Tokkymov M.A.

Find the smallest natural number that starts with 2016 (in decimal notation) and is divisible by 2017.

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Let the number $n$ have $k+4$ digits. Then $2016 \cdot 10^{k} \leq n < 2017 \cdot 10^{k}$. Since $n$ is divisible by 2017, then $n \leq 2017 \cdot 10^{k} - 2017$. Therefore, $2017 \leq (2017 - 2016) \cdot 10^{k} = 10^{k}$, which means $k \geq 4$. Therefore, the smallest such number is $20170000 - 4 \cdot 2017$.

## Answer

20161932. 

![](https://cdn.mathpix.com/cropped/2024_05_06_43e7ff204cbd87ef9854g-20.jpg?height=163&width=1301&top_left_y=0&top_left_x=0)

Author: Tolpogo A.K.

On the plane, two parabolas are given: $y = x^2$ and $y = x^2 - 1$. Let $U$ be the set of all points on the plane lying between the parabolas (including points on the parabolas themselves). Does there exist a segment of length greater than $10^6$ that is entirely contained in $U$?

## Solution

The tangent to the first parabola at the point $(a, a^2)$ has the equation $y = 2a(x - a) + a^2$. The points of intersection of this line with the second parabola are $(a - 1, (a - 1)^2 - 1)$ and $(a + 1, (a + 1)^2 - 1)$. The segment $AB$ lies entirely in $U$, and the square of its length, equal to $2^2 + 16a^2$, can be arbitrarily large.

## Answer

Yes, it exists.