9.5. Find all natural $n$ such that from sticks of lengths $1,2, \ldots, n$ (all numbers from 1 to $n$ exactly once) an equilateral triangle can be formed. All sticks must be used.
Answer. All natural $n \geq 5$, the remainders of which when divided by 6 are $0, 2, 3$ or 5.

Solution. The sum of the lengths of all sticks, equal to $\frac{n(n+1)}{2}$, must be divisible by 3, so the product $n(n+1)$ must be divisible by 6, taking into account the mutual simplicity of the factors, we get that one of them is divisible by 2, and one (possibly the same) - by 3. By considering the four possible cases and noting that $n \leq 4$ do not work, we obtain the $n$ specified in the answer. We will show that for each of them, the required construction is possible. Note that six arbitrary consecutive numbers $k, k+1, k+2, k+3, k+4, k+5$ can be divided into three pairs of numbers with the same sum: $\{k, k+5\},\{k+1, k+4\},\{k+2, k+3\}$. Therefore, if we show how to form a regular triangle for $n=5,6,8,9$, then by adding the necessary number of consecutive sixes of numbers to these numbers and the corresponding number of pairs of sticks to each side of the regular triangle, we will obtain a way to form a regular triangle for all $n$ specified in the answer.

For $n=5,6,8,9$ the problem is solved as follows: 1) $n=5,1+4=2+3=5$, 2) $n=6,1+6=2+5=3+4$, 3) $n=8,1+2+3+6=4+8=5+7$, 4) $n=9,1+2+3+4+5=6+9=7+8$.

Grading criteria. From considerations of divisibility by 3, the required variants for $n$ modulo 6 are found: 2 points.

Proof that $n \geq 5: 1$ point.

Construction of examples. The idea of sixes: 3 points. Examples for $n=5,6,8,9: 1$ point.

10th grade

Each problem is worth 7 points.