## Task 4 - 150934

Determine all possibilities to represent the number 60 as the sum of at least two consecutive natural numbers.
1. Case:

The number of summands is odd, i.e., $2k+1$ for a positive integer $k$. If $n$ is the existing middle summand, the sum can be written as

$$
(n-k)+(n-(k-1))+\cdots+(n-1)+n+(n+1)+\cdots+(n+(k-1))+(n+k)=n \cdot(2k+1)
$$

Here, $n \geq k$ must hold so that the smallest summand $n-k$ is a non-negative integer. For this sum to equal 60, $2k+1$ must be an odd divisor of 60, and $n=\frac{60}{2k+1}>k$ must hold.

This yields the following solutions:

For $2k+1=3$, $n=20>1$, so $60=19+20+21$ and

for $2k+1=5$, $n=12>2$, so $60=10+11+12+13+14+15$.

For the only other odd divisor $2k+1=15$ of 60, we get $n=4$, which is less than $k=7$. This would yield the sum

$$
60=(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9+10+11
$$

which includes negative summands and thus is not a solution in the sense of the problem.

2. Case:

The number of summands is even, i.e., $2k$ for a positive integer $k$. In particular, the sum then contains $k$ odd summands, and is thus only even (like 60) if $k$ is also even, i.e., can be written as $2m$ with a positive integer $m$. The sum thus consists of $4m$ summands.

Let the smallest of these summands be denoted by $n-(2m-1)$, where $n \geq 2m-1$ is a natural number. Then

$$
60=(n-(2m-1))+\cdots+(n+(2m-1))+(n+(2m))=4mn+2m=2m(2n+1)
$$

In particular, $2n+1$ is an odd divisor of 60, so $2m=\frac{60}{2n+1} \leq n+1$ holds. This yields the following solution:

For $2n+1=15$, $2m=4 \leq 8$ and we get $60=4+5+6+7+8+9+10+11$.

Otherwise, $2n+1 \leq 5$, so $n \leq 2$, but $2m=\frac{60}{2n+1} \geq 12$, so we again only get integer, but not natural, solutions.