$4-$ $[$ Evenness and Oddness $]$

Find all natural $n$, for which ( $n+1$ )! is divisible by the sum $1!+\ldots+n!$.

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Let $n>2$ and $(n+1)!=k(1!+\ldots+n!)$. Note that $kn((n-1)!+n!)=n \cdot(n-1)!+n \cdot n!=n!+n \cdot n!=(n+1)!)$

Dividing the equation by $k$, we get $(k-1)!(k+1)(k+2) \cdot \ldots \cdot n=1!+\ldots+n!$. However, the left side of this equation is even, while the right side is odd. Contradiction.

## Answer

$n=1,2$.

Given a strip of cells (one cell wide), infinite in both directions. Two cells on the strip are traps, between them are $N$ cells, one of which is occupied by a grasshopper. On each move, we call out a natural number, after which the grasshopper jumps this number of cells to the left or to the right (at its choice). For which $N$ can we call out numbers in such a way as to guarantee driving the grasshopper into one of the traps, regardless of its initial position between the traps and the directions it chooses to jump? (We always see where the grasshopper is sitting.)

## Solution

We number the cells in order so that the traps receive numbers 0 and $N+1$.

Let $N+1=2^{k}$. We will call the distance $d$ to the nearest trap each time. The highest power of two that divides $d$ is less than $k$. If the grasshopper jumps in the opposite direction and does not fall into the trap, it will remain between the traps, and the distance to one of the traps will become $2d$, and to the other $2^{k}-2d$. Both of these distances are divisible by a higher power of two. Therefore, sooner or later, the distances will be divided by $2^{k}$, which means the grasshopper will fall into the trap.

Let $N+1=2^{k} t$, where $t$ is odd and greater than 1. We declare all cells with numbers divisible by $t$ to be traps. Suppose the grasshopper is not in a trap. Note that it is not standing exactly halfway between two traps, since all numbers of the form $1/2(a t+b t)$ are either non-integers or multiples of $t$. Therefore, for any number called out, the grasshopper has a jump that does not lead to a trap.

## Answer

For $N=2^{k}-1$, where $k$ is a natural number.