Kosukhin O.n.

Sasha found that there were exactly $n$ working digit buttons left on the calculator. It turned out that any natural number from 1 to 99999999 can either be entered using only the working buttons, or obtained as the sum of two natural numbers, each of which can be entered using only the working buttons. What is the smallest $n$ for which this is possible?
Let's show that the conditions of the problem are met if the buttons with digits $0,1,3,4,5$ remain functional.

Indeed, any digit from 0 to 9 can be represented as the sum of some two "functional" digits. Let the number from 1 to 99999999 that we want to obtain consist of digits $a_{1}, a_{2}, \ldots, a_{8}$ (some of them, including the leading ones, may be zeros). Represent each of them as the sum of two "functional" digits: $a_{1} = b_{1} + c_{1}, a_{2} = b_{2} + c_{2}, \ldots, a_{8} = b_{8} + c_{8}$. Then the number composed of "functional" digits $b_{1}, b_{2}, \ldots, b_{8}$ and the number composed of "functional" digits $c_{1}, c_{2}, \ldots, c_{8}$ give the desired number when summed.

Suppose that the desired result can be achieved for some set of four "functional" digits. Let $a$ be some odd digit. Among the numbers from 1 to 99999999, there will be one that ends in $a$ and contains "non-functional" digits in its decimal representation. Then this number can be represented as the sum of two numbers, whose representations contain only "functional" digits. Therefore, for each odd digit $a$, there will be two "functional" digits (possibly the same) whose sum ends in $a$. On the other hand, it is not difficult to see that among all the sums of pairs of four "functional" digits, there can be no more than four odd numbers. Therefore, one of the digits $1,3,5,7,9$ will not appear at the end of such sums. Contradiction.

## Answer

5. 

![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-01.jpg?height=249&width=2060&top_left_y=1676&top_left_x=-1)

Author: Gasikov S.B., Folklore

Does there exist a quadratic trinomial $f(x) = a x^{2} + b x + c$ with integer coefficients and $a$ not divisible by 2014, such that all numbers $f(1), f(2), \ldots, f(2014)$ have different remainders when divided by 2014?

## Solution

Let $f(x) = 1007 x^{2} + 1008 x = 1007 x(x+1) + x$. Since the product $x(x+1)$ is an even number for all natural $x$, $1007 x(x+1)$ is divisible by 2014 for all such $x$. Therefore, $f(x)$ gives the same remainder when divided by 2014 as $x$. Thus, all numbers $f(1), f(2), \ldots, f(2014)$ have different remainders when divided by 2014.

## Answer

Exists.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-01.jpg?height=241&width=1224&top_left_y=2561&top_left_x=37)

Each of two regular polyhedra $P$ and $Q$ is cut by a plane into two parts. One part of $P$ and one part of $Q$ are attached to each other along the plane of the cut. Can the resulting polyhedron be a regular polyhedron different from the original ones, and if so, how many faces can it have?

## Solution

Let $R$ be the resulting polyhedron. Clearly, the part of polyhedron $P$ contains at least one vertex $A$ not lying in the plane of the cut. The polyhedral angle of polyhedron $P$ at this vertex will also be a polyhedral angle of polyhedron $R$; this means that polyhedra $P$ and $R$ are similar. Similarly, $Q$ is also similar to them. Moreover, if at least one edge of polyhedron $P$ emanating from $A$ does not intersect (even at another vertex!) the plane of the cut, then it will also be an edge in $R$. Then in similar polyhedra $P$ and $R$, the edges are equal, and therefore, the polyhedra are equal, which contradicts the condition.

Thus, the part of $P$ that entered $R$ is a pyramid with vertex $A$. Similarly, the part of $Q$ that entered $R$ is a pyramid with vertex $B$. Therefore, at least half of the faces in $R$ adjoin the same vertex. This excludes the dodecahedron and the icosahedron. If our polyhedra are cubes, then triangular pyramids are cut off from $P$ and $Q$, and the resulting polyhedron has no more than five vertices, which is impossible.

The remaining cases of the octahedron and the tetrahedron are possible, as shown in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-02.jpg?height=448&width=1972&top_left_y=760&top_left_x=40)

## Answer

Yes; 4 or 8 faces.