Final Round

Answer: yes, he can. If Bob cuts a piece of the apple into $20$ pieces, then the total number of the pieces increases by $19$. If Bob cuts a piece of the apple into $14$ pieces, then the total number of the pieces increases by $13$. So, if Bob makes $x$ cuts into $20$ pieces and $y$ cuts into $14$ pieces, then the apple ($1$ piece) is cut into exactly $1+19x+13y$ pieces.

It remains to show that there exist nonnegative integer numbers $x$ and $y$ such that $1+19x+13y=1!+2!+3!+\cdots +1013!+2014!$. This equality is equivalent to the equality $19x+13y=2!+3!+4!+\cdots +1013!+2014!$.

We divide the summands in the right-hand side of the last equality into some groups: $$\begin{array}{rl}2!+3!+4!+\cdots +1013!+2014!& =(2!+3!+4!+5!)+(6!+8!)+\\ & (7!+9!+10!)+(11!+12!)+(13!+14!+\cdots +1013!+2014!).\end{array}$$ We show that the sum of the numbers in each group is divisible either by $13$ or by $19$. Indeed, $2!+3!+4!+5!=152=8\cdot 19$, $6!+8!=6!(1+7\cdot 8)=6!\cdot 57=6!\cdot 3!\cdot 19$, $7!+9!+10!=7!(1+8!\cdot 9+8!\cdot 9!\cdot 10)=7!\cdot 793=7!\cdot 61!\cdot 13$, $11!+12!=11!(1+12)=11!\cdot 13$, and the last sum $13!+14!+\cdots +1013!+2014!$ is divisible by $13$ since each summand of this sum is divisible by $13$.

Therefore, the right-hand side of the equality has the form $19a+13b$. Thus Bob can cut the apple into $1!+2!+3!+\cdots +1013!+2014!$ pieces (it is sufficient to make $a$ cuts into $20$ pieces and $b$ cuts into $14$ pieces).