## Task 6B - 231246B

Determine whether there exists a natural number $k$ with $k \geq 1$ and $k$ natural numbers $a_{1}, a_{2}, \ldots, a_{k}$, which are not necessarily pairwise distinct, such that

$$
a_{1}+a_{2}+\ldots+a_{k}=1984 \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{k}}=1
$$

If this is true, provide such natural numbers $a_{1}, a_{2}, \ldots, a_{k}$.
Here, it makes sense to break down into powers of 2, as their quotients can be added easily.

By continuously dividing $1984: 2$, we get the product $2^{6} \cdot 31=64 \cdot 31$. The multiplication $32+64 \cdot 30=1952$ results in a sum of the quotients $\frac{1}{32}+\sum_{1}^{30} \frac{1}{64}=\frac{1}{2}$. The difference of 1984-1952=32, when broken down into $4 \cdot 8=32$, forms the sum of the quotients $\sum_{1}^{4} \frac{1}{8}=\frac{1}{2}$, and thus $k=30+1+4=35$.

$$
\sum_{1}^{30} 64+\sum_{1}^{1} 32+\sum_{1}^{4} 8=1984
$$

and

$$
\sum_{1}^{30} \frac{1}{64}+\sum_{1}^{1} \frac{1}{32}+\sum_{1}^{4} \frac{1}{8}=1
$$