Konagin S.V.

Find $x_{1000}$, if $x_{1}=4, x_{2}=6$, and for any natural $n \geq 3$, $x_{n}$ is the smallest composite number greater than $2 x_{n-1}-x_{n-2}$.
We will prove by induction that $x_{n}=1 / 2 n(n+3)$.

Base. For $n=3,4$, the formula is correct: $2 x_{2}-x_{1}=8$, so $x_{3}=9$; $2 x_{3}-x_{2}=12$, so $x_{4}=14$.

Induction step. $2 x_{n}-x_{n-1}=2 \cdot 1 / 2 n(n+3)-1 / 2(n-1)(n+2)=1 / 2(n+1)(n+4)-1$. By the condition, $x_{n+1}$ is the first composite number greater than

$1 / 2(n+1)(n+4)-1$. But the number $1 / 2(n+1)(n+4)$ is composite. Indeed, if $n$ is odd, then $1 / 2(n+1)(n+4) = (n+4) \cdot 1 / 2(n+1)$. Each of the factors is an integer greater than 2. Similarly, the case of even $n$ is considered. Thus, $x_{n+1}=1 / 2(n+1)(n+4)$.

Substituting $n=1000$, we get $x_{1000}=500 \cdot 1003$.

## Answer

501500. 

Author: Kanel-Belov A.Ya.

A unit square is divided into a finite number of smaller squares (which may have different sizes). Can the sum of the perimeters of the squares intersecting the main diagonal be greater than 1993? (If a square intersects the diagonal at a single point, this is also considered an intersection.)

## Solution

Divide the unit square into four equal smaller squares. Squares that intersect the diagonal only at a vertex are called squares of the first level. Each of the remaining squares is divided into four smaller squares. Squares that intersect the diagonal only at a vertex are called squares of the second level. Again, each of the remaining squares is divided into four smaller squares, and so on (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_696cdbe703e904f1264dg-04.jpg?height=386&width=466&top_left_y=773&top_left_x=796)

Perform this operation 500 times. We obtain $2^{k}$ squares of the $k$-th level. The side length of each of them is $2^{-k}$. Therefore, their total perimeter is 4. Consequently, the total perimeter of all squares of all levels is $4 \cdot 500 > 1993$.

## Answer

It can.