Svyatov M.

On a plane, grasshopper Kolya and 2020 of his friends are sitting. Kolya is going to jump over each of the other grasshoppers (in any order) such that the starting and ending points of each jump are symmetric relative to the grasshopper being jumped over. We will call a point a finishing point if Kolya can reach it after the 2020-th jump. For what largest number $\$ N \$$ can there be an initial arrangement of grasshoppers for which there are exactly $\$ \mathrm{~N} \$$ different possible finishing points?
Evaluation. Method 1. Introduce a coordinate system on the plane so that Kolya is sitting at the point \$(0,0) \$. Let \$lvec \{\mathrm{a}\} \_1 \$$, \$|vec \{\mathrm{a}\} \_2 \$$, \ldots, \$ \mid vec\{\mathrm{a}\}\{2020\} \$$ be the radius vectors of the grasshoppers in the order of their jumping over Kolya. It is not difficult to find the radius vector of the finish point \$2(-|vec \{\mathrm{a}\} \_1+\mid \operatorname{vec}\{\mathrm{a}\} \_2 - |vec \{\mathrm{a}\}=3+\ldots$.

![](https://cdn.mathpix.com/cropped/2024_05_06_864d5071b377f8fb364bg-11.jpg?height=60&width=2012&top_left_y=1628&top_left_x=17)
(select 1010 terms at odd positions).

Method 2. Notice that if a grasshopper jumps over some point \$ A \$$ and then over some point \$ B \$$, it will ultimately move by the vector \$ 2$ loverrightarrow \{\mathrm{AB}\}=2$ (loverrightarrow \{\mathrm{OB}\} - loverrightarrow \{\mathrm{OA}\}) \$$. Each finish point can be associated with a sequence in which Kolya jumped over his friends to end up in this point after 2020 jumps - that is, a permutation of the numbers from 1 to 2020 (consider all grasshoppers except Kolya numbered). Suppose Kolya at some point is at point \$ 0 \$$ and is about to jump over grasshoppers located at points \$A, B, C (consider only three consecutive jumps). If Kolya makes these jumps in the order \$ABC\$, he will first move by the vector \$2loverrightarrow\{ОА \$$ and then in two jumps move by the vector \$2loverrightarrow \{\mathrm{BC}\}=2(loverrightarrow \{O C\} - loverrightarrow \{\mathrm{OB}\}) \$$, so in total, after three jumps, he will move by the vector \$2(loverrightarrow \{\mathrm{OA}\}+ loverrightarrow \{\mathrm{OC}\} - loverrightarrow \{\mathrm{OB}\}) \$$. If Kolya makes these jumps in the order \$CBA\$, he will, similarly, move by the vector \$2(loverrightarrow \{\mathrm{OC}\}+ loverrightarrow \{\mathrm{OA}\} - loverrightarrow \{\mathrm{OB}\}) \$$, ending up in the same point as in the previous case. Thus, in the sequence of Kolya's jumps, any two grasshoppers standing one apart can be swapped without changing the finish point. Therefore, the finish point is characterized not by the permutation of the numbers from 1 to 2020 (of which there are 2020!), but only by which numbers in such a permutation stand at odd positions. Accordingly, there cannot be more than \$C_ \{2020\} \wedge\{1010\} \$$ different finish points.

Example. Place the grasshoppers at points on the number line with coordinates 0 (at Kolya), 1, 3, 3², ..., $3^{2019}$. Prove that all sums of powers of three with coefficients $\$$ rt $1 \$$ are different. By adding to such a sum the constant sum $1+3+3^{2}+\ldots+3^{2019}$, we get a sum with coefficients 0 and 2. It corresponds to a 2020-digit ternary number consisting of zeros and twos, and all such numbers are different.

## Answer

For $\$ N=$ C_ \{2020\} \wedge\{1010\} \$$.

For any natural number $K$, there exist infinitely many natural numbers $T$ that do not contain zeros in their decimal representation and such that the sum of the digits of the number $K T$ is equal to the sum of the digits of the number $T$. Prove this.

## Solution

Let the number $K$ (in decimal notation) have $k$ digits. Then, in the role of $T$, we can take any number 999... 99, written with $n$ nines, where $n \geq k$; then both the sum of the digits of $T$ and the sum of the digits of $K T$ are equal to $9 n$ (in the number $K T=K\left(10^{n}-1\right)$, digits separated by $n$ places sum to 9; for example,

$743552 \cdot 99999999=743552 \cdot\left(10^{8}-1\right)=$

$=74355200000000-743552=74355199256448$

the proof for the general case is left to the reader).