## Task 29/82

Determine all possibilities to represent the number 1000 as a sum of consecutive natural numbers.
According to the problem statement, the following should hold:

$$
\sum_{i=0}^{n}(a+i)=1000=\sum_{i=0}^{n} a+\sum_{i=0}^{n} i=(n+1) a+\frac{n(n+1)}{2}=0.5(n+1)(2 a+n)
$$

Certainly, exactly one of the two factors is even; if $n$ is even, then $2 a+n$ is even and $n+1$ is odd, and if $n$ is odd, then $2 a+n$ is odd and $n+1$ is even. This implies that the even factor is divisible by 16 (since 2000 is divisible by 16).

From $2000=2^{4} \cdot 5^{3}$, we get the factorizations

$$
2000=16 \cdot 125 \quad ; \quad 2000=80 \cdot 25 \quad ; \quad 2000=400 \cdot 5
$$

Since $a \geq 1$, it follows that $2 a+n > n+1$. Therefore, the possible values for $n$ are $n_{1}=4, n_{2}=15$, and $n_{3}=24$. These yield $a_{1}=198, a_{2}=55$, and $a_{3}=28$. Indeed,

$$
\sum_{i=198}^{202} i=\sum_{i=55}^{70} i=\sum_{i=28}^{55} i=1000
$$

and due to the above arguments, there can be no further possibilities.