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34. Prove that there are infinitely many natural numbers $N$ such that the polynomial $x^{8}+N x^{4}+1$ can be factored into factors, each of which is a polynomial of the fourth degree with integer coefficients. | 34. Answer: for example, any $N$ of the form $4 k^{4}-8 k^{2}+2$.
Then
\[
\begin{aligned}
& x^{8}+\left(4 k^{4}-8 k^{2}+2\right) x^{4}+1= \\
& \quad=\left(x^{4}-2 k x^{3}+2 k^{2} x^{2}-2 k x+1\right)\left(x^{4}+2 k x^{3}+2 k^{2} x^{2}+2 k x+1\right)
\end{aligned}
\] | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,265 |
35. There is a pile of 660 stones on the table. In one move, you can split any of the existing piles into two smaller ones. At the same time, the sizes of any two piles on the table at the same time must differ by less than a factor of two. What is the maximum number of piles that can result?
## 9th Grade
## Plot 1
... | 35. Answer: 30
Example: Let's go in reverse - we will combine two piles into one without violating the conditions of the problem. Let's say we initially had piles (two of each kind): $15,15,16,16, \ldots, 29,29$. It is easy to see that the condition is not violated, and the sum is 660. We will combine the two smallest... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,266 |
38. Let $L$ be the point of intersection of lines $A P$ and $C M$, and $S$ be the point of intersection of lines $A N$ and $C Q$. Prove that $L S \| P Q$. | 38. Note that $C I$ is the bisector of angle $P C Q$. Indeed, $P I=I Q$, so point $I$ lies on the midpoint of the arc $P Q$ of the circumcircle of triangle $A I C$. Let $T$ be the point of intersection of the lines $A P$ and $C Q$. By Thales' theorem, the required parallelism is equivalent to the equality $T L / T P = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,269 |
40.
$$
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x}+\frac{z^{2}}{z^{2}+2 x y} \geqslant 1
$$ | 40. Since $2 x y \leqslant x^{2}+y^{2}$, we get
$$
\begin{aligned}
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x} & +\frac{z^{2}}{z^{2}+2 x y} \geqslant \\
& \geqslant \frac{x^{2}}{x^{2}+y^{2}+z^{2}}+\frac{y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{z^{2}}{x^{2}+y^{2}+z^{2}}=1
\end{aligned}
$$ | 1 | Inequalities | proof | Yes | Yes | olympiads | false | 21,271 |
41.
$$
\frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y}>6
$$ | 41. Transform the left side using $2 x y \leqslant x^{2}+y^{2}$, multiplying the numerator and denominator by 2:
$$
\begin{aligned}
& \frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y} \geqslant \\
& \geqslant 2\left(\frac{x^{2}+2 y^{2}+2 z^{2}}{2 x^... | 6 | Inequalities | proof | Yes | Yes | olympiads | false | 21,272 |
42.
$$
\frac{x^{3}}{x^{3}+2 y^{2} \sqrt{z x}}+\frac{y^{3}}{y^{3}+2 z^{2} \sqrt{x y}}+\frac{z^{3}}{z^{3}+2 x^{2} \sqrt{y z}} \geqslant 1
$$ | 42. Solution. Let $a=x \sqrt{x z}, b=y \sqrt{y x}, c=z \sqrt{z y}$ and substitute into the inequality from the first part (taken for variables $\mathrm{a}, \mathrm{b}, \mathrm{c}$). We get
$$
\begin{aligned}
& \frac{(x \sqrt{x z})^{2}}{(x \sqrt{x z})^{2}+2 y \sqrt{y x} \cdot z \sqrt{z y}}+\frac{(y \sqrt{y x})^{2}}{(y ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 21,273 |
43.
$$
\frac{x^{3}}{x^{3}+2 y^{2} z}+\frac{y^{3}}{y^{3}+2 z^{2} x}+\frac{z^{3}}{z^{3}+2 x^{2} y} \geqslant 1
$$
## Plot 3
There is a pile of $n$ stones on the table. In one move, you can split any of the existing piles into two smaller ones. At the same time, the sizes of any two piles on the table must be "similar... | 43.
Solution. Divide the first fraction by $z^{3}$, the second by $x^{3}$, the third by $z^{3}$, and let $\frac{x}{z}=a, \frac{y}{x}=b, \frac{z}{y}=c$. Then the first fraction transforms (considering the identity $a b c=1$) as follows:
$$
\frac{x^{3}}{x^{3}+2 y^{2} z}=\frac{a^{3}}{a^{3}+\frac{2}{c^{2}}}=\frac{a^{3}}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 21,274 |
44. Let similar sizes be those that differ by no more than $\sqrt{2}$ times. Prove that it is impossible to make 3 piles from the original pile. | 44. Let's consider two numbers $x$ and $y$ that resulted from the first step. We will assume that $x \geqslant y$, then $x \leqslant \sqrt{2} y$. Moreover, since $\sqrt{2}$ is irrational, $x < \sqrt{2} y$. Let's verify that it is no longer possible to make the next move. Indeed, suppose we divided $x$ into $x_{1}$ and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,275 |
45. Let similar sizes be those that differ by no more than a factor of two. Prove that then any pile can be divided into piles of one stone. | 45. Let's consider the reverse process: assume we initially have several piles and we can take two piles and combine them together, adhering to the rule of similar sizes. Now we need to ensure that from $n$ unit piles, we can form one pile of size $n$ in this way. We can do this as follows: we will simply take the two ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,276 |
46. Let similar sizes be those that differ by less than a factor of 2. Prove that a pile of 660 stones can be divided into 30 piles. | 46. Let's go in reverse - we will combine two piles into one without violating the conditions of the problem. Let's say we initially had piles (two of each type): $15,15,16,16, \ldots, 29,29$. It is easy to see that the condition is not violated, and the sum is 660. We will combine the two smallest piles (both 15), the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,277 |
49. Given a triangle $D E F$. A circle passing through vertices $E$ and $F$ intersects sides $D E$ and $D F$ at points $X$ and $Y$ respectively. The bisector of angle $\angle D E Y$ intersects $D F$ at point $Y^{\prime}$, and the bisector of angle $\angle D F X$ intersects $D E$ at point $X^{\prime}$. Prove that $X Y \... | 49. The sought parallelism is equivalent to the equality $\frac{D X^{\prime}}{X^{\prime} X}=\frac{D Y^{\prime}}{Y^{\prime} Y}$ (Thales' theorem). By the property of the angle bisector, $\frac{D X^{\prime}}{X^{\prime} X}=\frac{D F}{F X}, \frac{D Y^{\prime}}{Y^{\prime} Y}=\frac{D E}{E Y}$. But triangles $D E Y$ and $D F ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,280 |
50. Let $M$ be the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$ (we choose the arc that does not contain point $C$), and $N$ be the midpoint of the arc $BC$ (we choose the arc that does not contain point $A$). Prove that $MN \parallel PQ$. | 50. Note that the center of the circumcircle $A I C$ is the midpoint of the arc $A C$ of the circumcircle $A B C$ (this is precisely the lemma about the trilemma). Therefore, $I Z$ is the bisector of angle $\angle B$. But as we proved in part (1), $P Q \perp I Z$, so $P Q \perp B I$. But $M N$ is also perpendicular to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,281 |
53. Let $P=7$, and each number in the second row is three times larger than its neighbor in the first row. Prove that the numbers in the two central cells are divisible by 7. | 53. Let the number in the corner be $x$. Then its neighbor from the second row is $3 x$ (here and throughout modulo 7). Now we can get the neighbor from the first row: it is $-2 x \equiv 5 x$, and its neighbor from the second row is $3 \cdot 5 x \equiv x$. From the condition on the sum, we get the next number in the fi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,284 |
54. Let each number in the second row be $K$ times greater than its neighbor in the first row. Prove that each number in this table, when multiplied by $K$, gives the same remainder when divided by $P$ as the sum of the numbers adjacent to it in the column. | 54. We will prove by induction on the row number $i$. We will prove two statements:
$P(i)$ - what needs to be proven according to the condition,
$Q(i)$ - there exists a number $r_{i}$ such that after multiplying the first row by it, the numbers will give the same remainders when divided by $P$ as the $i$-th row (respe... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,285 |
56. Let similar sizes be those that differ by no more than $\sqrt{2}$ times. Prove that then no pile can be divided into three piles. | 56. Let's consider two numbers $x$ and $y$ that resulted from the first step. We will assume that $x \geqslant y$, then $x \leqslant \sqrt{2} y$. Moreover, since $\sqrt{2}$ is irrational, $x < \sqrt{2} y$. Let's verify that it is no longer possible to make the next move. Indeed, suppose we divided $x$ into $x_{1}$ and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,287 |
57. Let similar sizes be those that differ by no more than a factor of two. Prove that then any pile can be divided into piles of one stone each. | 57. Let's consider the reverse process: assume we initially have several piles and we can take two piles and combine them together, adhering to the rule of similar sizes. Now we need to ensure that from \( n \) unit piles, we can form one pile of size \( n \) in this way.
We will do this as follows: we will simply tak... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,288 |
58. Let similar sizes be those that differ by no more than $k$ times. Prove that for any $k<2$ there exists a number $N_{k}$ such that no pile can be divided into more than $N_{k}$ piles. | 58. Let's say, starting from some number, we obtained some multiset $A$, the minimum of which is equal to $m$. Then all its elements are $<2 m$.
Let's run the process in reverse, trying to obtain the original number from the set $A$. The first move we will combine two numbers, both $\geqslant m$, and get some sum $x \... | 2([\log_{\frac{2}{k}}(2)]+1) | Combinatorics | proof | Yes | Yes | olympiads | false | 21,289 |
61. Let $X$ be the intersection point of circles $w_{1}$ and $w_{2}$ such that $X$ and $B$ lie on opposite sides of line $A C$. Prove that then point $X$ lies on the altitude $B H$ of triangle $A B C$. | 61. Consider the intersection point of line $B H$ and circle $w_{1}$, let's call it $X$ and prove that it lies on $w_{2}$. Let $\alpha, \beta, \gamma$ be the angles of triangle $A B C$. Then $\angle H A C=90^{\circ}-\gamma$, and $\angle A H X=\gamma$. Since $A H C X$ is a cyclic quadrilateral, $\angle A C X=\angle A H ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,292 |
62. Let $Y$ be the intersection point of circles $w_{1}$ and $w_{2}$ such that points $Y$ and $B$ lie on the same side of line $A C$. Prove that point $Y$ lies on the median $B M$. | 62. Let $P$ be the point of intersection of the median $B M$ and the circle $w_{1}$. Note that the circle $w_{1}$ is symmetric with respect to the perpendicular bisector of $A C$. Therefore, the point $P^{\prime}$, which is the image of $P$ under the reflection about the perpendicular bisector of $A C$, also lies on th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,293 |
64. Can the degree of the polynomial $P$ be determined by $A_{P}$? | 64. Answer: no.
For example, $A_{x}=A_{x^{3}}$. | no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,295 |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and less than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,296 |
1.2. Prove that there exists $n$ such that $a\left(n, 10^{100}\right) \leqslant 1+10^{-100}$. | Solution. Consider all possible ways to write the number $1+10^{-100}$ as a sum of positive rational fractions with the denominator $10^{200}$ (not necessarily in lowest terms). Clearly, this number is finite, and we denote it by $n$ - by writing down all corresponding sets on cards, we can verify that such a set of ca... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,299 |
1.3. Prove that $a(2,4)<\sqrt{3}$. | Solution. Consider, for example, the following pair of sets:
$$
\left(\frac{21}{34}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right) \text { and }\left(1, \frac{13}{34}, \frac{13}{68}, \frac{13}{102}\right) .
$$
The sum in each is $\frac{347}{204}$, which is even less than 1.71, while $\sqrt{3}=1.73 \ldots$
Let's check... | \frac{347}{204}<\sqrt{3} | Inequalities | proof | Yes | Yes | olympiads | false | 21,300 |
1.4. Is the sequence $a(2, k)-a(1, k)$ bounded? | Solution. Answer: no. Let's compute $a(1, k)$. As already noted, for any $l \leqslant k$ there must be a card where the $l$-th largest number is not less than $\frac{1}{l}$. If there is only one card, then the sum on this card is, therefore, not less than $1+\frac{1}{2}+\ldots+\frac{1}{k}$. On the other hand, the card ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,301 |
3.1. Let $a=4, b=9$. Prove that the desired pair will be found. | Solution. By Fermat's Little Theorem $4^{p-1}+9^{p-1} \equiv 2$, but at the same time $4^{\frac{p-1}{2}}+9^{\frac{p-1}{2}}=2^{p-1}+3^{p-1} \equiv 2$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,306 |
3.4. Prove that the desired pair will be found if $a=2, b=3$, and $\frac{p-1}{2}$ is prime. | # Solution.
Lemma 2. Let $q>2$ be a prime, and $u k<q$ such that for any $x=1,2, \ldots, q-1$, the number $x$ and the remainder of $k x$ modulo $q$ have different parities. Then $k=q-1$.
Proof of Lemma 1. Substituting $x=1$, we get that $k$ is even, and thus $k x$ is always even, meaning the remainder of $k x$ has th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,309 |
2.4. Prove that for any connected figure consisting of 100 cells, two non-overlapping copies of it can be placed in a triangle with a side length of 85.
(Samoilov V.S., Tesler A.A.)
## Plot 3. | 2.4. Consider a connected figure made of $\mathrm{k}$ cells. Consider a graph on the cells (edges connect neighboring cells). This graph is connected, which means it has at least $k-1$ edges. Therefore, there are at least $k-1$ pairs of sides of triangles adjacent to each other, and thus not on the boundary. Therefore,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,323 |
3.3. Let $f$ be a cubic polynomial. In the expression $\pm f(x+1) \pm f(x+2) \pm \ldots \pm f(x+n)$, the signs were arranged such that the result is a constant. Prove that $n$ is divisible by 4. | 3.3. Since the cube must shrink, the number of + and - is the same. Next, let's look at the coefficient of the final difference at $x^{\wedge} 2$. It is equal to $3(+-1+-2+-. .+-n)=3 n(n+1) / 2$, so either $n$ is divisible by 4, or $n+1$ is divisible by 4. The latter is impossible, since $n$ is even. | n | Algebra | proof | Yes | Yes | olympiads | false | 21,324 |
4. Prove that if an ordered triple $(x, y, z)$ can be obtained from an ordered triple $(a, b, c)$ by multiple applications of the specified operations, then the same can be done in one operation.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation... | Solution:
Let's try to get $(x, y, z)$ in one operation. The required result is given by the polynomial
$$
P(t)=x+\frac{y-x}{b-a}(t-a)+\frac{z-\left(x+\frac{y-x}{b-a}(z-a)\right)}{(c-a)(c-b)}(t-a)(t-b)
$$
(This is interpolation by Newton: the first term sets the value $x$ at point $a$, the second, while preserving i... | proof | Logic and Puzzles | other | Yes | Yes | olympiads | false | 21,325 |
1. Let the garden have the shape of an equilateral triangle with a side of 1. Prove that Georgy Konstantinovich will have enough hedgehogs with a total length of $\sqrt{3}$. | # Solution:
Draw the medians to their point of intersection. The length of each median is $\sqrt{3} / 2$, and since the medians intersect each other in the ratio $2: 1$, the total length of the segments of the medians to the points of intersection is $\sqrt{3}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,326 |
3. Let the garden have the shape of an equilateral triangle with a side of 1. Prove that Georgy Konstantinovich will have to buy hedgehogs of length at least $\frac{3 \sqrt{3}}{4}$.
# | # Solution:
Let $M, N$ be the midpoints of sides $AB, AC$ of our equilateral triangle $ABC$, and $K$ be the midpoint of $MN$. Then the projections of the parts of the hedgehogs that have fallen inside the triangle $AMN$ onto the segment $AK$ must completely cover it - otherwise, Erik will slide perpendicular to $AK$ t... | \frac{3\sqrt{3}}{4} | Geometry | proof | Yes | Yes | olympiads | false | 21,327 |
1. Prove that if in a country any two schools are connected by exactly one direct route, then this country is harmonious.
# | # Solution:
Note that in a tree, any two vertices are connected by exactly one path, which is the shortest.
Take three distinct vertices $x, y$, and $z$. Denote the vertices on the path from $y$ to $x$ as $y=x_{0}, x_{1}, x_{2}, \ldots, x_{n}=x$, and the vertices on the path from $y$ to $z$ as $y=z_{0}, z_{1}, z_{2},... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,328 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,335 |
1.2. It is known that $A B$ is parallel to $F C$. Prove that $A X = B Y$.
# | # Solution.
Let's place new points $E^{\prime}$ and $F^{\prime}$ on the rays $B E$ and $C F$ such that $A D = B E^{\prime} = C F^{\prime}$. Then it is obvious that triangle $D E^{\prime} F^{\prime}$ is equilateral. However, when rotating by $60^{\circ}$ around $D$, points $E$ and $E^{\prime}$ transition to $F$ and $F^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,340 |
1.3. Let the initial triangle $A B C$ be fixed and its side equal to $a$, while the other points are not fixed. $F^{\prime}$ is a point different from $F$ such that triangle $D E F^{\prime}$ is equilateral. Find the geometric locus of points $F^{\prime}$ and its length. | # Solution.
This is a segment of length $2 a$. Let $A B C$ be the medial triangle of $A_{0} B_{0} C_{0}$, then $D$ runs along the segment $A C_{0}$, and $E$ runs along $B A_{0}$. The quadrilateral $D E F^{\prime} C_{0}$ is cyclic, with arcs $D E$ and $E F^{\prime}$ each 60 degrees. Therefore, the angle $F^{\prime} C_{... | 2a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,341 |
2.1. Solve the inequality in natural numbers: $n \leqslant n!-4^{n} \leqslant 4 n$. | # Solution.
We will prove by induction that $n!>4^{n}+4 n$. The base case $n=10$ is verified; the inductive step:
$$
(n+1)!>(n+1)\left(4^{n}+4 n\right)=4^{n+1}+16 n+(n-3)\left(4^{n}+4 n\right)>4^{n+1}+4(n+1)
$$
Thus, there are no solutions. | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,343 |
2.4. Solve the inequality in natural numbers: $n \leqslant n!-k^{n} \leqslant k n$.
# | # Solution.
Since the case of composite (in particular, even) $k$ has already been discussed in point 2, as in point 3, it remains to consider the case $n!=k^{n}+k^{2}$. Notice that the left side of the equation is divisible by $k-1$, while the right side is congruent to 2 modulo $k-1$ (and $k>3$). This is a contradic... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,346 |
3.1. How to compute $f_{5}$ in 5 lines, using only conjunctions and disjunctions (i.e., logical AND and OR)?
# | # Solution.
$$
\begin{aligned}
x_{6} & =x_{1} \vee x_{3} \\
x_{7} & =x_{2} \wedge x_{6} \\
x_{8} & =x_{3} \vee x_{5} \\
x_{9} & =x_{4} \wedge x_{8} \\
x_{10} & =x_{7} \vee x_{9}
\end{aligned}
$$ | x_{10}=x_{7}\veex_{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,347 |
3.2. Prove that for all $n \geqslant 4$ the variables $x_{1}$ and $x_{2}$ appear in at least two rows of any program computing $f_{n}$.
# | # Solution.
Notice that $f_{n}$ depends on all its variables, so $x_{1}$ and $x_{2}$ must appear in at least one row. Suppose such a row is exactly one: $x_{n+i}=b_{i}\left(x_{1}, x_{2}\right)$ for some $i$. Consider three functions $g_{1}\left(x_{3}, \ldots, x_{n}\right)=f\left(0,0, x_{3}, \ldots, x_{n}\right), g_{2}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,348 |
3.3. Let $C\left(f_{n}\right)$ be the minimum number of lines in a program computing $f_{n}$. Prove that $C\left(f_{n}\right) \geqslant C\left(f_{n-2}\right)+2$ for $n \geqslant 4$. | # Solution.
Lemma: Let for some variable $x_{n+k}$ the function $b_{k}\left(x_{i}, x_{j}\right)$ depend only on one variable (NVO $b_{k}\left(x_{i}, x_{j}\right)=h\left(x_{i}\right)$, where $b_{k}$ and $h$ can also be constants). Then we can reduce the program by one line.
Proof: Indeed, remove the $k$-th line, and r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,349 |
3.4. Let $M\left(f_{n}\right)$ be the minimum number of lines in a program computing $f_{n}$, where each line computes either a disjunction or a conjunction. Prove that $M\left(f_{n}\right) \geqslant M\left(f_{n-2}\right)+3$ for $n \geqslant 4$. | # Solution.
Let $x_{1}$ and $x_{2}$ participate in the computation of variables $x_{n+i}=b_{i}\left(x_{1}, y\right)$ and $x_{n+j}=b_{j}\left(x_{2}, z\right), i \neq j$ (where $y$ can be equal to $x_{2}$, and $z$ can be equal to $x_{1}$). Substitute $x_{2}=0$ and note that the resulting function does not depend on $x_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,350 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,353 |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and smaller than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,357 |
2. Vasya wrote down 20 consecutive natural numbers in some order. Prove that there will be two numbers standing next to each other that have at least one digit in common.
# | # Solution:
Among 20 consecutive numbers, there will be 10 from the same decade, and they are not single-digit; they share the same digit in the tens place - let's denote it as $N$.
In addition, each digit, including $N$, appears exactly twice among the last digits of 20 consecutive numbers; among them, it appears on... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,358 |
4. Ten consecutive four-digit numbers were written on the board. Then they were erased, replacing each of them with its smallest prime divisor. Prove that now there is a number greater than 10 on the board. | 4.
Assume the opposite. Let's list all prime numbers less than 10:
$2, 3, 5, 7$. Among 10 consecutive numbers, there are always 5 odd numbers, of which no more than two can be divisible by 3. No more than one odd number can be divisible by 5 or 7. Therefore, there will be at least one odd number that is not divisibl... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,364 |
5. The number 2015 was divided into 12 addends, after which all numbers that can be obtained by adding some of these addends (from one to nine) were written down. What is the minimum number of numbers that could have been written down? | 5.
Answer: 10
Example: Let's break down 2015 into 11 numbers of 155 and one number of 310. In this case, the sum of the addends can range from $155 \times 1$ to $155 \times 10$.
We will prove that it is impossible to get fewer than 10 different numbers: first, note that 2015 is not divisible by 12, so we will have ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,365 |
6. Given 2 piles of 50 stones each. Three players are playing: the first player can take 1 stone from one pile and 5 from the other on their turn, the second player can take 2 and 6, and the third player can take 3 and 7. The game ends when someone cannot make a move. Can it happen that the first player made the last m... | 6.
Assume the opposite. Initially, there are 100 stones, and in one round, the players take 24 stones from the piles, so the entire process will take no more than 4 rounds. In one round, no more than 18 stones disappear from one pile, which means that before the first player's move on the third round, there were no f... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,366 |
7. In the country, there are roads with one-way traffic. The King goes for a walk along one of the roads every day, disregarding traffic rules and not walking the same road two days in a row. The Duke, on even days of the month, drives along the road according to the rules, and on odd days, he drives against the rules ... | # 7.
Notice that if there is a cycle of roads in the country, the king will be able to travel along it indefinitely. Suppose there are no cycles, then let's look at the duke's route. In some months, there are 31 days, after the 31st comes the first day of the next month - that is, the duke will travel two days in a ro... | proof | Other | proof | Yes | Yes | olympiads | false | 21,367 |
1. In one city, there are many houses, in each house there are an equal number of apartments, in each apartment there are an equal number of aquariums, in each aquarium there are an equal number of fish, and each fish has an equal number of scales. The number of fish in each house is greater than the number of scales o... | Solution. Let the number of apartments in the building be denoted by K, the number of aquariums in an apartment by A, the number of fish in an aquarium by P, and the number of scales on a fish by S. Then, according to the condition, K$\cdot$A$\cdot$P > A$\cdot$P$\cdot$S. By reducing the positive A$\cdot$P, we get that ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,368 |
2. Vlad numbered the cells of a chessboard (from 1 to 64) in some order. Gosha did the same with his chessboard, but the numbering turned out to be different. Could it be that the cells of Vlad's board are connected by a knight's move if and only if the cells of Gosha's board with the same numbers are connected by a ki... | Solution. It cannot. For example, a corner cell of Vlad's board is connected to two other cells. Then the cell with the same number on Goshi's board should also be connected to exactly two cells, but any cell on a chessboard can be connected by a king's move to no fewer than three other cells. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,369 |
3. Ekaterina told Andrei a certain integer. Prove that Andrei will be able to write 10 consecutive integers on the board, and then erase one of them, so that the sum of the nine remaining numbers is equal to the number Ekaterina told him. | Solution 1. Write down the sum of any NINE consecutive numbers, for example, $1+2+3+4+5+6+7+8+9=45$. Then add 1 to the largest number: $1+2+3+4+5+6+7+8+10=46$. Then add 1 to the next largest number: $1+2+3+4+5+6+7+9+10=47$. Then add 1 to the next largest number, and so on until all numbers have increased by 1: $2+3+4+5... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,370 |
4. Inside a square, four points are marked and nine segments are drawn (see figure). The lengths of all segments are 1. The segments drawn from the points to the sides of the square are perpendicular to them. Find the length of the side of the square. | Solution. Let the side of the square be X. Then the distance
from the left side of the square to the left bottom point is (X-1)/2, and the distance from the right side of the square to the ri... | 2.6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,371 |
5. It is known that the divisors of each "non-square" number can be divided into pairs so that the products of the divisors in each pair are equal. For example, $18=1 \cdot 18=2 \cdot 9=3 \cdot 6$. But do there exist "non-square" numbers, all divisors of which can be divided into triples so that the products of the div... | Solution. They do not exist. Suppose such a number is found. Let's denote it by $\mathrm{N}$, the number of its divisors by $\mathrm{k}$, and their product by $\Pi$. Then $\Pi=\mathrm{N} k / 2$, since the product of divisors in each of the $\mathrm{k} / 2$ pairs equals N. Let the product of divisors in each triplet be ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,372 |
6. The sum of four real numbers is 10, and the sum of their squares is 30. Prove that some two of them differ by no more than 1. | Solution. Let the numbers be $a, b, c, d$. Then $a+b+c+d=10, a^{2}+b^{2}+c^{2}+d^{2}=30$. From this, $(a+b+c+d)^{2}-a^{2}-b^{2}-c^{2}-d^{2}=2 a b+2 a c+2 a d+2 b c+2 b d+2 c d=100-30=70$. Then $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3a^{2}+3b^{2}+3c^{2}+3d^{2}-2ab-$ $2ac-2ad-2bc-2bd-2cd=90-70=20$. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,373 |
7. In the security company "OOO", 2015 employees work. Several rapid response teams (several people in each) have been formed from them, and any two teams have at least one common employee. Prove that all employees of the company "OOO" can be arranged around a Very Secure Object along a circle of length 1 km in such a ... | Solution. Suppose the opposite. Let's denote the maximum length to which all rapid response groups can be stretched as m (by our assumption $m < 1 / 3$). Among all possible arrangements for which the condition of stretching all groups to $m$ is satisfied, we choose the one in which the number of groups stretched exactl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,374 |
2. There are 1000 crows sitting on a wire. At the end of each minute, every third (third, sixth, ninth, and so on) crow flies away.
a) Which crows, by their initial count, will remain on the wire in the end?
b) How many minutes will pass before the crows stop flying away? | Solution. a) It is clear that at the end there will be no more than two ravens left, and the first and second never fly away. Therefore, it will be precisely these two. Answer: 1 and 2.
b) It is easy to see that the number of ravens flying away in the next minute is one-third of their number, rounded down. Thus, it on... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,376 |
3. We will call a number consisting of identical digits beautiful. Can any five-digit number be represented as the sum of beautiful numbers of pairwise different lengths?
# | # Answer: No.
Solution (1). If the required sum contains a number with at least five digits, then the result is no less than 11111. Otherwise, it is no more than $9+99+999+$ $9999=11106$. Therefore, none of the numbers from 11107 to 11110 can be represented in this way.
Solution (2). The required representation of a ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,377 |
4. Seven natives from several tribes are sitting in a circle by the fire. Each one says to the neighbor on their left: “Among the other five, there are no members of my tribe.” It is known that the natives lie to foreigners and tell the truth to their own. How many tribes are represented around the fire? | Solution. If there are at least 4 natives from one tribe, then two of them sit next to each other, and one of them will lie to the other, although they should tell the truth. If there is only one native from a certain tribe, then he tells the truth to his left neighbor, although he should lie. Therefore, each tribe has... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,378 |
6. Vasya and Petya thought of two different numbers, and each of them turned out to have as many prime divisors as composite ones. Can Vasya's and Petya's numbers have common divisors greater than 1? | Solution. Let the number $n$ have more than two prime divisors, counting multiplicity. Then for each prime $p$ dividing $n$, the number $n / p$ is composite (and all these numbers are distinct). Also, $n$ divides itself, so there are more composite divisors than prime ones. Therefore, there cannot be more than two prim... | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,379 |
7. In the corners of a square courtyard stand four houses, in which hooligans who are friends with each other live. Starting from January 1, 2017, every day, two hooligans from neighboring houses would quarrel and remain in a quarrel forever. By January 1, 2018, for the first time, there was no one left to quarrel with... | Solution. Divide the hooligans into two gangs: each will unite hooligans from opposite houses. Let there be $X$ hooligans in one and $Y$ in the other. To make sure all necessary hooligans quarrel, $X \cdot Y$ quarrels are required, and this equals 365. This number has only two prime factors: 5 and 73. Therefore, there ... | 78or366 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,380 |
1. In two examples, we replaced letters with digits (the same digits with the same letters, different ones with different ones). It is known that ДВАЖДЫ+ДВА
is divisible by 13. Prove that ТРИЖДЫ+ТРИ is also divisible by 13. | # Solution:
Notice that DVADVА is divisible by 1001, and 1001 is divisible by 13, so ZHDY is divisible by 13, which means TRICTRI + ZHDY is divisible by 13, so TRIZHDY + TRI is also divisible by 13, and so on. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,381 |
2. There is a number 1201201201201. In one move, you can swap two adjacent digits, but it is forbidden to swap digits on the positions where they have already been swapped. Additionally, 0 cannot be placed in the first position. Two players take turns, the one who cannot make a move loses. Who will win with correct pla... | # Solution:
First, swap the second and third digits. After this move, no matter how you play, exactly 10 more moves will occur, so the first player will win. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,382 |
4. At a round table, 10 people were sitting. Each of them was either a knight (who always tells the truth) or a liar (who always lies). One day, each of them accused four other people, who were not sitting next to them, of lying. Prove that there was such a person who accused the person sitting opposite them of lying.
... | # Solution:
Notice that there is at least one knight and at least one liar. Therefore, there will be a pair of adjacent people where one is a knight and the other is a liar. A knight could accuse only liars of lying, and since he did not accuse his neighbor, there must be at least 5 liars. Similarly, there must be at ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,384 |
5. Is it possible to divide a triangle into convex quadrilaterals in such a way that no three of the quadrilateral vertices lie on the same straight line?
# | # Solution:
The sum of the angles of quadrilaterals is $180 * 2 \mathrm{k}$, where $\mathrm{k}$ is the number of quadrilaterals. At the same time, it is equal to $180+180 * 2 n$, where n is the number of additional vertices we marked to cut the triangle. That is, $2 \mathrm{k}=2 \mathrm{n}+1$ - a contradiction. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,385 |
6. Prove that a number of the form $3^{\mathrm{n}} * 7^{\mathrm{k}}$ cannot be written using only odd digits for natural numbers $\mathrm{n}$ and $\mathrm{k}$.
# | # Solution:
Assume the opposite. We will look at the remainders when divided by 20. The power of three gives remainders 3, 9, 7, and 1. The power of seven gives remainders 7, 9, 3, and 1. Since the remainders coincide, any remainder we can get from our product, we can also get from the power of three. That is, it will... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,386 |
7. On a 5 by 5 board, a certain number of bishops and knights were placed. Then, they counted how many times each bishop is attacked, and all the obtained numbers were added together. What is the maximum sum that could have been obtained?
# | # Solution:
Answer: 80.
The example is constructed through a chessboard coloring - bishops on one color, rooks on the other.
Evaluation: First, place bishops everywhere. The total sum will be 64. Now, in each cell, write the number of points that can be gained by replacing a bishop with a knight. We will get the fol... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,387 |
3. Once, on the last day of the month, while looking at the wall calendar, Kostya said with surprise: "Look, in this month the number of B-days (that's what Kostya called Tuesdays and Sundays), S-days (Wednesdays and Saturdays) and P-days (Mondays and Fridays) are the same!" "Nothing surprising," replied his friend And... | Solution: Let's call a month good if $B=P=C$.
Notice that over the course of a week (regardless of which day it starts), $B=C=P=2$.
Therefore, any 28-day month is good (it has 4 weeks, $B=P=C=8$).
In a 29-day month, there are 4 weeks and one extra day. Since for four weeks $B=P=C$, the extra day (the 29th) must not ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,389 |
6. There are 8 gold coins, one of which is counterfeit, and 8 silver coins, one of which is also counterfeit. The gold coins weigh 20 g each, the silver coins weigh 10 g each, and the counterfeit coins are 1 g lighter than the genuine ones. There are two-pan scales without weights. Only gold coins are allowed on the le... | # Solution:
First: 3333 vs CCCCCCCC, find four real gold («>» means the coins on the left scale, «=» means those that did not participate in the weighing) Second: 33 (real) vs $\operatorname{CCCC}($ find 4 real silver) Third: 3 (real) vs CC (from the suspicious ones) (find +2 real silver)
Fourth: 3 (real) vs CC (one ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,392 |
7. At a round table sit knights, who always tell the truth, and liars, who always lie. Counterintelligence reported that one of those sitting at the table might also be a spy, whom everyone considers a knight, although he always lies. When questioned, 12 of those sitting at the table stated, "My right neighbor is a lia... | Solution: If there is no spy, the statement "my neighbor is a liar" means a knight is next to a liar or a liar is next to a knight, i.e., "alternation." Since the table is round, the number of such alternations is even. If there is a spy, the sequence "L-S-R" will give two "my neighbor is a liar" responses instead of o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,393 |
1.1. It is known that $f(f(x))=x^{4}+4 x^{3}+8 x^{2}+8 x+4$. Find all such polynomials $f(x)$. | Solution. Let $f(x)=a x^{2}+b x+c$. Then $f(f(x))=a^{3} x^{4}+2 a^{2} b x^{3}+\ldots$ From this, $a=1, b=2$. Finally, let's look at the coefficient of $x^{2}$ and determine that $c=1$. Indeed,
$$
\left(x^{2}+2 x+1\right)^{2}+2\left(x^{2}+2 x+1\right)+1=x^{4}+4 x^{3}+8 x^{2}+8 x+4
$$ | f(x)=x^2+2x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,394 |
1.2. Let $q<0$, and let $f(x)=x^{2}+p x+q$. Prove that the polynomial $f(f(\ldots f(x) \ldots))$ has a root. | Solution. We will prove that $f(f(\ldots f(x) \ldots))$ (composition taken $n$ times) has a positive root. We will prove this by induction on $n$. Base case: $n=1$. Indeed, $f(0)=q<0$, and the branches of the parabola are directed upwards. Inductive step: let $x_{0}$ be the positive root for $f(f(\ldots f(x) \ldots))$ ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,395 |
1.3. Prove that there does not exist a polynomial $f(x)$ such that
$$
f(f(\ldots f(x) \ldots))=x^{243}-3 x^{162}-4 x^{81}+13
$$
( $f$ is applied more than once). | Solution (1). Note that $243=3^{5}$, which means that $f(x)$ is a polynomial of the third degree, and the composition is applied 5 times - there are simply no other options. Suppose the degree of $f$ is divisible by 3, but $f$ contains monomials with exponents not divisible by three. Note that the same is true for any ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,396 |
1.4. The polynomial $f(x)$ satisfies the equation
$$
f(f(x))+f(-f(-x))=f(-f(x))+f(f(-x))
$$
Prove that $f(-x)=f(x)$. | Solution. Consider the polynomial $h(x)=f(x)-f(-x)$. Then the condition can be rewritten as $f(f(x))-f(-f(x))=f(f(-x))-f(-f(-x))$ or $h(f(x))=h(f(-x))$. Note that $h(x)$ is an odd polynomial, i.e., $h(x)=2 a_{2 n+1} x^{2 n+1}+\ldots$, where $a_{2 n+1}$ is half the leading coefficient of $h$, which is also the coefficie... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,397 |
3.1. Let $N=8$. Prove that it is possible to achieve no more than one black chip remaining. | Solution. We will show that if there are more black chips, their number can be reduced. This is obvious if they stand every other (we perform the first operation) or next to each other (then, if not all chips are black, we find a triplet BCC and perform the second operation). If black chips stand every two - also (we t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,402 |
3.3. Let $N=15$ and only the second operation is allowed. Prove that from any arrangement, less than $10^{4}$ others can be obtained. | Solution. We will paint the positions in three colors; the parity of the number of blacks on all positions changes simultaneously. With a fixed set of parities, there are exactly $2^{12}$ arrangements, and with a fixed set up to inversion $-2^{13}<10000$. | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,403 |
3.4. Let the number of chips $N=1000$ and exactly one of them is black (both operations are allowed again). Can we obtain a configuration with only black chips? | Solution. No. Let $a_{i}=0$ if there is a white chip at the $i$-th position and $a_{i}=1$ otherwise. Consider the sum $\sum a_{i} 2^{i}$ modulo 5 - it is well-defined, since the remainders modulo 5 repeat with a period of 4, and 1000
is divisible by 4. The first operation adds or subtracts from the sum $2^{k}+$ $2^{k+2... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,404 |
1. Once, two elves Ilsa and Elsa, and two dwarves Bilyn and Dilyn gathered. One of them gave something to their human friend Vasily, after which each of them spoke:
- The gift was a sword.
- I didn't give anything!
- Ilsa gave a necklace.
- Bilyn gave a sword.
It is known that the elves lie when talking about dwarves... | Solution. The phrase "I didn't give anything" could not have been said by the elf, so it must have been said by the dwarf. Since this phrase is about a gift, it is a lie, and this dwarf must have given something to Vasily. Then the phrase "Ilse gave the necklace" is a lie, and it could not have been said by the elf. Th... | Dilyngavethesword | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,405 |
2. A hundred mages and a hundred alchemists gathered together to turn iron into gold. Each mage took three magic wands. For the experiment to be successful, a hundred people from those gathered must sit at a round table, and each mage sitting at the table must point their magic wands at two neighbors and the person sit... | Solution. From condition 1, it follows that if, for example, two mages sit next to each other, then two alchemists must follow them, then two mages again, and so on. This means that alchemists and mages either alternate (...AMAMAM...), or sit in pairs (...AAMMAMAM...). In both cases, the second condition is violated: e... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,406 |
3. Nine natural numbers are given. The first is written only with ones, the second only with twos, ..., the ninth only with nines. Can one of these numbers be equal to the sum of all the others? | Solution. No, because in this case the sum of all numbers should be even, but it is odd. Another possibility is to try all options for the last digit. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,407 |
4. The results of the fishing competition are summed up based on the points in three categories (float, feeder, spinner), and in each category, the first three places, the next three, the next three, and so on are awarded. The first place gives 1 point, the second place gives 2 points, and so on (the fewer points, the ... | Solution. Let $\mathrm{N}$ participants have overtaken him. This means that each of them scored no more than 5. Therefore, all together they ($\mathrm{N}+1$ people) scored no more than $5 \mathrm{~N}+6$. On the other hand, in this sum, there are only 9 ones and 9 twos, and the remaining $(3 \mathrm{~N}+3)-18$ results a... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,408 |
6. A $3 \times 3$ grid is filled with numbers from 1 to 9, as shown on the right. Can two more $3 \times 3$ grids be filled with the same numbers so that the following condition is met: any two numbers can be adjacent (by side) in no more than one of the three
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8... | Solution. Let's count the number of "neighborhoods". In each square, there are 12, so in three squares, there are 36. But each number has no more than 8 neighbors, so the total is no more than $9 \cdot 8 / 2 = 36$. Equality means that each pair of neighbors appears somewhere, i.e., each number has exactly 8 neighbors.
... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,410 |
7. A digit (not the first or the last) was erased from a power of two, then the two parts were shifted. Prove that the result cannot be another power of two. | Solution. Let the original number be $\mathrm{N}$, and the obtained number be $\mathrm{M}$. If the erased digit is neither the second nor the third, then $\mathrm{M}$ differs from $\mathrm{N}/10$ by less than $1 \%$, meaning it is greater than $\mathrm{N}/8$ but less than $\mathrm{N}/16$.
If the erased digit is not th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,411 |
1. Timur thought of a three-digit number and told it to Anton, who then wrote down three numbers that result from replacing one of the digits of the original number with 1 (for example, if the number 215 was thought of, Anton would write down 115, 215, 211). The sum of all the numbers written down by Anton turned out t... | Solution: Notice that from the number abc, we get the sum $1 b c + a 1 c + a b 1 = 111 + 2^{\star} a b c$. From the equation $1243 = 111 + 2 * a b c$, it follows that $a b c = 566$. | 566 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,412 |
3. Once, on the last day of the month, while looking at the wall calendar, Kostya said with surprise: “Look, in this month the number of B-days (that's what Kostya called Tuesdays and Sundays), S-days (Wednesdays and Saturdays) and P-days (Mondays and Fridays) are the same!” “Nothing surprising,” replied his friend And... | Solution: Let's call a month good if $B=P=C$.
Notice that over the course of a week (regardless of which day it starts), $B=C=P=2$.
Therefore, any 28-day month is good (it has 4 weeks, $B=P=C=8$).
In a 29-day month, there are 4 weeks and one extra day. Since for four weeks $B=P=C$, the extra day (the 29th) must not ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,413 |
5. There are 8 gold coins, one of which is counterfeit, and 8 silver coins, one of which is also counterfeit. The gold coins weigh 20 g each, the silver coins weigh 10 g each, and the counterfeit coins are 1 g lighter than the genuine ones. There are two-pan scales without weights. Only gold coins are allowed on the le... | # Solution:
First: 3333 vs CCCCCCCC, find four real gold («>» means the coins on the left pan, «=» means those that did not participate in the weighing)
Second: 33 (real) vs $\mathrm{CCCC}($ find 4 real silver)
Third: 3 (real) vs $\mathrm{CC}$ (from the suspicious ones) (find +2 real silver)
Fourth: 3 (real) vs $\m... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,415 |
6. At a round table sit knights, who always tell the truth, and liars, who always lie. Counterintelligence reported that one of those sitting at the table might also be a spy, whom everyone considers a knight, although he always lies. When questioned, 12 of those sitting at the table stated, "My right neighbor is a lia... | Solution: If there is no spy, the statement "my neighbor is a liar" means a knight is next to a liar or a liar is next to a knight, i.e., "alternation." Since the table is round, the number of such alternations is even. If there is a spy, the sequence "L-S-R" will give two "my neighbor is a liar" responses instead of o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,416 |
7. One hundred points are arranged in a circle, next to which numbers $1, 4, 9, 16$, ..., 10000 (the squares of the first 100 natural numbers) are written in blue pen. Each pair of points is connected by a segment, on which the sum of the two numbers at the ends of the segment is written in red pen. Which is greater: t... | Solution: Each of the blue numbers is included in 50 odd red sums (when added to 50 numbers of the opposite parity) and in 49 even red sums (when added to the remaining 49 numbers with the same parity). Therefore, its contribution to the odd sums is greater. Thus, the sum of the odd red numbers is greater than the sum ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,417 |
1. Find all sequences with natural non-repeating terms, for which $a_{n}$ is divisible by $a_{a_{n}}$ for all $n$. | Solution. Only $a_{n}=n$, because if there exists $k \neq a(k)$, we get an infinite decreasing sequence of iterations $a(k), a(a(k))$., which is impossible. | a_{n}=n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,418 |
2. A tournament of the game "Rock, Paper, Scissors" ${ }^{3}$ is held according to the Olympic system, ${ }^{4}$ with 16 players registered. The organizers have determined who will play against whom in the first match, which winners of which matches will play against each other in the next, and so on. Each participant ... | Solution. Let's consider the general case with $2^{n}$ players. To uniquely draw the tournament bracket, it is sufficient to know the figure of the tournament winner and whether the right or left player won in each of the $2^{n}-1$ matches. Indeed, then the figures of the winners of the left and right semi-finals are u... | 3\cdot2^{15} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,419 |
3. A line intersects sides $A B$ and $B C$ of square $A B C D$ at points $X$ and $Y$, and the extensions of sides $A D$ and $C D$ at points $Z$ and $T$. Prove that triangles $C X Z$ and $A Y T$ have the same area. | Solution. Note that angle $Z X A$ is equal to angle $Y T C$, as alternate interior angles when $A B$ and $C D$ are parallel and $Z T$ is the transversal. Therefore, the right triangle $Z X A$ is similar to the right triangle $Y T C$ by the acute angle. Let $\frac{Z X}{Y T}=k$. Then if $A H$ is the altitude of triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,420 |
7. Natural numbers $a, b, c, d$, $e$ and $f < a$ are such that $a b d + 1$ is divisible by $c$, $a c e + 1$ is divisible by $b$, and $b c f + 1$ is divisible by $a$. Prove that if $d / c < 1 - e / b$, $\operatorname{mod} d / c < 1 - f / a$. | Solution. Consider the expression $a b d+a c e+b c f+1$. It is obvious that it is divisible by $a, b$ and $c$, and therefore by $a b c$, since $a, b$ and $c$ are pairwise coprime.
Multiplying the inequality $d / cb d+c e$. Thus, $a b c \geqslant a b d+a c e+1$. Also, $a b c>b c f$, since $a>f$. Therefore, $a b d+a c e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,421 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,423 |
4. In the equality, digits were replaced with letters: identical digits with identical letters, and different ones with different letters. The result was
$$
Y!\cdot P!\cdot A!=K P O C C
$$
What could the original equality have looked like? Find all the options and prove that there are no others.
$N!$ denotes the pro... | Solution. If C is not 0, then the left side does not exceed $24 \cdot 6 \cdot 2<10000$.
Then $\mathrm{C}=0$, which means there are two digits on the left that are not less than 5. That is, the left side is not less than $1! \cdot 5! \cdot 6!$, and it cannot be greater due to the five-digit nature. This results in the ... | 1!\cdot5!\cdot6!=7200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,430 |
5. Vasisualiy Lokhankin has 8 identical cubes. Engineer Ptiburdikov took a cube and wrote all the numbers from 1 to 6 on its faces, one number per face; then he did the same with the other cubes. Is Vasisualiy now guaranteed to be able to arrange these cubes into a $2 \times 2 \times 2$ cube such that the sum of the vi... | Solution. No. Let 1 be opposite 4, 2 be opposite 5, and 3 be opposite 6 on each die. Then at each vertex, faces with different remainders when divided by 3 meet, so the sum of the visible numbers on one die is always divisible by 3. Therefore, the total sum of the visible numbers is divisible by 3. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,431 |
6. In the cells of an infinitely long tape, natural numbers are recorded in order. Vasisuaili Lokhankin and Engineer Ptiburdikov took out cardboard rectangles marked with cells of the same size. Each cut out several cells in one row of their rectangle. At the same time, Vasisuaili cut out one more hole than Ptiburdikov... | Solution. Let's arrange it somehow so that the sum of Ptitburdukov's numbers is greater than the sum of Vasisualiy's numbers (for example, so that one of the numbers visible to Ptitburdukov is greater than the sum of all the numbers visible to Vasisualiy). Now let's move both figures one cell to the right, the differen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,432 |
7. On the table lie several piles, each containing an equal number of stones. Vasisualiy Lokhankin and Engineer Ptiburdukov are playing the following game. On a turn, one must move several stones (but not all) from one pile to another, but they must move more stones than were moved on the previous turn (on the first tu... | Solution. Obviously, when the move number exceeds the total number of stones, the game ends, i.e., there is no draw.
After the first player's move, the situation will be « $n-x, n+x, n, \ldots, n »$ If there is now a winning move in $2 x+1$ or more stones, the second player makes it. Otherwise, he transfers $2 x$ ston... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,433 |
1. The lines containing the sides of a certain quadrilateral are given by the equations $\mathrm{y}=\mathrm{ax}+\mathrm{b}$, $\mathrm{y}=\mathrm{ax}+\mathrm{c}, \quad \mathrm{y}=\mathrm{dx}+\mathrm{b}, \quad \mathrm{y}=\mathrm{dx}+\mathrm{c}$. Find the coordinates of the point of intersection of the diagonals of the co... | Solution. Obviously, the quadrilateral is a parallelogram. Two opposite vertices of it (corresponding to the intersections of the pairs of lines $\mathrm{y}=\mathrm{ax}+\mathrm{b}, \mathrm{y}=\mathrm{dx}+\mathrm{b}$ and $\mathrm{y}=\mathrm{ax}+\mathrm{c}, \mathrm{y}=\mathrm{dx}+\mathrm{c}$) have coordinates (o,b) and (... | (0,\frac{b+}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,434 |
2. A 13-digit display shows the number 1201201201201. Robots C3PO and R2D2 take turns rearranging its digits. In one move, they can swap two adjacent digits, but it is forbidden to swap digits on positions that have already been swapped by either robot. Additionally, a zero cannot be placed in the first position. The p... | Solution. C3PO will win. It is enough for him to swap the first zero with the two on his first move. After this, in any case, 10 more moves will be made - one can freely perform permutations in all pairs of adjacent positions, except for the 2nd and 3rd (which have already been swapped) and the 1st and 2nd (swapping wh... | C3PO | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,435 |
3. Let's call a pair of natural numbers correct if the larger number of the pair is divisible by the smaller one. Do there exist 46 different natural numbers that form exactly 1000 correct pairs? | Solution. Yes. For example, the numbers 2, $2^{2}, 2^{3}, \ldots, 2^{45}$ form $45 \cdot 44 / 2=990$ correct pairs. As the 46th number, we can take $3 \cdot 2^{10}$, which will form 10 correct pairs with the others. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,436 |
4. Is it possible to fill all cells of an infinite grid with distinct natural numbers so that the numbers in any pair of adjacent cells differ by no more than 2015? | Solution. No. It is clear that in any square of size NxN, the numbers differ by no more than $4029 \mathrm{~N}$. At the same time, it contains $\mathrm{N}^{2}$ different numbers. For $\mathrm{N}>4030$, this is impossible. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,437 |
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