k-8\ge N+m-k. $$ Now, $2n-4>5N-4n \iff 6n>5N+4 \iff 6n\ge 5(N+1)$, so if $n\ge 5(N+1)/6$, there will always be a solution to $a+2b+3c=d$.
Unless I've made a slip-up some place, that should complete the proof.
This is pure plagiarism of Einar Rodland's answer and also Hagen von Eitzen's comment on a duplicate I asked, so I am indebted to them. But I wanted to really understand what was happening and couldn't quite follow their line of reasoning, so I decided to write things out mostly so I could make sense of it.
$S=\{166,167,...,999\}$ has $834$ elements and also if $a,b,c\in S$ are distinct elements:
$$a+2b+3c\geq 168+2\cdot 167+3\cdot 166= 1000>d$$
for any $d\in S$ this means there is a set $S$ with $n=834$ elements still not satisfying our condition.
I will prove that when $|S|=835$ we always have a solution. Take $c<b<k$ the three smallest elements in $S$ and consider $m=2b+3c$. We need to find $a$ and $d$ distinct elements in $S'=S \setminus\{c,d\} $ such that $a+m=d$. We notice $a$ and $d$ exist if and only if $S'$ and $S'+m$ have an intersection.
Let us suppouse by way of contradiction, they are disjoint. In this case:
$$|S'\dot{\cup}(S'+m)|=|S'|+|S'+m|=2|S'|=1666$$
We will find lower and upper bounds for $\max (S'\dot{\cup}(S'+m))$. A lower bound is $k+1665$ because $k$ is the lowest term and clearly there are $1665$ numbers above it before we reach the maximum. An upper bound on the other hand is $999+m$. therefore:
$$k+1665\leq \max(S'\dot{\cup}(S'+m))\leq 999+2b+3c$$ $$c+1665\leq 999+2b+3c\Rightarrow 666\leq2(b+c)\Rightarrow 666\leq 4b\Rightarrow 167\leq b$$
But this would mean there is only one element smaller than $167$ and so $|S|$ could have at most $1+(999-167+1)=834$ elements. But this is absurd, as we considered $S$ with $|S|=835$. Thus $S'$ and $S'+m$ are not disjoint and we have found our four elements with $a+2b+3c=d$.
