Solution. From $\mathrm{x}+(\mathrm{x}+1)+\ldots+(\mathrm{x}+\mathrm{k}-1)^{\prime}=2005$ we get $\mathrm{k}(2 \mathrm{x}+\mathrm{k}-1)=1 \cdot 2 \cdot 5 \cdot 401$. Therefore, $\mathrm{k}=2,5,10$ respectively $2005=1002+1003$, $2005=399+400+401+402+403$ and $2005=196+\ldots+205$.
1. Represent the number 2005 as a sum of consecutive natural numbers, in all possible ways.