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8.4. In a chess tournament, each of the 10 players played one game against each other, and Petya came in last place (scored fewer points than any other participant). Then two players were disqualified, and all points earned in matches against them were annulled, and these two players were removed from the table. It tur... | Answer: 4 points.
Solution. In a tournament with 10 players, played in a single round, $\frac{10 \cdot 9}{2}=45$ points are distributed. Therefore, there will be a player who scores no more than $45: 10=4.5$ points. This means that the player who finished in absolute last place scored no more than 4 points. Similarly,... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. In the notebook, all irreducible fractions with the numerator 15 are written, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15} \cdot$ How many such fractions are written in the notebook? | Answer: 8 fractions.
Solution. We look for all suitable irreducible fractions of the form $\frac{n}{15}$. Since $\frac{1}{16} < \frac{15}{n}$, then $\frac{15}{225} > \frac{15}{n}$, and $n > 225$. Therefore, $225 < n < 240$. The fraction $\frac{n}{15}$ is irreducible, meaning $n$ is not divisible by 3 or 5. It is not d... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. In the Banana Republic, parliamentary elections were held in which all residents participated. All those who voted for the "Mandarin" party love mandarins. Among those who voted for other parties, $90\%$ do not like mandarins (the rest do like them). What percentage of the votes did the "Mandarin" party receive in... | Answer: $40 \%$.
Solution. Let $x \%$ be the percentage of votes for the party "Mandarin". Then the percentage of votes for the other parties is $(100-x) \%$. One tenth of $(100-x) \%$ love mandarins, so we get the equation $x+(100-x) / 10=46$. Solving this, we find that $x=40$.
## Criteria.
5 points. A correct equa... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.5. In an isosceles triangle \(ABC\), the angle \(A\) at the base is \(75^\circ\). The bisector of angle \(A\) intersects side \(BC\) at point \(K\). Find the distance from point \(K\) to the base \(AC\), if \(BK=10\). | Answer: 5.
Solution. In triangle $ABC$, $\angle B = (180^0 - 75^0 - 75^0) = 30^0$. Drop perpendiculars from point $K$: $KH$ to side $AB$, and $KN$ to side $AC$. In the right triangle $HBK$, the hypotenuse $BK$ is 10. Angle $B$ is 30 degrees, and the side opposite to it is half the hypotenuse, so $HK = 5$. Triangles $A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Given a triangle $A B C$. A line parallel to $A C$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively, and the median $A M$ at point $Q$. It is known that $P Q=3$, and $Q T=5$. Find the length of $A C$. | Answer: $AC=11$.
Solution. First method. Draw a line through point $Q$ parallel to $BC$ (where $N$ and $L$ are the points of intersection of this line with sides $AB$ and $AC$ respectively, see Fig. 9.3a). Since $AM$ is the median of triangle $ABC$, then $LQ=NQ$. Moreover, $PT \| AC$, so $PQ$ is the midline of triangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take? | Answer: 91.
Solution. Example. Consider nine numbers equal to 91, and the number 182. Their sum is 1001.
Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \times 11 \times 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On th... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are nuts in three boxes. In the first box, there are six nuts fewer than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer. | Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is defined by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Misha and Masha had the same multi-digit integer written in their notebooks, ending in 9876. Masha placed a plus sign between the third and fourth digits from the right, while Misha placed a plus sign between the fourth and fifth digits from the right. To the surprise of the schoolchildren, both resulting sums tur... | Solution: Let the written number have the form $\overline{x 9876}$, where $x$ is also some natural number. Then Misha got the sum $x+9876$, and Masha got the sum $10 x+9+876$. From the equality $x+9876=10 x+9+876$, we find that $x=999$.
Answer: 9999876 and there is no other number.
Recommendations for checking:
| is... | 9999876 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Answer. 190. Solution 1. Among the lines intersecting $y=20-x$, there is the line $y=x$. Moreover, the picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=2... | Solution 2. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. Let the line $y=x$ intersect the line $y=20-x$ at point $A$, then $x_{A}=10$. The remaining 18 lines are divided into pairs, intersecting the line $y=20-x$ at symmetric points $B$ and $C$ (triangles $M A B$ ... | 190 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. The sum of the digits of a natural number $a$ is to the sum of the digits of the number $2a$ as 19 to 9. Prove that the decimal representation of the number $a$ has at least 29 digits. | Solution. From the condition, it follows that the sum of the digits of the number $2a$ is divisible by 9. Therefore, the number $2a$ itself, and consequently the number $a$, and the sum of its digits, are divisible by 9. In addition, the sum of the digits of the number $a$ is divisible by 19. Thus, it is divisible by 1... | 29 | Number Theory | proof | Yes | Yes | olympiads | false |
Problem 2. On a plane, 100 points are marked. It turns out that on two different lines a and b, there are 40 marked points each. What is the maximum number of marked points that can lie on a line that does not coincide with a and b? | Answer: 23. Solution: Lines $a$ and $b$ have no more than one common point. If such a point exists and is marked, then together on lines $a$ and $b$ there are 79 marked points; otherwise, there are 80. Therefore, outside lines $a$ and $b$, there are no more than 21 marked points.
Take an arbitrary line $c$ that does n... | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A row of 101 numbers is written (the numbers are not necessarily integers). The arithmetic mean of all the numbers without the first one is 2022, the arithmetic mean of all the numbers without the last one is 2023, and the arithmetic mean of the first and last numbers is 51. What is the sum of all the writte... | Answer. 202301. Solution. Let the sum of all numbers be $S$, the first number be $a$, and the last number be $b$. According to the conditions, $(S-a) / 100=2022, \quad(S-b) / 100=2023$, $(a+b) / 2=51$, from which we get $S-a=2022 \cdot 100, S-b=2023 \cdot 100, a+b=51 \cdot 2$. Adding these three equations, we get $2 S=... | 202301 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In an online store, two types of New Year's gifts are sold. The first type of gift contains a toy, 3 chocolate, and 15 caramel candies and costs 350 rubles. The second type of gift contains 20 chocolate and 5 caramel candies and costs 500 rubles. Eugene wants to buy an equal number of caramel and chocolate candies, ... | Answer: 3750 rubles.
Solution. Consider integer values $m$ and $n$ - the quantities of purchased gift sets of candies of the 1st and 2nd types, respectively. These quantities must satisfy the conditions of the problem: the total number of caramel and chocolate candies in them is the same, and this number is the smalle... | 3750 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The school dance studio calculated that this year they have already performed the dance "Circle Dance" 40 times, and in each performance, exactly 10 people participated, and any two dancers performed together no more than once. Prove that the studio has at least 60 dancers. | Solution. First solution. According to the problem, any two dancers could meet at no more than one performance. We will consider such two dancers as a pair. Consider any performance: from 10 participants of the performance, no more than $\frac{10 \cdot 9}{2}=45$ pairs can be formed (the number of "handshakes," $C_{10}^... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$. | Answer: $A C=13$.
Solution. (Fig. 5). First, we will prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. For what least natural $\mathrm{k}$ does the quadratic trinomial
$$
y=k x^{2}-p x+q \text { with natural coefficients } p \text{ and } q \text{ have two }
$$
distinct positive roots less than 1? | Solution. First, let's estimate the coefficient $\mathrm{k}_{\text{from below. Let }} 0k^{2} x_{1}\left(1-x_{1}\right) x_{2}\left(1-x_{2}\right) \geq 1 \Rightarrow k>4 \text{. Thus, we have obtained the }$
$$
$$
lower bound. To show the accuracy of the estimate, it is sufficient to construct an example with $k=5: y=5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The magpie-crow was cooking porridge, feeding her chicks. The third chick received as much porridge as the first two combined. The fourth chick received as much as the second and third. The fifth chick received as much as the third and fourth. The sixth chick received as much as the fourth and fifth. The seventh chi... | Answer: 40 g.
Solution. Let the first chick receive a grams of porridge, the second - b grams, Then the others received $\mathrm{a}+\mathrm{b}, \mathrm{a}+2 \mathrm{~b}, 2 \mathrm{a}+3 \mathrm{~b}, 3 \mathrm{a}+5 \mathrm{~b}, 0$ grams respectively. In total, they received $8 \mathrm{a}+12 \mathrm{~b}$ grams of porridg... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Among the numbers from 1 to $10^{23}$, which are more numerous - those with a two-digit sum of digits or those with a three-digit sum? Answer: those with a three-digit sum. | Solution. The number $10^{23}$ has a digit sum of 1, so we will not consider it. We will write numbers as 23-digit numbers, adding leading zeros. We will call digits "complementary" if their sum is 9. To each number with a two-digit digit sum, we will correspond a number by replacing each digit with its "complementary"... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.2. Petya has 25 coins, each with a denomination of $1, 2, 5$, or 10 rubles. Among these coins, 19 are not two-ruble coins, 20 are not ten-ruble coins, and 16 are not one-ruble coins. How many five-ruble coins does Petya have? | Answer: 5.
Solution. Since among Petya's coins, 19 are not two-ruble coins, Petya has a total of 6 two-ruble coins. Similarly, it follows that Petya has 5 ten-ruble coins and 9 one-ruble coins.
Thus, Petya has $25-6-5-9=5$ five-ruble coins. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Anton guessed a three-digit number, and Lesha is trying to guess it. Lesha sequentially named the numbers 109, 704, and 124. Anton noticed that each of these numbers matches the guessed number in exactly one digit place. What number did Anton guess? | Answer: 729.
Solution. Note that the first and third numbers have a common hundreds digit 1, the first and second have a common tens digit 0, and the second and third have a common units digit 4.
Suppose the first digit of the guessed number is 1. Then the first and third numbers no longer have common digits with the... | 729 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.8. In a grove, there are trees of four species: birches, firs, pines, and aspens. There are a total of 100 trees. It is known that among any 85 trees, there will be trees of all four species. Among what minimum number of any trees in this grove will there definitely be trees of at least three species? | Answer: 69.
Solution. Suppose there are no more than 15 birches in the grove. Then there are at least 85 other trees, and according to the problem's condition, among them, there must be trees of all four types. This is a contradiction. Therefore, there must be at least 16 birches in the grove. Similarly, we can conclu... | 69 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. By September 1st, Vlad bought himself several ballpoint and gel pens. He noticed that if all the pens he bought were gel pens, he would have paid 4 times more than he actually did. And if all the pens were ballpoint pens, the purchase would have cost him 2 times less than the actual price. How many times m... | Answer: 8.
Solution. If all the pens were gel pens, their price would be 4 times the actual price, which in turn is 2 times more than if all the pens were ballpoint pens. Therefore, gel pens cost $4 \cdot 2=8$ times more than ballpoint pens. Consequently, one gel pen is 8 times more expensive than one ballpoint pen. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.6. Vanya received three sets of candies as a New Year's gift. In the sets, there are three types of candies: lollipops, chocolate, and jelly. The total number of lollipops in all three sets is equal to the total number of chocolate candies in all three sets, as well as the total number of jelly candies in all... | Answer: 29.
Solution. There are more lollipops than jelly candies in the first set by 7, and in the second set by 15. Since the total number of each type of candy is the same in all sets, and there are 0 lollipops in the third set, there must be $7+15=22$ jelly candies in the third set.
Similarly, there are more loll... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.7. Jerry the mouse decided to give Tom the cat a square cake $8 \times 8$ for his birthday. In three pieces marked with the letter "P," he put fish, in two pieces marked with the letter "K," he put sausage, and in one piece, he added both, but did not mark it (all other pieces are without filling). Jerry also... | # Answer: 5.
Solution. Let's call a piece with fish and sausage a coveted piece.
According to the problem, in any $6 \times 6$ square, there are at least 2 pieces with fish. In any such square, at least one known piece with fish is already included; consider the squares that contain only 1 known piece of fish (all of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. Three merchants: Foma, Yerema, and Julius met in Novgorod. If Foma gives Yerema 70 gold coins, then Yerema and Julius will have the same amount of money. If Foma gives Yerema 40 gold coins, then Foma and Julius will have the same amount of money. How many gold coins should Foma give Yerema so that they bot... | Answer: 55.
Solution. From the first condition, it follows that Yuliy has 70 more coins than Yeremy. From the second condition, it follows that Foma has 40 more coins than Yuliy. Therefore, Foma has $40+70=110$ more coins than Yeremy. For them to have an equal amount of money, Foma must give Yeremy half of this differ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.4. In a coastal village, 7 people go fishing every day, 8 people go fishing every other day, 3 people go fishing every three days, and the rest do not fish at all. Yesterday, 12 people went fishing, and today, 10 people are fishing. How many people will go fishing tomorrow? | Answer: 15.
Solution. Let's calculate how many times in total they fished yesterday and today. 7 people who fish every day fished 2 times each, i.e., a total of 14 times. 8 people who fish every other day fished exactly 1 time each, i.e., a total of 8 times. Therefore, these 15 people fished a total of $14+8=22$ times... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.6. Four girls and eight boys came to take a class photo. The children approach the photographer in pairs and take a joint photo. Among what minimum number of
photos will there definitely be either a photo of two boys, or a photo of two girls, or two photos of the same children? | Answer: 33.
Solution. Suppose at some point there is neither a photo of two boys, nor a photo of two girls, nor two photos of the same children. Then, on each photo, there is a boy and a girl, and on different photos, there are different pairs. However, the total number of possible pairs consisting of a boy and a girl... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.8. A natural number $n$ is called good if 2020 gives a remainder of 22 when divided by $n$. How many good numbers exist? | Answer: 10.
Solution. In the subsequent solution, the expression of the form $a^{k}$ - the number $a$ raised to the power of $k$ - is the number $a$ multiplied by itself $k$ times. We will also consider $a^{0}=1$. For example, $3^{2}=3 \cdot 3=9$, and $2^{0}=1$.
Since 2020 gives a remainder of 22 when divided by $n$,... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. The pages in a book are numbered as follows: the first sheet is two pages (with numbers 1 and 2), the second sheet is the next two pages (with numbers 3 and 4), and so on. The hooligan Petya tore out several consecutive sheets from the book: the first torn page has the number 185, and the number of the las... | Answer: 167.
Solution. Since any leaf ends with a page number that is an even number, the number of the last torn-out page is either 158 or 518. But 158 does not work, since 158 < 185. Therefore, the last page ends with the number 518. Now let's calculate the number of torn-out pages. Among the pages from 1 to 518, th... | 167 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
 = d$, then $(x + y) \vdots d$, i.e., the distance between the cities is divisible by all the calculated GCDs.
Now suppose the distance between the cities (let's call it $S$) is divisible by some n... | 39 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. Carlson and Little Man have several jars of jam, each weighing an integer number of pounds.
The total weight of all the jars of jam that Carlson has is 13 times the total weight of all the jars that Little Man has. Carlson gave Little Man the jar with the smallest weight (from those he had), after which t... | Answer: 23.
Solution. All variables in the solution will be natural numbers, since the weights of all jars are integers by condition.
Let Little have a total of $n$ pounds of jam initially, then Karlson had $13 n$ pounds of jam. Let Karlson give away his smallest jar with $a$ pounds of jam. Then, by condition, $13 n-... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. A square was cut into four equal rectangles, and from them, a large letter P was formed, as shown in the figure, with a perimeter of 56.
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. Four children were walking along an alley and decided to count the number of firs planted along it.
- Anya said: "There are a total of 15 firs along the alley."
- Borya said: "The number of firs is divisible by 11."
- Vера said: "There are definitely fewer than 25 firs."
- Gena said: "I am sure that their... | Answer: 11.
Solution. Let $N-$ be the number of elms along the alley.
Suppose Genya told the truth, and $N$ is divisible by 22. But then $N$ is also divisible by 11, i.e., Borya also told the truth. But according to the problem, one of the boys was wrong. Therefore, Genya must have been wrong, but Borya was right. So... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. In a class, there are 20 students. Thinking about which girls to send a Valentine's card to on February 14, each boy made a list of all the girls in the class he finds attractive (possibly an empty list). It is known that there do not exist three boys whose lists have the same number of girls. What is the ... | Answer: 6.
Solution. Let the number of girls in the class be $d$. According to the condition, there are no three boys whose lists match in the number of girls, so there can be a maximum of 2 empty lists, 2 lists with one girl, 2 lists with two girls, ..., 2 lists with $d$ girls. This means that the number of lists, an... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. At the ball, ladies and gentlemen arrived - in total less than 50 people. During the first dance, only a quarter of the ladies were not invited to dance, and 2/7 of the total number of gentlemen did not invite anyone. How many people came to the ball? (For the dance, a certain gentleman invites a certain l... | Answer: 41.
Solution. Let the number of ladies be $n$, and the number of gentlemen be $m$. We will count the number of pairs who danced. On one hand, it is $\frac{3}{4} n$, and on the other hand, $\frac{5}{7} m$. By equating, we find the ratio
$$
\frac{n}{m}=\frac{20}{21}
$$
Since the fraction on the right side is i... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For three real numbers $p, q$, and $r$, it is known that
$$
p+q+r=5, \quad \frac{1}{p+q}+\frac{1}{q+r}+\frac{1}{p+r}=9
$$
What is the value of the expression
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r} ?
$$ | Answer: 42.
Solution. Multiply the two given equalities, we get
$$
5 \cdot 9=\frac{p+q+r}{p+q}+\frac{p+q+r}{q+r}+\frac{p+q+r}{p+r}=\left(1+\frac{r}{p+q}\right)+\left(1+\frac{p}{q+r}\right)+\left(1+\frac{q}{p+r}\right)
$$
Subtract 3 from both sides of the equation, we get
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r}... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. Masha wrote on the board in ascending order all natural divisors of some number $N$ (the very first divisor written is 1, the largest divisor written is the number $N$ itself). It turned out that the third from the end divisor is 21 times greater than the second from the beginning. What is the largest valu... | Answer: 441.
Solution. The second divisor from the beginning is the smallest prime divisor of the number $N$, let's denote it as $p$. The third divisor from the beginning is either $p^{2}$ or the second largest prime divisor of the number $N$, let's denote it as $q$.
Case 1. The third divisor from the beginning is $p... | 441 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. To 30 palm trees in different parts of an uninhabited island, a sign is nailed.
- On 15 of them, it is written: "Exactly under 15 signs, treasure is buried."
- On 8 of them, it is written: "Exactly under 8 signs, treasure is buried."
- On 4 of them, it is written: "Exactly under 4 signs, treasure is burie... | Answer: 15.
Solution. Suppose the treasure is not buried under at least 16 plaques. Then there are two plaques with different inscriptions under which there is no treasure. According to the condition, the inscriptions on both should be true, but they contradict each other. Contradiction.
Therefore, the treasure is no... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.5. Buratino has many coins of 5 and 6 soldi, more than 10 of each type. Coming to the store and buying a book for $N$ soldi, he realized that he could not pay for it without receiving change. What is the greatest value that the natural number $N$ can take if it is no more than 50? | Answer: 19.
Solution. It is easy to check that with $N=19$ change is necessary.
Notice that numbers from 20 to 24 do not meet the condition, as they can be paid for without change: $N=20=4 \cdot 5, 21=3 \cdot 5+6, 22=2 \cdot 5+2 \cdot 6, 23=5+3 \cdot 6, 24=4 \cdot 6$.
It is clear that then numbers from 25 to 50 also... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. At a ball, 29 boys and 15 girls arrived. Some boys danced with some girls (no more than once in each pair). After the ball, each person told their parents how many times they danced. What is the maximum number of different numbers the children could have mentioned? | Answer: 29.
Solution. The largest possible number that could have been named is 29 (in the case where there is a girl who danced with all the boys), and the smallest is 0. Thus, we have that the number of different numbers named is no more than 30.
We will prove that it cannot be exactly 30. Suppose the opposite, tha... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. How many pairs of natural numbers $a$ and $b$ exist such that $a \geqslant b$ and
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{6} ?
$$ | Answer: 5.
Solution. From the condition, it follows that $\frac{1}{6}=\frac{1}{a}+\frac{1}{b} \leqslant \frac{2}{b}$, from which $b \leqslant 12$. Also, $\frac{1}{6}=\frac{1}{a}+\frac{1}{b}>\frac{1}{b}$, so $b>6$. Therefore, $b$ can take values from 7 to 12 inclusive.
Using $\frac{1}{a}=\frac{1}{6}-\frac{1}{b}=\frac{... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Yura has $n$ cards, on which numbers from 1 to $n$ are written. After losing one of them, the sum of the numbers on the remaining cards turned out to be 101. What number is written on the lost card? | Answer: 4.
Solution. Suppose that $n \leqslant 13$. Then $1+2+\ldots+n=\frac{n(n+1)}{2} \leqslant 91101$, a contradiction.
Therefore, $n=14$, and the missing number is $1+2+\ldots+14-101=105-101=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 10.4. In the central cell of a $21 \times 21$ board, there is a chip. In one move, the chip can be moved to an adjacent cell by side. Alina made 10 moves. How many cells can the chip end up in? | Answer: 121.
Solution. We will paint the entire board in a checkerboard pattern so that the central cell of the board is black. With each move of the chip to an adjacent cell, the color of the cell on which the chip is standing will change. After an odd number of moves, the chip will always be on a white cell, and aft... | 121 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. Oleg has four cards, on each of which natural numbers are written on both sides (a total of 8 numbers are written). He considers all possible quadruples of numbers, where the first number is written on the first card, the second on the second, the third on the third, and the fourth on the fourth. Then, fo... | Answer: 21.
Solution. Let the numbers on one card be $a$ and $b$, on another card - $c$ and $d$, on the third card - $e$ and $f$, and on the fourth card - $g$ and $h$. According to the problem, the sum of 16 terms of the form $a c e g + a c e h + \ldots + b d f h$ equals 330. Note that this sum is also obtained by exp... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Inside a circle, 16 radii of the circle and 10 concentric circles, whose centers coincide with the center of the circle, are drawn. Into how many regions do the radii and circles divide the circle? | Answer: 176.
Solution. 10 circles divide the circle into 10 rings and one smaller circle, a total of 11 parts. 16 radii divide each of the 11 parts into 16 more. In total, there are $11 \cdot 16=176$ regions. | 176 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. Along a road, 25 poles are standing in a row. Sometimes a goldfinch lands on one of the poles, and immediately a goldfinch flies away from one of the adjacent poles (if there was at least one goldfinch on the adjacent poles at that moment). Also, no more than one goldfinch can sit on each pole.
Initially... | Answer: 24.
Solution. First, we will show that it is impossible to occupy all the poles. Suppose this happened. Consider the last tit that sat down. Since it occupied the last unoccupied pole, there must have been an occupied pole next to it. Consequently, the tit sitting on that pole must have flown away. Contradicti... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. A natural number $n$ is called interesting if $2 n$ is a perfect square, and $15 n$ is a perfect cube. Find the smallest interesting number.
Answer: 1800. | Solution. Let's factorize the number $n$ into prime factors. For a number to be a square, it is necessary that in this factorization all prime numbers appear in even powers, and for a number to be a cube, it is necessary that all prime numbers appear in powers divisible by 3.
Let's see what power of two divides $n$. F... | 1800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. Senya has three straight sticks, each 24 centimeters long. Senya broke one of them into two pieces so that he could form the outline of a right-angled triangle with the two pieces of this stick and the two whole sticks. How many square centimeters is the area of this triangle? | Answer: 216.
Solution. Let Senya break one of the sticks into parts of lengths $a$ and $24-a$, which we will call small, while sticks of length 24 will be called large. From four sticks, Senya formed a triangle, so one of the sides consists of two sticks, and the other two sides consist of one stick each.
Let's see w... | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. At the call of the voivode, 55 soldiers arrived: archers and swordsmen. All of them were dressed either in golden or black armor. It is known that swordsmen tell the truth when wearing black armor and lie when wearing golden armor, while archers do the opposite.
- In response to the question "Are you wea... | Answer: 22.
Solution. To the first question, affirmative answers will be given by archers in gold armor and archers in black armor, that is, all archers.
To the second question, affirmative answers will be given by archers in gold armor and swordsmen in gold armor, that is, all soldiers in gold armor.
To the third q... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 11.7. For what least natural $x$ is the expression
$$
\sqrt{29+\sqrt{x}}+\sqrt{29-\sqrt{x}}
$$
an integer? | Answer: 400.
Solution. If we square the given expression, we get $58+2 \sqrt{29^{2}-x}$. If the initial expression was an integer, then this number is a perfect square. We need to find the smallest $x$, i.e., we need to make this expression the largest possible perfect square. Notice that $58+2 \sqrt{29^{2}-x}<58+2 \s... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There were 2013 empty boxes. In one of them, 13 new boxes (not nested inside each other) were placed. Thus, there became 2026 boxes. In one of them, 13 new boxes (not nested inside each other) were placed again, and so on. After several such operations, there became 2013 non-empty boxes. How many boxes are there in ... | # Solution.
After each operation, the number of non-empty boxes increases by 1. Initially, there were no non-empty boxes. Therefore, a total of 2013 operations were performed. With each operation, 13 new boxes are added. Thus, we have
$2013 + 13 \times 2013 = 2013 \times 14 = 28182$.
Recommendations for checking.
A... | 28182 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. An electronic clock displays time from 00.00.00 to 23.59.59. How much time during a day does the clock display exactly four digit 3s? | 3. If the display shows ab.cd.mn, then $a \neq 3$, so there are 5 cases where one of the digits $b, c, d, m, n$ is not 3, while the others are 3. a) On the display ab.33.33, where $b \neq 3$. There are 21 such sets.
b) On the display a3.c3.33, where c is not 3. Here a can take any of the three values 0, 1, or 2, and c... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Around a round table, $2 n(n>5)$ people are sitting - knights and liars. Liars give a false answer to any question, while knights give a true answer. Each of them knows who is a knight and who is a liar. Each of them answered two questions: "Who is your left neighbor?", "Who is your right neighbor?". A wise man, who... | 5. Two adjacent knights (R), like two adjacent liars (L), give answers "R" and "R", while adjacent R and L give answers "L" and "L". If the number of L is less than \( n-1 \), then in a group of consecutive R, the outermost R can be replaced by L, and we get the same set of answers with a different number of R. Also, L... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed as an integer number of centimeters. For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
(A. Magazinov) | Answer. $N=102$.
Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $(201-N)$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot b... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4 In the thirtieth kingdom, there are three types of coins in circulation: bronze rubles, silver coins worth 9 rubles, and gold coins worth 81 rubles. From the treasury, which contains an unlimited supply of each type of coin, a certain amount was issued with 23 coins, which is less than 700 rubles. Find this amount,... | Solution. Since the issued amount is less than 700 rubles, the number of gold coins is less than $700: 81$, which means no more than 7. Then the number of bronze and silver coins together is at least $23-7=16$. Note that the number of bronze coins cannot exceed 8: otherwise, we can replace 9 bronze coins with 1 silver ... | 647 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5 In some cells of a $10 \times 10$ table, crosses are placed such that each of them is the only one either in its row or in its column. What is the maximum number of crosses that can be in such a table? Justify your answer. | Solution. We will place crosses in all cells of the first row and the first column, excluding the top-left cell - a total of 18 crosses. Then each cross in the first row is unique in its column, and each cross in the first column is unique in its row. Therefore, it is possible to place 18 crosses.
We will show that if... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the sides $B C$ and $A D$ of the convex quadrilateral $A B C D$, their midpoints - points $M$ and $N$ respectively - are marked. Segments $M N$ and $A C$ intersect at point $O$, and $M O=O N$. It is known that the area of triangle $A B C$ is 2017. Find the area of quadrilateral $A B C D$. | Answer: 4034.
## Solution:
Let $A C$ and $M N$ intersect at point $O$ (see the first figure on the right), $S_{\triangle A B C}=2017=S$.
We will prove that $S_{\triangle A D C}=S_{\triangle A B C}$, then $S_{A B C D}=2 S$.
There are various ways to reason.
$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
 | Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3.1. As a homework exercise, Tanya was asked to come up with 20 examples of the form $*+*=*$, where $*$ should be replaced with different natural numbers (i.e., a total of 60 different numbers should be used). Tanya loves prime numbers very much, so she decided to use as many of them as possible, while still ... | # Answer: 41.
Solution. Note that in each example, instead of asterisks, three odd numbers cannot be used, i.e., at least one even number must be used. There is exactly one even prime number, which is 2. Therefore, among the 60 different numbers in the examples, at least 19 even composite numbers will be used, and the... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
 | Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H ... | 101 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5.1. In the cells of a $12 \times 12$ table, natural numbers are arranged such that the following condition is met: for any number in a non-corner cell, there is an adjacent cell (by side) that contains a smaller number. What is the smallest number of different numbers that can be in the table?
(Non-corner c... | Answer: 11.
Solution. First, let's provide an example with the arrangement of 11 different numbers.
| 1 | 2 | 3 | 4 | 5 | 6 | 6 | 5 | 4 | 3 | 2 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 3 | 4 | 5 | 6 | 7 | 7 | 6 | 5 | 4 | 3 | 2 |
| 3 | 4 | 5 | 6 | 7 ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6.1. Even natural numbers $a$ and $b$ are such that $\operatorname{GCD}(a, b) + \operatorname{LCM}(a, b) = 2^{23}$. How many different values can $\operatorname{LCM}(a, b)$ take? | Answer: 22.
Solution. Note that LCM $(a, b) \vdots$ GCD $(a, b)$, therefore
$$
2^{23}=\text { GCD }(a, b)+\text { LCM }(a, b) \vdots \text { GCD }(a, b) .
$$
From this, it follows that GCD $(a, b)$ is a natural divisor of the number $2^{23}$.
At the same time, GCD $(a, b) \neq 1$ (since $a$ and $b-$ are even number... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8.1. Different positive numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+b c=115 \\
b^{2}+a c=127 \\
c^{2}+a b=115
\end{array}\right.
$$
Find $a+b+c$. | Answer: 22.
Solution. Subtract the third equation from the first and transform:
$$
\begin{gathered}
\left(a^{2}+b c\right)-\left(c^{2}+a b\right)=0 \\
a^{2}-c^{2}+b c-a b=0 \\
(a-c)(a+c)+b(c-a)=0 \\
(a-c)(a+c-b)=0
\end{gathered}
$$
By the condition $a \neq c$, therefore $b=a+c$. Now add the first two equations:
$$
... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.9. In a class, there are $m$ students. During September, each of them went to the swimming pool several times; no one went twice on the same day. On October 1st, it turned out that all the numbers of swimming pool visits by the students were different. Moreover, for any two of them, there was definitely a day when t... | Answer. $m=28$.
Solution. For each natural $n$, let $X_{n}=$ $=\{1,2, \ldots, n\}$. To each student, we associate the set of all days when he went to the pool (this will be a subset of $X_{30}$). Thus, we have obtained a set of $m$ (according to the condition, non-empty) subsets of $X_{30}$. The condition is equivalen... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1. Given two natural numbers. One number was increased by 3, and the other was decreased by 3. As a result, their product increased by 600. By how much will the product decrease if the opposite is done: the first number is decreased by 3, and the second is increased by 3? | Answer: 618.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=600$, so $a-b=203$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 203+9=618$. | 618 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 900 meters left to reach it. Whe... | Answer: 1500.
Solution: When Tolya has traveled 900 meters, Dima will have traveled 600 meters, so at the moment when Tolya is 900 meters from the intersection, Dima will be 1200 meters from the intersection. According to the Pythagorean theorem, the distance between the boys is 1500 meters. | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1. Petya and Masha take candies from a box in turns. Masha took one candy, then Petya took 2 candies, Masha - 3 candies, Petya - 4 candies, and so on. When the number of candies in the box became less than needed for the next turn, all the remaining candies went to the one whose turn it was to take candies. ... | Answer: 110.
Solution. Since $1+3+5+7+9+11+13+15+17+19=100101$, then Masha got the last candy. Then Petya took for himself $2+4+6+8+10+12+14+16+18+20=2(1+2+3+4+5+6+7+8+9+10)=110$ candies. | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1. An ant, starting from point A, goes $1+\frac{1}{10}$ cm north, then $2+\frac{2}{10}$ cm west, then $3+\frac{3}{10}$ cm south, then $4+\frac{4}{10}$ cm east, then $5+\frac{5}{10}$ cm north, then $6+\frac{6}{10}$ cm west, and so on. After 1000 steps, the ant is at point B. Find the distance between points A... | Answer: 605000.
Solution. Let's divide 1000 steps into quartets. After each quartet, the ant will move southeast relative to its current position, by a distance equal to the diagonal of a square with side 2.2, i.e., $\sqrt{2.2^{2}+2.2^{2}}=\sqrt{9.68}$. After 250 such quartets, the ant will be at a distance of $250 \c... | 605000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
It is known that the equations $x^{2}+(2 a-5) x+a^{2}+1=0$ and $x^{3}+(2 a-5) x^{2}+\left(a^{2}+1\right) x+a^{2}-4=0$ have common roots. Find the sum of these roots. | Answer: 9.
Solution. The left side of the second equation is obtained from the left side of the first equation by multiplying by $x$ and adding the expression $a^{2}-4$. Therefore, $a^{2}-4=0$, from which $a=2$ or $a=-2$. If $a=2$, then the first equation has no roots. Therefore, $a=-2$, and then we get that the first... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1. Grisha thought of such a set of 10 different natural numbers that their arithmetic mean is 16. What is the maximum possible value of the largest of the numbers he thought of? | Answer: 115.
Solution: The sum of the given numbers is $10 \cdot 16=160$. Since all the numbers are distinct, the sum of the 9 smallest of them is no less than $1+2+\cdots+9=45$. Therefore, the largest number cannot be greater than $160-45=115$. This is possible: $(1+2+\cdots+9+115): 10=16$. | 115 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1. Roma decided to create his own multiplication table. The rows correspond to the numbers 12, $13,14, \ldots, 60$, and the columns - to the numbers $15,16,17, \ldots, 40$. In the cells of the table, he wrote the products of the pairs of numbers from the row and column. How many of these products will be eve... | Answer: 962.
Solution. Note that the product of two numbers is odd if and only if both factors are odd, and even in all other cases. In total, the table contains 49$\cdot$26 products. Note that among the numbers from 12 to 60, there are 24 odd numbers, and among the numbers from 15 to 40, there are 13 odd numbers. The... | 962 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1. In triangle $A B C$, the bisector $A L$ and the median $B M$ are drawn. It turns out that $A B=2 B L$. What is the measure of angle $B C A$, if $\angle L M A=127^{\circ}$? | Answer: $74^{\circ}$.
Solution.
By the property of the angle bisector, $\frac{A C}{C L}=\frac{A B}{B L}=\frac{2}{1}$, so $A C=2 C L$. Since $M$ is the midpoint of $A C$, then $A M=$ $M C=C ... | 74 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. For the Day of the Russian Flag, the seller decided to decorate the store window with 10 horizontal strips of fabric in three colors. At the same time, he follows two conditions:
1) strips of the same color should not hang next to each other;
2) each blue strip must hang between a white and a red one.
In how many w... | 5. Let's call the white and red strips of fabric green. Denote by $T_{n}$ the number of ways to decorate the shop window with $n$ strips of fabric. The first strip can only be a green strip. If the second strip is blue, then the following strips must be such that the first one is green. The total number of ways to hang... | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.7. On the board, 99 numbers were written, none of which are equal. In a notebook, $\frac{99 \cdot 98}{2}$ numbers were written—all differences between two numbers from the board (each time subtracting the smaller number from the larger one). It turned out that the number 1 was written exactly 85 times in the notebook... | Answer. $d=7$.
Solution. We will prove that $d \geqslant 7$. All numbers on the board can be divided into chains of numbers of the form $a, a+1, a+2, \ldots, a+t$ such that numbers from different chains do not differ by exactly 1. Such a partition is not difficult to construct by connecting any two numbers that differ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.9. Find the largest number $m$ such that for any positive numbers $a, b$, and $c$, the sum of which is 1, the inequality
$$
\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}} \geqslant m
$$
holds.
(l. Emelyanov) | Answer. $m=1$.
First solution. First, we prove that $m=1$ satisfies the requirements of the problem. Notice that $ab + c = ab + c(a + b + c) = (c + a)(c + b)$. Therefore,
\[
\begin{aligned}
& \sqrt{\frac{ab}{c + ab}} + \sqrt{\frac{bc}{a + bc}} + \sqrt{\frac{ca}{b + ca}} = \\
& = \sqrt{\frac{ab}{(c + a)(c + b)}} + \sq... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.5. Snow White entered a room where 30 chairs were arranged around a round table. Some of the chairs were occupied by dwarfs. It turned out that Snow White could not sit down without having someone next to her. What is the minimum number of dwarfs that could have been at the table? Explain how the dwarfs should have b... | Answer: 10.
Solution: If there were three consecutive empty chairs at the table in some place, Snow White could sit down in such a way that no one would sit next to her. Therefore, in any set of three consecutive chairs, at least one must be occupied by a dwarf. Since there are 30 chairs in total, there cannot be fewe... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Let $p(x)=2 x^{3}-3 x^{2}+1$. How many squares of integers are among the numbers $p(1), p(2), \ldots, p(2016)$? | # Answer: 32.
## Solution:
Notice that $p(x)=(x-1)^{2}(2 x+1)$.
For an integer $x(1 \leq x \leq 2016)$, the number $p(x)=(x-1)^{2}(2 x+1)$ is a perfect square of an integer either when $x=1(p(1)=0)$, or (for $x \geq 2$) when the number $2 x+1$ is a perfect square.
Note that for $x \geq 2$, the inequality holds: $5 ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded?
保留源文本的换行和格式，所以翻译结果如下：
... | Answer: She could.
## Solution:
Here is an example of a code table:
| A | Б | К | М | П | У | Ш |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this:
10101010100010001100110111001110111110 - a total of 38 characters. Decoding is una... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. In the train, there are 18 identical cars. In some cars, exactly half of the seats are free, in some others - exactly one third of the seats are free, and in the rest, all seats are occupied. At the same time, in the entire train, exactly one ninth of all seats are free. In how many cars are all seats occupied? | Answer: in 13 carriages.
Solution. Let the number of passengers in each carriage be taken as a unit. We can reason in different ways.
First method. Since exactly one ninth of all seats in the train are free, this is equivalent to two carriages being completely free. The number 2 can be uniquely decomposed into the su... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number ... | Answer: increased by $80 \%$.
Solution. First method. If the amount of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this, $70 \%$ is due to the increase in Antonovka, and $50 \%$ is due to the increase in Grushovka. Therefore, the increase due to White Naliv wo... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.6. In each cell of a $5 \times 5$ square, exactly one diagonal has been drawn. A vertex of a cell is free if it is not the endpoint of any of the drawn diagonals. Find the maximum possible number of free vertices. | Answer: 18 vertices.
Solution. Example. See Fig. 7.6a. On each of the six horizontal lines, three vertices are free.
Estimate. The total number of cell vertices: $6 \cdot 6=36$. Let's select nine cells that do not share any vertices (see Fig. 7.6b). They contain all 36 vertices. In each cell, a diagonal is drawn, so ... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On Eeyore's Birthday, Winnie-the-Pooh, Piglet, and Owl came to visit. When Owl left, the average age in this company decreased by 2 years, and when Piglet left, the average age decreased by another 1 year. How many years older is Owl than Piglet?
Answer: Owl is 6 years older than Piglet. | Solution. Let the average age of those who remained after Piglet be x, and Piglet's age be y. Then $2x + y = 3(x + 1)$, which means $y = x + 3$. Let Owl's age be K. Then $3(x + 1) + K = 4(x + 3)$, which means $K = x + 9$. Therefore, Owl is older than Piglet by $(x + 9) - (x + 3) = 6$ years.
Criteria. If the solution i... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.3. In the park, all bicycle paths run from north to south or from west to east. Petya and Kolya simultaneously started from point $A$ and rode their bicycles at constant speeds: Petya - along the route $A-B-C$, Kolya - along the route $A-D-E-F-C$ (see fig.), and both spent 12 minutes on the road. It is known that Kol... | Answer: 1 minute.
Solution. Draw the segment $D H$, as shown in Fig. 2. Kolya travels 1.2 times faster than Petya, so it would take him $12 / 1.2=10$ minutes to travel the route $A-B-C$. The difference in time
. Then each of them said: "Among my neighbors, there is a liar." What is the maximum number of people sitting at the table who can say: "Among my neighbors, there is a k... | # Answer: 8.
Solution. Note that two liars cannot sit next to each other (otherwise, each of them would be telling the truth). Therefore, no liar can say the second phrase.
On the other hand, 3 knights also cannot sit next to each other (otherwise, the middle one would have lied by saying that he has a neighbor who i... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Five consecutive natural numbers were written on the board, and then one number was erased. It turned out that the sum of the remaining four numbers is 2015. Find the smallest of these four numbers. | Answer: 502
Solution. If the smallest number is K, then the sum of all numbers is not less than $\mathrm{K}+(\mathrm{K}+1)+(\mathrm{K}+2)+ (\mathrm{K}+3)=4 \mathrm{~K}+6$ and not more than $\mathrm{K}+(\mathrm{K}+2)+(\mathrm{K}+3)+(\mathrm{K}+4)=4 \mathrm{~K}+9$. Therefore, $4 \mathrm{~K}+6 \leq 2015 \leq 4 \mathrm{~K... | 502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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