1. a) Can 10 consecutive natural numbers be found whose sum equals 2016:

b) Can 7 consecutive natural numbers be found whose sum equals 2016.
Answer: a) No; b) Yes.

Solution: a) Suppose that $a+(a+1)+(a+2)+\cdots+(a+9)=10 a+45=$ $2016 \Leftrightarrow 10 a=1971$. Since the last equation is impossible for natural $a$, the answer is "no".

b) $285+286+287+288+289+290+291=2016$.

Evaluation criteria. Correct solution - 20 points; in part a) for the proof - 11 points: in part b) for the example - 9 points.