8.4. We have $n$ sticks of lengths $1,2, \ldots, n$. Can a square be formed from these sticks, and if not, what is the minimum number of sticks that can be broken in half to form a square a) for $n=12;$ b) for $n=15?$
Answer: a) 2 sticks; b) It is possible. Solution. a) Since the sum $1+2+\ldots+12=78$ is not divisible by 4, it is impossible to form a square. The side of the square would be $\frac{78}{4}=19.5$. If only one stick is broken, its parts might end up on two different sides of the square, but the other two sides cannot have a non-integer length. Therefore, at least two sticks must be broken. With two broken sticks, a square can be formed. For example, if we break the sticks of length 1 and 3 in half, the square can be formed as follows: $\left(\frac{1}{2}+12+7\right),\left(\frac{1}{2}+11+8\right),\left(\frac{3}{2}+10+2+6\right),\left(\frac{3}{2}+9+5+4\right)$. b) For $n=15$, a square can be formed, for example, as follows: $(15+14+1),(13+12+5),(11+10+9),(8+7+6+4+3+2)$.

## Grade 9