4. Zadatak

Vrijedi

$$p+(p+1)+\ldots+(p+2012)+(p+2013)=$$ $$=\underbrace{(p+\ldots+p)}_{2014}+(1+2+\ldots+2013)=$$ $$=2014\cdot p+2013\cdot2014:2=$$ $$=1007\cdot2p+2013\cdot1007=$$ $$=1007\cdot(2p+2013)$$

Dakle, zbroj $p+(p+1)+\ldots+(p+2012)+(p+2013)$ je djeljiv s $1007$ pa je složen.