Prove that it is possible to make 0 out of 11 distinct natural numbers less than 191

Question

I am practising for a maths competition and am stuck on this question.

Define $\mathbb{N}=\{1,2,...\}$ and let $X$ be a subset of $\mathbb{N}$ containing $11$ distinct natural numbers strictly less than $191$. Show that it is possible to obtain the number $0$ just from addition and subtraction of elements of $X$.

I have attempted a number of approaches, from creating a group from the elements of $X$ as integers with addition, but this is not generally closed under addition (or rather showing it is is the same question as what is being asked) to generalising to a subset of $a$ elements strictly less than $b$ and finding a condition on $a$ and $b$ such that it is possible to arrive at $0$ ($11$ and $191$ seem suspiciously specific to me). Nothing I have tried has yielded any results.

Answer 1

Start with the remark: Any combination of sums and subtractions between elements of a set $S\subset \mathbb N$ can be written as the difference of two sums. Thus, $$s_1-s_2-s_3+s_7+s_9+s_{10}-s_{11}=(s_1+s_7+s_9+s_{10})-(s_2+s_3+s_{11})$$ for example. So, from the start, we are looking for two disjoint subsets of $S$ with the same sum.

Now, fix a subset $S\subset \{1, \cdots, 191\}$ with $|S|=11$. We seek to prove that $S$ must have two non-empty disjoint subsets with the same sum.

The smallest possible sum a non-empty subset of $\{1, \cdots, 191\}$ might have is $1$.

The largest, given that the subset has $11$ or fewer elements, is $191+190+\cdots +181=2046$.

Hence, in principle, there are only $2046$ possible sums one can make from subsets of $S$. But $S$ has $2^{11}=2048$ subsets of which all but $1$ is non-empty. Since $2047>2046$ there must be two distinct non-empty subsets of $S$ with the same sum. Removing any shared elements doesn't change the fact that the sums are the same, so we must have two disjoint subsets $S_1, S_2$ with the same sum. Now, subtracting the sum of the elements in $S_2$ from the sum of the elements in $S_1$ yields $0$. And we are done.

Might add: $\{1, 2, 4, 8,16,32,64, 128\}$ has $8$ elements and no two disjoint subsets have the same sum. It's clear that you can't add a $9^{th}$ to that particular set unless you go up to $256$. This may suggest that you can rule out sets with size $9$ or $10$ as well, but I have not thought that through. Conway-Guy sets might do better, I am not sure. I note that the general question, replacing $191$ with a general cap $N$, is open.

Update: The Conway Guy set $\{77,117,137, 148,154,157,159,160,161\}$ has $9$ elements and has distinct subset sums (this is not obvious but, if I have not introduced any errors, it should follow from the work of Conway and Guy). Their set with $10$ elements has maximal element $301$, though, so I think $10$ is impossible. As I say, however, the general problem remains open. It is not known that the Conway Guy set is optimal in general. See A005318 for more details.