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10.3. Given natural numbers $1,2,3, \ldots, 999$. In one move, it is allowed to replace any two numbers with their difference. Prove that after 998 moves (the last one), an even number will remain. | 10.3. Initially, there is an even number (500) of odd numbers. On each step, the parity of the number of odd numbers is preserved. Therefore, before the last move, there will be two odd numbers or two even numbers, and an even number will remain. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,332 |
10.4. In triangle $A B C$, a point $P$ is taken on side $A B$ such that $B P = B C$ $(A B > B C)$. The bisector $B M$ of triangle $A B C$ intersects the circumcircle of the triangle at point $N$. Prove that quadrilateral $A P M N$ is cyclic. | 10.4 In triangle $A B C$, a point $P$ is taken on side $A B$ such that $B P = B C$ ( $A B > B C$ ). The bisector $B M$ of triangle $A B C$ intersects the circumcircle of the triangle at point $N$. Prove that the quadrilateral $A P M N$ is cyclic.
In triangles $P B M$ and $C B M$, the sides $P B$ and $B C$, and the sid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,333 |
10.5. Uncle Chernomor assigns 9 or 10 of his thirty-three bogatyrs (knights) to duty each evening. What is the smallest number of days after which it can happen that all bogatyrs have been on duty the same number of times? | # 10.5. 7.
Let \( m \) and \( n \) be the number of days when 9 and 10 heroes were on duty, respectively. Let \( k \) be the number of days each hero was on duty. Then \( 9m + 10n = 33k \).... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,334 |
11.5. Given an infinite grid plane. A teacher and a class of 30 students play a game, taking turns - first the teacher, then each of the students in turn, then the teacher again, and so on. In one move, a player can paint a unit segment that is a boundary between two adjacent cells. Segments cannot be painted more than... | Answer: They will not be able to.
Solution: The teacher will choose a square $K$ of size $100 \times 100$ and will color segments of its boundary if possible. Suppose that before her $n$-th move, all these segments are colored. Then $n \leqslant 401$, since there are 400 segments on the boundary. By this point, no mor... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,335 |
11.6. In the tetrahedron \( SABC \), the lengths of all six edges are distinct. Point \( A' \) in the plane \( SBC \) is symmetric to point \( S \) with respect to the perpendicular bisector of segment \( BC \). Points \( B' \) in the plane \( SAC \) and \( C' \) in the plane \( SAB \) are defined similarly. Prove that... | Solution. We will denote the circumcircle of triangle $XYZ$ as $(XYZ)$. Let $\omega$ be the circumsphere of the tetrahedron $SABC$. It intersects the plane $SAB$ along the circle $(SAB)$. The point $C'$ lies on the circle $(SAB)$, and therefore on the sphere $\omega$. Similarly, we find that the points $A'$ and $B'$ li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,336 |
11.7. Find all permutations $\left(a_{1}, a_{2}, \ldots, a_{2021}\right)$ of the numbers $1,2, \ldots, 2021$ such that for any natural numbers $m, n$ satisfying the condition $|m-n|>20^{21}$, the inequality holds:
$$
\sum_{i=1}^{2021} \operatorname{GCD}\left(m+i, n+a_{i}\right)<2|m-n|
$$
(A permutation ( $a_{1}, a_{2... | Answer. The identity permutation, that is, $a_{i}=i$.
Solution. Consider the permutation $\left(a_{1}, a_{2}, \ldots, a_{2021}\right)$, for which the condition of the problem is satisfied. Let $d_{i}=i-a_{i}$. Obviously, the sum of all numbers $d_{i}$ is zero. Suppose not all $d_{i}$ are 0, in such a case there exist ... | a_{i}=i | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,337 |
11.8. Each of 100 girls has 100 balls; among these 10000 balls, there are 100 balls of 100 different colors. Two girls can exchange, passing one ball to each other. They want to achieve that each girl has 100 balls of different colors. Prove that they can achieve this with a series of exchanges such that any ball parti... | Solution. Lemma. Let $k$ be a natural number, and each of the hundred girls has $k$ balls, and in total they have $k$ balls of each of 100 colors. Then the girls can each choose one of their balls so that all 100 balls are of different colors.
Proof. We will say that a girl is friends with a color if she has a ball of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,338 |
1. Without using calculators or other computational tools, determine which of the two numbers is greater:
$$
\frac{1+\sqrt{5}}{2} \text { or } \sqrt[6]{18} ?
$$ | 1. Answer: $\sqrt[6]{18}$.
Let's raise both numbers to the sixth power. To find $\left(\frac{1+\sqrt{5}}{2}\right)^{6}$, we can use formulas for simplified multiplication, sequentially cubing and squaring. However, let's consider a simpler method from a computational standpoint.
First, note that the number $x=\frac{1... | \sqrt[6]{18} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,339 |
2. There are four old Soviet copper coins with denominations of 1 kopek, 2 kopeks, 3 kopeks, and 5 kopeks. The weight of each coin in grams should be numerically equal to its denomination. It is known that one of the coins is defective and differs in weight from the normal one. How can you determine the defective coin ... | 2. For the first weighing, place coins 2 and 3 on one pan, and coin 5 on the other. If balance is achieved, the counterfeit coin is 1. For the second weighing, place 1 and 2 on one pan, and 3 on the other. If balance is achieved, the counterfeit coin is 5.
Assume that balance was not achieved in either weighing. Then ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,340 |
4. Is it possible to form 16 three-digit numbers using only three different digits such that all 16 numbers give different remainders when divided by 16? (It is not necessary for each three-digit number to consist of different digits.) | 4. Answer: no, it cannot.
Let's show that at least one of the remainders cannot be obtained in this way.
First of all, if all three digits have the same parity, then the remainders obtained from the numbers also have the same parity. Therefore, for three digits, we will consider two cases: EEI and III (two digits are... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,341 |
5. An isosceles triangle \(ABC\) with vertex \(B\) is inscribed in a circle of radius \(R\). A circle of radius \(r\) is inscribed in it. Another circle of radius \(r\) is inscribed in the segment bounded by the chord \(AB\) and the smaller arc. Find the ratio \(R: r\).
For a complete solution to each problem, 7 point... | 5. Answer: $\frac{8}{3}$.
Let $a=AB=BC, AC=2b$. Then the height dropped to the base is $h=\sqrt{a^{2}-b^{2}}$. The area is $S=bh$, and the semiperimeter is $p=a+b$. Therefore, $r=\frac{S}{p}=\frac{b \sqrt{a^{2}-b^{2}}}{a+b}$.
By the Law of Sines, $a=2R \sin \hat{A}=\frac{2Rh}{a}$, from which $R=\frac{a^{2}}{2 \sqrt{a... | \frac{8}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,342 |
10.5. We will call a non-empty (finite or infinite) set $A$, consisting of real numbers, complete if for any real numbers $a$ and $b$ (not necessarily distinct and not necessarily in $A$) such that $a+b$ is in $A$, the number $ab$ is also in $A$. Find all complete sets of real numbers.
(N. Agakhanov) | Answer. There is only one such set: the set $\mathbb{R}$ of all real numbers.
First solution. Let $A$ be a complete set. Since it is non-empty, we can choose an element $a \in A$. Then $a+0=$ $=a \in A$, so $a \cdot 0=0 \in A$. Since $(-x)+x=0 \in A$, we now get that $(-x) \cdot x=-x^{2} \in A$ for all real $x$. By th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,343 |
10.6. Inside an isosceles trapezoid \(ABCD\) with bases \(BC\) and \(AD\), there is a circle \(\omega\) with center \(I\), touching segments \(AB\), \(CD\), and \(DA\). The circumcircle of triangle \(BIC\) intersects side \(AB\) again at point \(E\). Prove that line \(CE\) is tangent to circle \(\omega\).
(B. Obukhov) | Solution. Note that $I$ lies on the axis of symmetry of the trapezoid, so $\angle I C D=\angle I B A$. Using the cyclic nature of quadrilateral $C B E I$, we get $\angle I C D=\angle I B A=\angle I B E=\angle I C E$. Since line $C D$ is tangent to circle $\omega$, then the line $C E$, symmetric to it with respect to $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,344 |
10.8. Find all pairs of distinct real numbers \(x\) and \(y\) such that \(x^{100} - y^{100} = 2^{99}(x - y)\) and \(x^{200} - y^{200} = 2^{199}(x - y)\).
(I. Bogdanov) | Answer. $(x, y)=(2,0)$ and $(x, y)=(0,2)$.
Solution. For convenience, let's make the substitution $x=2a$ and $y=2b$. Then from the condition we have $(2a)^{100}-(2b)^{100}=2^{99} \cdot (2a-2b)$ and
$(2a)^{200}-(2b)^{200}=2^{199} \cdot (2a-2b)$. Dividing both equations by powers of two, we get $a^{100}-b^{100}=a^{200}... | (x,y)=(2,0)(x,y)=(0,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,345 |
9.5. There are 23 students in the class. During the year, each student in this class celebrated their birthday once, to which some (at least one, but not all) of their classmates came. Could it be that every two students in this class met at such celebrations the same number of times? (It is assumed that at each party,... | Answer. Yes, it could.
First solution. Let's provide an example of how this could happen. Arrange the students in a circle. Suppose that on each student's birthday, all classmates except the next one in the clockwise direction came. Then any two students $A$ and $B$ met at all celebrations except two: the one to which... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,346 |
9.6. Let's call a non-empty (finite or infinite) set $A$, consisting of natural numbers, complete if for any natural numbers $a$ and $b$ (not necessarily distinct and not necessarily in $A$) such that $a+b$ is in $A$, the number $ab$ is also in $A$. Find all complete sets of natural numbers.
(N. Agakhanov) | Answer. The set of all natural numbers, as well as the sets $\{1\},\{1,2\},\{1,2,3\}$, and $\{1,2,3,4\}$.
Solution. First, let's check that the sets $\{1\},\{1,2\}$, $\{1,2,3\},\{1,2,3,4\}$, and the set of all natural numbers are complete. For the last set, this is obvious; for the first four, note that if natural num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,347 |
9.7. In a white $2016 \times 2016$ table, some cells are painted black. We will call a natural number $k$ successful if $k \leqslant 2016$, and in each of the grid squares with side $k$ located in the table, exactly $k$ cells are painted. (For example, if all cells are black, then only the number 1 is successful.) What... | Answer: 1008 numbers.
Solution. Consider an arbitrary coloring of the table. Let there be at least two successful numbers, and let $a$ be the smallest of them, and $b$ be the largest.
Divide $b$ by $a$ with a remainder: $b = qa + r$, where $0 \leq r < a$. If $r \neq 0$, then the number $qa$ is also successful, and $a... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,348 |
9.8. Given a convex quadrilateral $A B C D$, in which $\angle D A B=$ $=90^{\circ}$. Let $M$ be the midpoint of side $B C$. It turns out that $\angle A D C=\angle B A M$. Prove that $\angle A D B=\angle C A M . \quad$ (E. Bakayev) | First solution. On the extension of segment $AB$ beyond point $A$, mark point $K$ such that $AB = AK$ (see Fig. 1). Then $AM$ is the midline in triangle $BCK$, hence $AM \parallel CK$. Therefore, $\angle BKC = \angle BAM = \angle ADC$. This implies that quadrilateral $AKDC$ is cyclic.
Again, using the parallelism of $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,349 |
9.6. Ten-digit natural numbers $a, b, c$ are such that $a+b=c$. What is the maximum number of their 30 digits that can be odd?
(I. Bogdanov)
# | # Answer. 29.
Solution. Note that if $a+b=c$, then all three numbers $a, b$, $c$ cannot be odd simultaneously. Therefore, among them, there is at least one even number, and the last digit of this number will also be even. Thus, among the 30 digits, there is at least one even digit, and the number of odd digits is no m... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,350 |
9.8. Given a trapezoid $A B C D$ with bases $A D$ and $B C$. It turns out that the point of intersection of the medians of triangle $A B D$ lies on the bisector of angle $B C D$. Prove that the point of intersection of the medians of triangle $A B C$ lies on the bisector of angle $A D C$. (A. Kuznetsov) | The first solution. Let $P$ and $Q$ be the points of intersection of the medians of triangles $ABD$ and $ABC$ respectively, and let $K$ be the midpoint of $AB$.
We complete triangle $ABD$ to form parallelogram $ALBD$ (see Fig. 2); then $CL = CB + BL = BC + AD$. The diagonals of this parallelogram intersect at point $K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,351 |
9.9. In an alphabet of $n>1$ letters, a word is any finite sequence of letters in which any two adjacent letters are different. A word is called good if it is impossible to erase all letters from it, except for four, so that the remaining sequence is of the form $a a b b$, where $a$ and $b$ are different letters. Find ... | Answer: $2 n+1$.
First solution. Let's call the length of a word the number of letters in it. Let $a_{1}, a_{2}, \ldots, a_{n}$ be the letters of the alphabet. Then it is not hard to verify that the following word is good:
$$
a_{n} a_{n-1} \ldots a_{2} a_{1} a_{2} a_{1} a_{2} a_{3} \ldots a_{n}
$$
It remains to show... | 2n+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,352 |
9.10. Vitya wrote down $n$ different natural numbers in his notebook. For each pair of numbers from the notebook, he wrote their least common multiple on the board. Could it happen for some $n>100$ that $\frac{n(n-1)}{2}$ numbers on the board are (in some order) consecutive terms of a non-constant arithmetic progressio... | # Answer. No.
First solution. Let's call a set of $n$ numbers in the notebook $k r a-$ beautiful if it results in the required set of least common multiples. Suppose a beautiful set of $n>100$ numbers exists. Choose from all such sets the set with the smallest sum of numbers.
Notice that if the difference of the resu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,353 |
10.6. Petya chose a natural number $n$ and wrote the following $n$ fractions on the board:
$$
\frac{0}{n}, \frac{1}{n-1}, \frac{2}{n-2}, \frac{3}{n-3}, \ldots, \frac{n-1}{n-(n-1)}
$$
Let the number $n$ be divisible by a natural number $d$. Prove that among the written fractions, there will be a fraction equal to the ... | Solution. Let $n=k d$. Then the fraction on the board is
$$
\frac{n-k}{k}=\frac{k d-k}{k}=d-1
$$
which is what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,354 |
10.7. Four identical triangles are used to form a convex quadrilateral. Is it true that this quadrilateral must have parallel sides? | Answer. Incorrect
Solution. Let's provide one of the possible examples. From two equal isosceles (but not equilateral) triangles, we can form a kite by attaching them to each other along their equal sides, as shown in Fig. 3. Each of these two isosceles triangles can be divided into two equal triangles by a height. As... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,355 |
10.8. Given a $1000 \times 1000$ chessboard. A figure called a cheetah, from any cell $x$, attacks all cells of a $19 \times 19$ square centered at $x$, except for the cells that are in the same row or column as $x$. What is the maximum number of non-attacking cheetahs that can be placed on the board?
(I. Bogdanov) | Answer: 100000.
Solution: Divide the board into $100^{2}$ squares of size $10 \times 10$. We will show that in each square, no more than 10 leopards can stand without attacking each other - from this it will follow that the total number of leopards cannot exceed $100^{2} \cdot 10=100000$.
Consider an arbitrary square... | 100000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,356 |
10.10. Given a triangle $A B C$, in which $\angle A=\angle C=30^{\circ}$. On its sides $A B, B C$, and $A C$, points $D, E$, and $F$ are chosen respectively such that $\angle B F D=\angle B F E=60^{\circ}$. The perimeter of triangle $A B C$ is $p$, and the perimeter of triangle $D E F$ is $p_{1}$. Prove that $p \leqsla... | Solution. Let $\angle A F D=\alpha$. Since angle $B D F$ is external for triangle $A D F$, then $\angle B D F=\angle D A F+\angle A F D=30^{\circ}+\alpha$. Also, angle $B F A$ is external for triangle $B F C$, so $60^{\circ}+\alpha=\angle B F A=\angle F B E+\angle F C B$. Therefore, $\angle F B E=30^{\circ}+\alpha=\ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,358 |
7.1. The numerator and denominator of a positive fraction are natural numbers. If the numerator is increased by 3 and the denominator by 2, the value of the fraction decreases. Provide an example and show how this could happen. | Answer: for example, $\frac{8}{3}$.
Solution. $\frac{8+3}{3+2}=\frac{11}{5}=2 \frac{1}{5}<2 \frac{2}{3}=\frac{8}{3}$.
There are other examples as well: the condition of the problem is satisfied by any fraction that is greater than $\frac{3}{2}$.
Grading criteria. «+» A correct example is provided and it is shown tha... | \frac{8}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,359 |
7.3. There are 30 logs, the lengths of which are 3 or 4 meters, and their total length is one hundred meters. How many cuts can be made to saw all these logs into pieces 1 meter long? (Each cut saws exactly one log). | Answer: 70.
Solution. First method The total length of the logs is 100 meters. If it were one log, 99 cuts would be needed. Since there are 30 logs, 29 cuts have already been made. Therefore, $99-29=70$ cuts remain to be made.
Second method Let's find the number of logs of each type. If all were 3-meter logs, their t... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,361 |
7.4. Four sevenths of the class went on a trip. When 7A and 7B went to the museum, and 7V and $7 \Gamma$ went to have lunch at a cafe, Mary Ivanovna calculated that there were 15 seventh-graders more in the museum than in the cafe. And when in the evening 7A and 7V went to the park, and 7B and 7G went to the theater, M... | # Answer: No.
Solution. Let $k$ be the number of schoolchildren who went to the cafe, then there were $k+15$ schoolchildren in the museum. Therefore, the total number of seventh-graders is $k+(k+15)=2k+15$.
If $p$ seventh-graders went to the park, then $p+8$ seventh-graders went to the theater, and the total number o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,362 |
7.5. Draw six rays so that they intersect at exactly four points, with three rays at each point. Mark the starting points of the rays with bold dots. | Answer: see, for example, Fig. 7.5.
There are other examples as well. In particular, some rays may have a common vertex or the vertex of one ray may lie on another.
Criteria for checking.
“+” A correct example is provided
“士” Several examples are provided, among which there is a correct one
“-” The problem is not ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,363 |
7.6. Five girlfriends, Sonya, Tanya, Lena, Galia, and Vika, were born in five cities: Riga, Penza, Kazan, Belgorod, and Moscow. Each of them loves candies produced in one of these cities. It is known that no one loves candies produced in their hometown. Sonya loves candies from Riga. Tanya is from Riga, and her favorit... | Answer: Sonya
Solution. From the condition, we can determine who likes which candies:
Sonya - from Riga;
Tanya - from Penza;
Vika - from Moscow;
Galina - from Belgorod; therefore, Lena - from Kazan.
It is known that no one likes candies produced in their hometown. Let's mark this condition in the table, as well a... | Sonya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,364 |
11.1. Let's call the main divisors of a composite number $n$ the two largest natural divisors of $n$ different from $n$. Composite natural numbers $a$ and $b$ are such that the main divisors of $a$ coincide with the main divisors of $b$. Prove that $a=b$.
(A. S. Golyanov) | Solution. Let $n>k$ be the main divisors of the number $a$; then $a / n$ and $a / k$ are the two smallest divisors of the number $a$, greater than one. Let $p$ be the smallest prime divisor of the number $a$, and $q$ be the smallest prime divisor of $a$ other than $p$ (if such exists). Then $a / n = p$. Next, $a / k$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,365 |
11.2. On a plane, the graphs of the functions $y=\sin x$ and $y=\tan x$ are drawn, as well as the coordinate axes. How can one construct a straight line using a compass and a ruler that is tangent to the sine graph both above the x-axis $(O x)$ and below (and possibly has other points of intersection)?
(A. Kuznetsov) | Solution. We will look for a tangent passing through the origin. The tangent to the sine graph at the point $\left(x_{0}, \sin x_{0}\right)$ has the equation $y=\left(x-x_{0}\right) \cdot \cos x_{0}+\sin x_{0}$. This line passes through the origin if and only if $0=-x_{0} \cdot \cos x_{0}+\sin x_{0}$, which is equivale... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 16,366 |
11.3. In the plane, a fixed acute-angled triangle $ABC$ is given with the largest side $BC$. Let $PQ$ be an arbitrary diameter of its circumcircle, with point $P$ lying on the smaller arc $AB$, and point $Q$ lying on the smaller arc $AC$. Points $X, Y$, and $Z$ are the feet of the perpendiculars dropped from point $P$ ... | Solution. Note that $\angle P A Q=90^{\circ}$, since $P Q$ is the diameter of the circle $(A B C)$. Let $M$ and $N$ be the midpoints of segments $A P$ and $A Q$ respectively. Since $\angle A Z P=90^{\circ}=\angle A X P$, the quadrilateral $A Z X P$ is inscribed in a circle with center at point $M$, hence $\angle P Z X=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,367 |
9.1. The square root of the number 49 can be extracted using this "formula": $\sqrt{49}=4+\sqrt{9}$. Are there other two-digit numbers whose square roots can be extracted in a similar manner and are integers? List all such two-digit numbers. | Answer: Yes, they exist: 64 and 81.
Solution. Consider all two-digit numbers that are squares of integers. The square roots of 16, 25, and 36 cannot be extracted in the specified manner, as the square roots of their last digits are not integers. The numbers 49, 64, and 81 are solutions.
The answer to the problem will... | 6481 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,369 |
9.2. All natural numbers from 1 to 1000 inclusive are divided into two groups: even and odd. In which group is the sum of all digits used to write the numbers greater, and by how much? | Answer. The sum of the digits of odd numbers is 499 more.
Solution. The sum of the digits of the number 1 is equal to the sum of the digits of the number 1000; the other numbers can be divided into pairs: $2-3,4-5,6-7,8-9, \ldots, 998-999$. In each pair, the unit digit of the odd number is 1 more than that of the even... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,370 |
9.3. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly 2022 in this case? | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 678$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 675=2700=678+2022$.
Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $a$. Their produc... | 678 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,371 |
9.4. Instead of the ellipsis signs, insert such numbers so that the expression $\left(x^{2}+\ldots \times x+2\right) \times(x+3)=$ $(x+\ldots) \times\left(x^{2}+\ldots \times x+6\right)$ becomes an identity. | Answer. $\left(x^{2}+3 x+2\right)(x+3)=(x+1)\left(x^{2}+5 x+6\right)$.
Solution. Let the unknown coefficients be $a, b, c$ respectively: $\left(x^{2}+a x+2\right)(x+3)=(x+b)\left(x^{2}+c x+6\right)$ and bring the polynomials in the left and right parts to the standard form: $x^{3}+(a+3) x^{2}+(3 a+2) x+6=x^{3}+(b+c) x... | (x^{2}+3x+2)(x+3)=(x+1)(x^{2}+5x+6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,372 |
9.5. In an acute-angled triangle $A B C$, the median $A M$ and the altitude $B H$ are drawn. A perpendicular line from point $M$ to line $A M$ intersects ray $H B$ at point $K$. Prove that if $\angle M A C=30^{\circ}$, then $A K=B C$. (B. Obukhov) | First solution. Since $\angle A H K = \angle A M K = 90^{\circ}$, points $A, H, M$, and $K$ lie on the circle $\omega$ with diameter $A K$ (see Fig. 1). By the problem's condition, the chord $HM$ of this circle subtends the angle $\angle M A H = 30^{\circ}$, so $H M = 2 R \sin 30^{\circ} = A K / 2$. On the other hand, ... | AK=BC | Geometry | proof | Yes | Yes | olympiads | false | 16,373 |
2. On a shelf, toy cars are lined up: a sedan, a bus, a truck, and a van. They are painted in different colors: red, blue, yellow, and green. It is known that the blue car is between the red and green cars; to the right of the yellow car stands the van; the bus is to the right of both the sedan and the van; the sedan i... | Solution. We read the given conditions and sequentially arrange the colors and cars.
The first sentence allows us to arrange three colors in two ways.
K С 3 or 3 С
The yellow car is to the left of these three, as a van is to its right.
. Let's move to the solution from the end. After the dandelions appear, there are 1... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,378 |
6. Participants of the summer physics and mathematics camp for schoolchildren received either an orange or a purple T-shirt. The number of participants in the physics group who received an orange T-shirt is equal to the number of participants in the mathematics group who received a purple T-shirt. Who is more - partici... | Solution 1. Camp participants are divided into four non-overlapping groups, by the color of their T-shirts and the subject they are studying. Let's denote the number of participants in each group as PhysOrange, PhysViolet, MathOrange, MathViolet. The numbers that need to be compared can be obtained as sums.
The number... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,379 |
8.1. Let all numbers $x, y, z$ be non-zero. Find all values that the expression
$$
\left(\frac{x}{|y|}-\frac{|x|}{y}\right) \cdot\left(\frac{y}{|z|}-\frac{|y|}{z}\right) \cdot\left(\frac{z}{|x|}-\frac{|z|}{x}\right)
$$
can take. | Solution: By the Pigeonhole Principle, among the numbers $x, y$, and $z$, there will be two numbers of the same sign. Then the corresponding bracket will be equal to 0, and the entire product will also be equal to 0.
Answer: Only the number 0.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,380 |
8.2. Petya and three of his classmates started a 100-meter race at the same time, and Petya came in first. After 12 seconds from the start of the race, no one had finished yet, and the four participants had run a total of 288 meters. When Petya finished the race, the other three participants had 40 meters left to run i... | Solution: The runners were supposed to run a total of 400 meters, and in 12 seconds they ran 288 meters, meaning that in 1 second the runners run $288: 12=$ 24 meters. When Petya finished, the runners had run $400-40=360$ meters, so from the start, $360: 24=15$ seconds had passed. Therefore, Petya runs at a speed of $1... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,381 |
8.3. Unknowns $x$, $y$, and $z$ are related by the equation $\frac{1}{x y}=\frac{y}{z-x+1}=\frac{2}{z+1}$. Prove that in this case, one of the unknowns is the arithmetic mean of the other two | Solution: From the equality $\frac{1}{x y}=\frac{2}{z+1}$, it follows that $z+1=2 x y$. Then, from the equality $\frac{1}{x y}=\frac{y}{z-x+1}$, we get that $2 x y-x=x y^{2}$. The unknown $x$ is not equal to 0 (otherwise, the first fraction does not exist), so the obtained equality can be reduced by $x$. This results i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,382 |
8.5. Let point $D$ be the midpoint of median $AF$ of triangle $ABC$, and $E$ be the point of intersection of line $CD$ with side $AB$. Prove that if $BD=BF=CF$, then $AE=DE$. | # Solution:
Method 1. 1) $\angle B D F = \angle B F D$, as the base angles of the isosceles triangle $B F D$. Therefore, $\angle A D B = 180^{\circ} - \angle B D F = 180^{\circ} - \angle D F B = \angle D F C$.
2) Triangles $A D B$ and $D F C$ are congruent by two sides and the included angle ( $A D = D F$ and $D B = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,384 |
8.6. All natural numbers from 1 to 20 were divided into pairs, and the numbers in each pair were added. What is the maximum number of the ten resulting sums that can be divisible by 11? Justify your answer. | Solution: The number 11 is the only number in the set that is divisible by 11, so adding it to any other number will disrupt divisibility by 11. Therefore, all 10 sums cannot be divisible by 11. One example where nine sums are divisible by 11 is as follows:
$(1,10),(2,20),(3,19),(4,18),(5,17),(6,16),(7,15),(8,14),(9,1... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,385 |
1. Seven friends entered a cafe and ordered 3 small cups of coffee, 8 medium cups, and 10 large cups. The volume of a small cup is half the volume of a medium cup, and the volume of a large cup is three times the volume of a small cup. How should the friends divide the cups of coffee among themselves so that everyone d... | Solution: Let's call the amount of coffee in a small cup a "norm". Then, in total, we have $3+8 \times 2+10 \times 3=49$ norms. Since there are seven friends, each should get 7 norms.
We divide as follows: 1 small + 2 large - 3 people; 2 medium + 1 large - 4 people.
Criteria. A correct example - 7 points. No addition... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,386 |
3. Do there exist positive numbers $a, b, c, d$ such that the numbers $d$ and $\sqrt{d}$ are the roots of the equations $a x+b=c$ and $\sqrt{a} x+\sqrt{b}=\sqrt{c}$, respectively? | Answer: No, they do not exist.
Solution: Let's prove it - Suppose such numbers do exist.
From the first equation, it follows that $a d=c-b$, from the second: $\sqrt{a} \sqrt{d}=\sqrt{a d}=\sqrt{c}-\sqrt{b}$.
Square the second equation, and equate $a d$ from these two equations. $(\sqrt{c}-\sqrt{b})^{2}=a d=c-b=(\sqr... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,388 |
4. In a football tournament, 12 teams participated. By September, they had played several games, and no two teams had played each other more than once. It is known that the first team played exactly 11 games. Three teams played 9 games each. One team played 5 games. Four teams played 4 games each. Two teams played only... | Answer: 5 games.
Solution. Let the first team K1 play 11 games - i.e., once with everyone. Teams K2 and K3 played 1 game each - these are games with team K1.
There are 9 teams left (K4-K12). Three of these teams (K4, K5, K6) played 9 games each. One of these games was with K1. And 8 with all teams K4-K12 (except them... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,389 |
5. Is it possible to fill all cells of a $7 \times 7$ table with numbers such that the sum of all neighbors (by side) of each number is equal to $1?$ | Answer: No, it cannot.
Solution. We look at the top left corner. In the cells "1" the sum of the numbers is 1. Then we move down the diagonal. In the cells "2" the sum of the numbers is, thus, 0 (together with
^{2}-2019\left(x^{2}+2 x\right)+2018=0$. Introduce the substitution $\boldsymbol{t}=\boldsymbol{x}^{2}+2 \boldsymbol{x}$, we get the equation: $t^{2}-2019 t+2018=0$, the roots of which are $t_{1}=1, t_{2}=2018$. Then $x^{2}+2 x-1=0$ or $x^{2}+2 x-2018... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,393 |
3. Each participant in the school charity event brought either one encyclopedia, or three fiction books, or two reference books. In total, 150 encyclopedias were collected. After the event, two bookshelves in the library were filled, with an equal number of books on each. On the first shelf, there was one fifth of all ... | Solution. Let $x$ be the fifth part of the reference books, $y$ be the seventh part of all fiction books. Then on the first shelf, there are $-x+y+150$ books, and on the second shelf, there are $-4x+6y$ books. From the condition of the problem, the number of books on the shelves is equal, therefore, $x+y+150=4x+6y$, fr... | 267 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,394 |
4. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $L$ and $K$ are marked, respectively, and $M$ is the intersection point of segments $A K$ and $C L$. It is known that the area of triangle $A M C$ is equal to the area of quadrilateral $L B K M$. Find the angle $A M C$. | Solution. From the equality of the areas of triangle $A M C$ and quadrilateral $L B K M$, it follows that the areas of triangles $A C K$ and $C B L$ are equal (see the figure), because in this case, the same area of triangle $C M K$ is added to the equal areas. We have $S_{A C K}=\frac{1}{2} A C \cdot C K \cdot \sin \a... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,395 |
5. In each cell of a $25 \times 25$ square table, one of the numbers 1 or -1 is written arbitrarily. Under each column, the product of all numbers in that column is written. To the right of each row, the product of all numbers in that row is written. Prove that the sum of the 50 written products cannot be zero. | Solution. Let's find the product of all 25 numbers written under each column and all 25 numbers written to the right of the rows. Since in this product each of the numbers in the square table appears twice, the product of these 50 products, each of which has 25 factors, will be positive, i.e., equal to 1. Since the pro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,396 |
# 1. Clone 1
In the box, there are candies of three types: chocolate, lollipops, and jelly. In total, there are 110 candies. If we don't count the jelly candies, there are 100 candies in the box. If we don't count the chocolate candies, the number is 20 more than the total number of candies without the lollipops. How ... | Answer: 40 chocolate, 60 lollipops, 10 jelly.
## Solution
Let the number of chocolate candies be C, jelly candies be J, and lollipops be L. According to the problem, $C+J+L=110, C+L=100$. From this, we get $J=10$. It is also known that $J+L=C+J+20$, then $L=C+20$. Substituting L into the first equation and considerin... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,397 |
# 2. Clone 1
The teacher wanted to write an example for calculation on the board:
$$
1,05+1,15+1,25+1,4+1,5+1,6+1,75+1,85+1,95=?
$$
but accidentally forgot to write one comma. After this, Kolya went to the board and, correctly performing all the operations, obtained an integer result. What is it? | Answer: 27
## Solution
Let's calculate the sum
$$
1.05+1.15+1.25+1.4+1.5+1.6+1.75+1.85+1.95=13.5
$$
Therefore, in order to get an integer, the decimal point should be omitted in the number with the fractional part 0.5. As a result, instead of the number 1.5, the number 15 will appear on the board, and the result wi... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,398 |
# 3. Clone 1
On an island, there live knights who always tell the truth, and liars who always lie. Before a friendly match, 30 islanders gathered in T-shirts with numbers on them—arbitrary natural numbers. Each of them said: “I have a T-shirt with an odd number.” After that, they exchanged T-shirts, and each said: “I ... | # Answer: 15
## Solution
Notice that each knight exchanged a shirt with an odd number for a shirt with an even number, while each liar exchanged a shirt with an even number for a shirt with an odd number. This means that each knight exchanged with a liar, so there are no more knights than liars. On the other hand, ea... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,399 |
# 4. Clone 1
The Hare and the Wolf are running in the same direction around a circle 200 meters in circumference at speeds of $5 \mathrm{~m} / \mathrm{c}$ and 3 m/s. It turns out that the distance between them is exactly the same as it was 40 seconds ago (the distance is measured along the shorter of the two arcs). Wh... | Answer: 40,60
## Solution
From the condition, it follows that the Hare runs faster than the Wolf. In addition, in 40 seconds, the Hare will run 200 meters, which is no more than 1 lap. We will assume that the Hare and the Wolf are moving in a circle clockwise. Let's consider two possible cases.
1) The Hare catches u... | 40,60 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 16,400 |
# 6. Clone 1
The figure shows a hexagon composed of identical equilateral triangles, each with an area of 10. Find the area of the shaded part.
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,403 |
# 7. Clone 1
On the table, there are 14 coins of 2 and 5 rubles (each denomination is present). Some of them are face up with their denomination (tails) side, and some are face up with the heads side. If every coin on the table is flipped, the sum of the visible denominations will become three times greater than it wa... | Solution. If we denote the sum of the visible denominations as $x$, then the sum of the denominations of the flipped coins will be $3x$, meaning the total amount of money on the table is $4x$, which is divisible by 4. Since both denominations are present, the total amount on the table can range from 31 to 67 rubles. Th... | 4or8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,404 |
# 8. Clone 1
The playing field for the game "Battleship" is a rectangle $1 \times 203$. Petya places single-cell detectors on this field, after which Vasya places a two-deck ship (rectangle $1 \times 2$). A detector triggers if the cell it is placed on is under the ship. What is the minimum number of detectors Petya s... | # Answer: 134
## Solution
The solution to this problem consists of two parts: an estimate, where we will prove that it is impossible to get by with fewer detectors, and an example that shows that the given number of detectors is sufficient.
Estimate.
Remark 1. If next to a detector there are two free cells, then in... | 134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,405 |
7.2. Cut a square into 9 triangles, eight of which are equal, and the ninth is different from them. | Solution. See fig.
Comment. A variant of cutting is provided - 7 points.
 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,407 |
7.3. When the passengers entered the empty tram, half of them took seats. How many passengers entered at the very beginning, if after the first stop their number increased by exactly $8 \%$ and it is known that the tram can accommodate no more than 70 people? | Answer: 50. From the condition, it follows that the number of passengers is divisible by 2 (since half took seats) and by 25 (8% is $2/25$ of the total number). Therefore, the initial number of passengers was divisible by 50, but it was less than 70, so it was 50.
Comment. Answer only - 0 points. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,408 |
7.4. Winnie-the-Pooh and Tigger are climbing two identical fir trees. Winnie-the-Pooh climbs up at half the speed of Tigger, but he descends three times faster than Tigger. Winnie-the-Pooh and Tigger started and finished at the same time. How many times faster does Tigger climb up than he descends? | Answer: one and a half times. Let the time it takes for the Tiger to climb up be $T$ minutes, and the time to climb down be $t$ minutes. Then, Winnie-the-Pooh takes $2 T$ minutes to climb up, and $-t / 3$ minutes to climb down. According to the condition, $T+t=$ $2 T+t / 3$, from which we get $t=1.5 T$. Comment. Only t... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,409 |
11.2. First solution. Let us number the rows with the numbers $1, \ldots, n$ from top to bottom, and the columns with the same numbers from left to right. We will denote a cell by a pair of the numbers of its row and column; in this case, we will consider that the cells of the diagonal with pluses have coordinates $(i,... | The second solution. Note that the sign in a cell changes if and only if the operation was applied to this cell an odd number of times. Suppose there are exactly $r$ rows and exactly $c$ columns to which the operation was applied an odd number of times (let's call them odd). Then the sign changed in exactly $r(n-c)$ ce... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,411 |
8.2. The price of a ticket to the stadium was 25 rubles. After the ticket prices were reduced, the number of spectators at the stadium increased by $50 \%$, and the revenue from ticket sales increased by $14 \%$. What is the new price of a ticket to the stadium after the price reduction? | Answer: 19 rubles.
Let's denote: $a$ - the number of spectators coming to the stadium; $x$ - the ratio of the new ticket price to the price of 25 rubles. The new revenue is: on the one hand $(25 \cdot x) \cdot a \cdot 1.5$, on the other hand $25 \cdot 1.14 \cdot a$. From the equality $(25 \cdot x) \cdot a \cdot 1.5 = ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,412 |
8.3. To the number 2014, a digit was added on the left and on the right. The six-digit number thus obtained became divisible by 36. Find all such six-digit numbers. | Answer: 220140, 720144, 320148.
A number is divisible by 36 if and only if it is divisible by 4 and by 9. According to the divisibility rule for 4, the digit $a$ must be appended to the right so that the two-digit number $4a$ is divisible by 4. There are three possible options: 0, 4, and 8. According to the divisibili... | 220140,720144,320148 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,413 |
8.4. In triangle $A B C$, the bisectors $A D$ and $C E$ were drawn. It turned out that $A E + C D = A C$. Find the angle $B$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: angle $B$ is equal to $60^{\circ}$.
Let $\alpha$ and $\gamma$ be the measures of angles $A$ and $C$ of triangle $ABC$. On side $AC$, we mark segment $AF$ equal to $AE$. From the condition, we get the equality of segments $CD$ and $CF$. Let the bisectors $AD$ and $CE$ intersect segments $EF$ and $FD$
^{4}+x^{4}=112$. | # Solution
Let's introduce the substitution $y=x+1, (y+1)^{4}+(y-1)^{4}=82, y^{4}+6 y^{2}-40=0, y^{2}=$ $-10, y^{2}=4, x_{1}=-3, x_{2}=1$.
## Grading Criteria
Only the answer is given - 0 points.
The fourth power is correctly raised in the solution process, but the roots are not found - -1 point.
The fourth power ... | x_{1}=-3,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,416 |
2. Vasya claims that the number of his friends who are good at mathematics is not less than $96.8\%$ and not more than $97.6\%$. It is known that the number of Vasya's friends does not exceed 150. Determine the largest number within the specified range that cannot be the number of Vasya's friends. | # Solution
Let $\mathrm{N}$ be the number of Vasya's friends. The number $x$ of Vasya's friends who are not good at math is in the range $0.024 N < x < 0.032 N$, then for $x=1, 31 \frac{1}{4} < N < 41 \frac{2}{3}$; for $x=2, 62 \frac{1}{3} < N < 83 \frac{1}{3}$; for $x=3, 93 \frac{3}{4} < N < 125$; for $x=4, 125 < N <... | 125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,417 |
4. Is it possible to arrange $2 \mathrm{n}$ black and several white chips in a circle so that each black chip has a diametrically opposite white chip and no two white chips are adjacent?
# | # Solution
Since each black chip corresponds to a diametrically opposite white chip and no two white chips are adjacent, the chips must alternate and there must be an equal number of each. There are $2 \mathrm{n}-1$ chips between a black and a white chip on a semicircle, so the outermost chips are of the same color, m... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,418 |
5. During the investigation of the stolen pie case, Bolvanshchik stated that the pie was stolen by the March Hare. The March Hare, in turn, testified that Sonya did not do it. Sonya stated that the pie was stolen by Bolvanshchik, and the March Hare could not have done it. It was later found out that the testimony of ex... | # Solution
If Bolvanshchik lied, then the conditions "March Hare did not steal," "Sonya did not steal," and "Bolvanshchik stole and March Hare did not steal" must all be true. These conditions are consistent with each other. Therefore, Bolvanshchik remains a suspect.
If March Hare lied, then the conditions "March Har... | Sonya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,419 |
1. At each vertex of a cube with an edge length of 1, there is the center of a sphere. All spheres are the same, and each touches three adjacent spheres. Find the length of that part of the spatial diagonal of the cube that lies outside the spheres. | Solution. Consider two spheres, the centers of which lie at the vertices of one edge. Since the spheres have the same radius and touch each other, the radius of each sphere is $\frac{1}{2}$. Consider the section $A B C_{1} D_{1}$. This section contains the diagonal $A C_{1}$, and each sphere cuts off a segment of lengt... | \sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,420 |
2. On a chessboard, a certain number of white and black pawns were placed. The number of white pawns is greater than the number of black pawns. The number of pawns standing on black squares is less than the number of pawns standing on white squares. Prove that there will be at least one white pawn standing on a white s... | Solution. We will prove by contradiction. Suppose that only black pawns are placed on white squares. Then the number of black pawns is greater than the number of all pawns placed on black squares, which is not less than the number of white pawns (we assumed that white pawns are placed only on black squares, and some bl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,421 |
3. Parabola
$$
f(x)=x^{2}+p x+q
$$
intersects the coordinate axes at three different points. Let's call these intersection points with the axes $A, B, C$. Suppose point $D$ has coordinates $D(0 ; 1)$. Prove that there exists a circle passing through points $A, B, C, D$. | Solution. According to the problem, the function $f(x)=x^{2}+p x+q$ has two roots $x_{1}, x_{2}$. Let's denote the points of intersection of the parabola with the axes as follows: $A\left(x_{1} ; 0\right), B\left(x_{2} ; 0\right), C(0 ; q)$. We need to consider two cases. The first case: $q<0$, point $C$ lies below the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,422 |
5. The computer generated several different natural numbers. For each pair of numbers, it determined the highest power of 2022 by which their difference is divisible. The option that the difference is only divisible by the zeroth power is also possible. It turned out that the computer got 2023 different answers. What i... | Solution.
Example. One of the possible sets is all numbers from 1 to $2022^{2023}$. The difference of any of them is divisible by 2022 no more than to the power of 2022, while all powers from 0 to 2022 are present.
Estimation. We will prove that it is impossible to select a larger number of numbers. Let's represent e... | 2022^{2023} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,424 |
1. Petya was exchanging stickers. He trades one sticker for 5 others. At first, he had 1 sticker. How many stickers will he have after 30 exchanges? | Answer: 121.
Solution: After each exchange, the number of Petya's stickers increases by 4 (one sticker disappears and 5 new ones appear). After 30 exchanges, the number of stickers will increase by $30 * 4 = 120$. Initially, Petya had one sticker, so after 30 exchanges, he will have $1 + 120 = 121$. | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,425 |
2. Vasya multiplied one four and 27 nines, while Petya multiplied 55 threes. Whose number turned out to be larger? Justify your answer. | Answer. Vasya's number is larger.
Solution. Since two threes in the product give 9, the product of 55 of Petya's threes is the same as the product of one three and 27 nines. Since the product of a three and 27 nines is less than the product of a four and 27 nines, Petya's number is smaller. | Vasya'islarger | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,426 |
3. Cut the figure (along the cell boundaries) into three equal (identical in shape and size) parts.
# | # Answer.
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,427 |
6. Three jumps of a two-headed dragon are equal to 5 jumps of a three-headed one. But in the time it takes for the two-headed dragon to make 4 jumps, the three-headed one makes 7 jumps. Which one runs faster? Justify your answer.
# | # Answer. Three-headed.
Solution. Consider the time it takes for a two-headed dragon to make 3*4=12 jumps. In this time, a three-headed dragon makes $3 * 7=21$ jumps. Since 12=4*3, 12 jumps of the two-headed dragon are equal to 4*5=20 jumps of the three-headed dragon. Thus, in the same amount of time, the three-headed... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,428 |
9.6. The sequence of numbers $a_{1}, a_{2}, \ldots, a_{2022}$ is such that $a_{n}-a_{k} \geqslant$ $\geqslant n^{3}-k^{3}$ for any $n$ and $k$ such that $1 \leqslant n \leqslant 2022$ and $1 \leqslant k \leqslant$ $\leqslant 2022$. Moreover, $a_{1011}=0$. What values can $a_{2022}$ take?
(N. Agakhanov) | Answer. $a_{2022}=2022^{3}-1011^{3}=7 \cdot 1011^{3}$.
Solution. Writing the condition for $n=2022, k=1011$ and for $n=1011, k=2022$, we get
$$
a_{2022}=a_{2022}-a_{1011} \geqslant 2022^{3}-1011^{3}
$$
and
$$
-a_{2022}=a_{1011}-a_{2022} \geqslant 1011^{3}-2022^{3}
$$
that is, $a_{2022} \geqslant 2022^{3}-1011^{3} ... | 2022^{3}-1011^{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,429 |
9.7. Petya divided a square grid $100 \times 100$ in some way into dominoes - grid rectangles $1 \times 2$, and in each domino, he connected the centers of its two cells with a blue segment. Vasya wants to divide the same square into dominoes in a second way, and in each of his dominoes, connect the two cells with a re... | Answer: Not necessarily.
First solution. Number the columns from left to right with numbers from 1 to 100. Let $a$ be the top row of the square, and $b$ be the row immediately below it. Suppose in Petya's tiling, these rows are occupied by vertical dominoes $a 1-b 1, a 2-b 2, \ldots, a 98-b 98$ and horizontal dominoes... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,430 |
9.8. In trapezoid $ABCD$, diagonal $BD$ is equal to base $AD$. Diagonals $AC$ and $BD$ intersect at point $E$. Point $F$ on segment $AD$ is chosen such that $EF \| CD$. Prove that $BE = DF$.
(A. Kuznetsov) | Solution. Since $A D \| C B$, triangles $E A D$ and $E C B$ are similar, and therefore $\frac{B E}{D E}=\frac{B C}{A D}$.
Extend triangle $B C D$ to form parallelogram $B C D K$ (see Fig. ??). Then triangles $D E F$ and $D B K$ are similar, so $\frac{D F}{D E}=\frac{D K}{D B}$. Finally, since $D K=B C$ and $D B=D A$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,431 |
9.9. On a plane, $N$ points are marked. Any three of them form a triangle, the angles of which in degrees are expressed by natural numbers. For what largest $N$ is this possible
$$
\text { (E. Bakayev) }
$$ | Answer: 180.
Fig. 1
First Solution. Example. First, we will show that the required is possible when $N=180$. Mark 180 points on the circle, dividing it into 180 equal arcs, each $2^{\circ}$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,432 |
9.10. Prove that there exists a natural number $b$ such that for any natural $n>b$ the sum of the digits of the number $n!$ is not less than $10^{100}$. ( | Solution. Let $a=10^{100}$. Denote by $s(m)$ the sum of the digits of the number $m$. Note the simple property $s(\ell)+s(m) \geqslant s(\ell + m)$, which is immediately clear if the numbers $\ell$ and $m$ are added in a column.
Lemma. Let $k$ be a natural number, and let the natural number $m$ be divisible by $10^{k}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,433 |
1. In the game "Sportlotto-Shish," the main prize is drawn according to the following rules. Each person in the studio independently writes down any number of different pairs of different integers from the set from 1 to 5. If some participants have written down the same pairs, these participants share the main prize. H... | Solution. A total of different pairs can be formed $5 * 4 / 2=10$. It is taken into account that the digits in the pair are different and the order of the digits within the pair does not matter. Each pair can be written down or not written down by the participant. A total of different sets can be formed $2^{10}=1024$. ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,434 |
2. In a row, 33 weights are arranged in ascending order. It is known that any four consecutive weights can be distributed on two scales so that equilibrium is achieved. The third weight weighs 9 g, and the ninth weighs 33 g. How much does the 33rd weight weigh? | Solution. Let the weights of the weights be $a_{1}<a_{2}<\cdots<a_{33}$. For all $k=1,2, \ldots, 30$ the equalities $a_{k}+a_{k+3}=a_{k+1}+a_{k+2}$, equivalent to $a_{k+3}-a_{k+2}=$ $a_{k+1}-a_{k}$, hold. Let $a_{4}-a_{3}=a_{2}-a_{1}=d$ and $a_{5}-a_{6}=a_{3}-a_{2}=\mathrm{c}$, then $a_{6}-$ $a_{5}=a_{4}-a_{3}=d, a_{7}... | 129 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,435 |
3. Show that the product of the sum of any two positive numbers and the sum of their reciprocals is not less than 4, and the product of the sum of any three positive numbers and the sum of their reciprocals is not less than 9. | Solution. The first statement follows from the Cauchy inequality $a^{2}+b^{2} \geq 2 a b$. The second relies on the first and on the Cauchy inequality: $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{(a+b)^{2}}{a b}+c \frac{a+b}{a b}+\frac{a+b}{c}+1 \geq 4+c \frac{2 \sqrt{a b}}{a b}+\frac{2 \sqrt{a b}}{c... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,436 |
4. Can the numbers 11, 12, and 13 be members (not necessarily consecutive) of one geometric progression | Solution. Suppose that $b_{k}=11, b_{l}=12, b_{n}=13 \quad-$ are terms of a geometric progression, that is, $b_{1} q^{k-1}=11, \quad b_{1} q^{l-1}=12, \quad b_{1} q^{n-1}=13$. Then $q^{k-1}=11 / 12, q^{l-n}=12 / 13$. Raise the first equality to the power of $l-n$, and the second to the power of $k-l$. After that, we ge... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,437 |
5. In triangle $A B C$, points $P$ and $Q$ are located on sides $A B$ and $B C$ respectively. Triangle $B P Q$ is acute, and $P M, Q N$ are its altitudes. Prove that if a circle can be circumscribed around quadrilateral $A P Q C$, then $M N \| A C$. | Solution: Since quadrilateral $A P Q C$ is inscribed, then $\angle B A C + \angle P Q C = 180^{\circ}$. But $\angle P Q M + \angle P Q C = 180^{\circ}$ (adjacent angles), so $\angle B A C = \angle P Q M$. A circle can be circumscribed around quadrilateral $P N M Q$ with diameter $P Q$, since $\angle P N Q = \angle P M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,438 |
1. Prove that if the equations
$$
x^{2}+p x+q=0 \text{ and } x^{2}+b x+c=0
$$
do not have roots, then the equation
$$
7 x^{2}+(2 p+3 b+4) x+2 q+3 c+2=0
$$
also does not have roots. | Since the equations $x^{2}+p x+q=0$ and $x^{2}+b x+c=0$ have no roots, then $x^{2}+p x+q>0$ and $x^{2}+b x+c>0$ for any $x$.
Notice that $\left(x^{2}+2 x+1\right) \geq 0$ for all $x$. Then
$7 x^{2}+(2 p+3 b+4) x+2 q+3 c+2=2\left(x^{2}+p x+q\right)+3\left(x^{2}+b x+c\right)+$
$+2\left(x^{2}+2 x+1\right)>0$ for any $x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,439 |
2. The least common multiple of seven natural numbers is 2012. Find the smallest possible sum of these numbers. | Solution: Let's factorize the number 2012 into prime factors $2012=2^{2} \cdot 503$.
One of the numbers, the LCM of which is 2012, must be divisible by $2^{2}=4$, and one (possibly the same) - by 503. If the same number is divisible by both 4 and 503, then it is divisible by 2012, and the sum of seven numbers is not l... | 512 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,440 |
3. Three lines intersect at one point 0. Outside these lines, a point M is taken and perpendiculars are dropped from it to them. The points $\mathrm{H}_{1}, \mathrm{H}_{2}$ and $\mathrm{H}_{3}$ are the bases of these perpendiculars. Find the ratio of the length of the segment OM to the radius of the circle circumscribe... | Solution
$2, \prime \prime$
First, consider two intersecting... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,441 |
3. The figure shows the graphs of the quadratic trinomials
$$
y=x^{2}+a x+b \quad \text { and } \quad y=x^{2}+c x+d
$$
Prove that $a^{2}-c^{2}>b-d$. | Solution:
Let $x_{1}$ and $x_{2}$ be the roots of the function $f(x)=x^{2}+a x+b$, and $x_{3}$ and $x_{4}$ be the roots of the function $g(x)=x^{2}+c x+d$. Note that $f(0)=b>d=g(0)>0$, and also $\left|x_{1}-x_{2}\right|>\left|x_{3}-x_{4}\right|$.
Since $\left|x_{1}-x_{2}\right|=\sqrt{D_{1}}=\sqrt{a^{2}-4 b}$ and $\le... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,443 |
4. Solve the inequality in integers
$$
\frac{x^{2}}{\sqrt{x-3 y+2}}+\frac{y^{2}}{\sqrt{-x+2 y+1}} \geq y^{2}+2 x^{2}-2 x-1 .
$$ | # Answer: $(0 ; 0),(2 ; 1)$.
## Solution.
Since $x, y$ are integers, then $x-3 y+2 \geq 1$ and $-x+2 y+1 \geq 1$, then $\frac{x^{2}}{\sqrt{x-3 y+2}}+\frac{y^{2}}{\sqrt{-x+2 y+1}} \leq y^{2}+x^{2} \quad \Rightarrow \quad y^{2}+2 x^{2}-2 x-1 \leq y^{2}+x^{2} \Leftrightarrow$ $x^{2}-2 x-1 \leq 0 \Leftrightarrow 1-\sqrt{... | (0;0),(2;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,444 |
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