SELECTION EXAMINATION

Let he choose at the beginning ${\alpha }_1$ from the seven pieces and each of them into seven pieces. Then he will have totally $7-{\alpha }_1+7{\alpha }_1=7+6{\alpha }_1$ pieces of paper. Suppose that in the next step he chooses ${\alpha }_2$ pieces of paper and cuts each of them into seven pieces. Then he will have totally $7+6{\alpha }_1-{\alpha }_2+7{\alpha }_2=7+6({\alpha }_1+{\alpha }_2)$ pieces of paper. If he continues this procedure $\kappa$ times, then he will have totally $7+6({\alpha }_1+{\alpha }_2+\cdots +{\alpha }_{\kappa })$ pieces of paper. Therefore we are looking for the value of $\kappa$ satisfying the equation $$7+6({\alpha }_1+{\alpha }_2+\cdots +{\alpha }_{\kappa })=2009\Rightarrow 6({\alpha }_1+{\alpha }_2+\cdots +{\alpha }_{\kappa })=2002,$$ which is absurd, because $2002$ is not divided by $6$. Hence it is not possible for him to have some time $2009$ pieces of paper.