This was asked on sci.math ages ago, and never got a satisfactory answer.

Given a number of sticks of integral length $ \ge n$ whose lengths

add to $n(n+1)/2$. Can these always be broken (by cuts) into sticks of

lengths $1,2,3, \ldots ,n$?You are not allowed to glue sticks back together. Assume you have an accurate measuring device.

More formally, is the following conjecture true? (Taken from iwriteiam link below).

Cutting Sticks Conjecture: For all natural numbers $n$, and any given sequence $a_1, .., a_k$ of

natural numbers greater or equal $n$ of which the sum equals

$n(n+1)/2$, there exists a partitioning $(P_1, .., P_k)$ of $\{1, .., n\}$

such that sum of the numbers in $P_i$ equals $a_i$, for all $1 \leq i \leq k$.Some links which discuss this problem:

**Answer**

This is not a solution, just something I found that might be relevant.

On the page linked to in the question, a reduction and various strategies are considered. I’ll briefly reproduce the reduction, both because I think it’s the most useful part of that page and perhaps not everyone will want to read that entire page, and also because I need it to say what I found.

Let a counterexample with minimal $n$ be given. If one of the sticks were of length $n$, we could use that stick as the target stick of length $n$ and cut the remaining sticks into lengths $1$ through $n-1$, since otherwise they would form a smaller counterexample. Likewise, if one of the sticks had length greater than $2n-2$, we could cut off a stick of length $n$ and the remaining sticks would all be of length $\ge n-1$, so again we could cut them into lengths $1$ through $n-1$ because otherwise they would form a smaller counterexample. Thus,

the lengths of the sticks in a counterexample with minimal $n$ must be $\gt n$ and $\lt 2n-1$.

Problem instances that satisfy these conditions for a potential minimal counterexample are called “hard” on that page; I suggest we adopt that terminology here.

The strategies discussed on that page include various ways of forming the target sticks in order of decreasing length. It was found that there are counterexamples both for the strategy of always cutting the next-longest target stick from the shortest possible remaining stick (counterexample $\langle11,12,16,16\rangle$) and for the strategy of always cutting the next-longest target stick from the longest remaining stick unless it already exists (counterexample $\langle10,10,12,13\rangle$), whereas if the stick to cut from was randomized, it was always possible to form the desired sticks up to $n=23$.

I’ve checked that all hard problem instances up to $n=30$ are solvable, and I found that they remain solvable independent of which stick we cut the target stick of length $n$ from. This is equivalent to saying that a problem instance for $n-1$ can always be solved if all stick lengths except one are $\gt n$ and $\lt 2n-1$ and one is $\lt n-1$, since all of these

instances can result from cutting a stick of length $n$ from a hard problem instance for $n$.

I thought that this might be generalized to the solvability of an instance being entirely determined by whether the sticks of length $\le n$ can be cut to form distinct integers, but that’s not the case, since it’s possible to leave only a few holes below $n$ such that the few remaining sticks above $n$ can’t fill them.

**Attribution***Source : Link , Question Author : deinst , Answer Author : joriki*