The integers from 1 to $n$ are written, one on each of $n$ cards. The first player removes one card. Then the second player removes two cards with consecutive integers. After that the first player removes three cards with consecutive integers. Finally, the second player removes four cards with consecutive integers. What is the smallest value of $n$ for which the second player can ensure that he completes both his moves?
Answer: $n=14$.

At first, let's show that for $n=13$ the first player can ensure that after his second move no 4 consecutive numbers are left. In the first move he can erase number 4 and in the second move he can ensure that numbers 8,9 and 10 are erased. No interval of length 4 is left.

If $n=14$ the second player can use the following strategy. Let the first player erase number $k$ in his first move, because of symmetry assume that that $k \leq 7$. If $k \geq 5$ then the second player can erase $k+1$ and $k+2$ and there are two intervals left of length at least 4: 1.. $(k-1)$ and $(k+3) . .14$, but the first player can destroy at most one of them. But if $k \leq 4$, then the second player can erase numbers 9 and 10 in his first move and again there are two intervals left of length at least 4: $(k+1) . .8$ and 11..14.

## 3 Geometry