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10.1. Find at least one four-digit number that has the following property: if the sum of all its digits is multiplied by the product of all its digits, the result is 3990. (I. Rubanov) | Solution. Note that $3990=2 \cdot 3 \cdot 5 \cdot 7 \cdot 19=1 \cdot 6 \cdot 5 \cdot 7 \cdot 19$ and $6+5+7+1=19$. Therefore, any four-digit number that contains one 1, one 6, one 5, and one 7 will work, for example, 1567.
Comment. To receive full credit, it is sufficient to provide any correct example. | 1567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. The hikers had several identical packs of cookies. At the midday break, they opened two packs and divided the cookies equally among all the participants of the hike. One cookie was left over, and the hikers fed it to a squirrel. At the evening break, they opened another three packs and also divided the cookies equ... | Solution: Let there be $n$ tourists in the group, and $m$ cookies in each pack (where $m$ and $n$ are natural numbers, and $n > 13$). According to the problem, the numbers $x = 2m - 1$ and $y = 3m - 13$ are divisible by $n$. Therefore, the number $3x - 2y = 23$ is also divisible by $n$. Since 23 is a prime number and h... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. When fully fueled, a motorboat can travel exactly 40 km upstream or exactly 60 km downstream. What is the greatest distance the motorboat can travel along the river if the fuel must be enough for the round trip, back to the starting point? Justify your answer. | # Solution:
Method 1. Let's launch a raft along with the boat, which will move with the current, and observe the boat from this raft. Then, no matter which direction the boat goes, upstream or downstream, it will be at the same distance from the raft by the time the fuel runs out. This means that the raft will be at t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. We took ten consecutive natural numbers greater than 1, multiplied them, found all the prime divisors of the resulting number, and multiplied these prime divisors (taking each exactly once). What is the smallest number that could have resulted? Fully justify your answer.
Solution. We will prove that among ten conse... | Answer: 2310.
## Municipal Stage of the All-Russian Mathematics Olympiad 2018-2019 Academic Year
10th Grade
Grading Criteria
| Score | Points for |
| :---: | :---: | :--- |
| 7 | Complete solution, a correct algorithm of actions for Zhenya is provided, and all conditions are verified. |
| 6 | Complete solution, a c... | 2310 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. How many 4-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: 1, $2,3, \ldots, 9$. That is, there are a total of 45 options. The ot... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.1. Sasha solved the quadratic equation $3 x^{2}+b x+c=0$ (where $b$ and $c$ are some real numbers). In his answer, he got exactly one root: $x=-4$. Find $b$. | Answer: 24
Solution. By Vieta's theorem $x_{1}+x_{2}=-b / 3$. In our case $x_{1}=x_{2}=-4$, hence $-b / 3=-8$ and $b=24$.
Remark. Another solution can be obtained by noticing that the equation should reduce to a perfect square $3(x+4)^{2}=0$. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find the sum
$$
\sqrt[7]{(-7)^{7}}+\sqrt[8]{(-8)^{8}}+\sqrt[9]{(-9)^{9}}+\ldots+\sqrt[100]{(-100)^{100}}
$$
(Each term is of the form $\left.\sqrt[k]{(-k)^{k}}\right)$ | Answer: 47
Solution. Note that $\sqrt[k]{(-k)^{k}}$ equals $k$ when $k$ is even, and equals $-k$ when $k$ is odd. Therefore, our sum is $-7+8-9+10-\ldots-99+100=(-7+8)+(-9+10)+\ldots+(-99+100)$. In the last expression, there are 47 parentheses, each of which equals 1. | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. For triangle $A B C$, the following is known: $A B=12, B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R-$ is the radius of the smallest circle in which this triangle can be placed.
Solution. Since segment $AC$ fits inside the circle, $2R \geqslant AC$. On the other hand, the circle constructed with $AC$ as its diameter covers the triangle $ABC$, since $\angl... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.1. Let's consider the number 616. The sum of its digits is 13, and the product of its digits is 36. What is the largest natural number whose sum of digits is 13 and the product of its digits is 36? | Answer: 3322111
Solution. In the decimal representation of this number, there cannot be any zeros. Let's list all the digits of this number that are greater than 1; their product is still 36. Since 36 can be factored into prime factors as $36=2 \cdot 2 \cdot 3 \cdot 3$, the possible sets of digits greater than 1 are: ... | 3322111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. On a plane, 55 points are marked - the vertices of a certain regular 54-gon and its center. Petya wants to paint a triplet of the marked points in red so that the painted points are the vertices of some equilateral triangle. In how many ways can Petya do this? | Answer: 72
Solution. Assume that our 54-gon is inscribed in a circle, and its vertices divide this circle into 54 arcs of length 1.
If one of the painted points is the center, then the other two painted points must be the ends of an arc of size $54 / 6=9$. There are 54 such possibilities.
Otherwise, all three painte... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 234^{234}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 234). How many of the listed numbers are perfect squares? (A perfect square is the square of an integer.) | # Answer: 71
Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is even, then $m^{k}$ is a perfect square. If $k$ is odd, then $m^{k}$ is a perfect square if and only if $m$ is a perfect square. Thus, the answer to our problem is the total number of even numbers and odd perf... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. On the coordinate plane, a parallelogram $O A B C$ is drawn, with its center located at the point $\left(\frac{19}{2}, \frac{15}{2}\right)$, and points $A, B$, and $C$ have natural coordinates. Find the number of such parallelograms. (Here, $O$ denotes the origin - the point $(0,0)$; two parallelograms with the sa... | Answer: 126.
Solution. Note that $O A B C$ is a parallelogram if and only if the point $E\left(\frac{19}{2}, \frac{15}{2}\right)$ is the midpoint of segments $O B$ and $A C$, and moreover, points $O, A, B, C$ do not lie on the same line.
$ | Answer: $90^{\circ}$.
Solution. Let $N$ and $M$ be the midpoints of segments $K C$ and $A C$, respectively. Then $M N$ is the midline of triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$, since quadrilateral $A B C D$ is cyclic.
Suppose points $M$ and $N$ lie on the same ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.6. Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $K$. It turns out that points $B, D$, as well as the midpoints of segments $A C$ and $K C$, lie on the same circle. What values can the angle $A D C$ take?
(G. Zhukov) | Answer: $90^{\circ}$.
Solution. Let $N$ and $M$ be the midpoints of segments $K C$ and $A C$, respectively. Then $M N$ is the midline of triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$, since quadrilateral $A B C D$ is cyclic.
Suppose points $M$ and $N$ lie on the same ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.7. Given a polynomial
$$
P(x)=a_{2 n} x^{2 n}+a_{2 n-1} x^{2 n-1}+\ldots+a_{1} x+a_{0}
$$
where each coefficient $a_{i}$ belongs to the interval $[100,101]$. For what minimal $n$ can such a polynomial have a real root? (I. Bogdanov, K. Sukhov) | Answer. $n=100$.
Solution. Let's call a polynomial that satisfies the condition of the problem beautiful. The polynomial $P(x)=100\left(x^{200}+x^{198}+\ldots+x^{2}+\right.$ $+1)+101\left(x^{199}+x^{197}+\ldots+x\right)$ is beautiful and has a root -1. Therefore, when $n=100$, the required is possible.
It remains to ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. "That" and "this", plus half of "that" and "this" - what percentage of three quarters of "that" and "this" would this be? | 3. "That" and "this", plus half of "that" and "this" - this is one and a half "that" and "this", which is twice as much as three quarters of "that" and "this", i.e., it constitutes $200 \%$ of them.
Answer: $200 \%$. | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In two groups, there is an equal number of students. Each student studies at least one language: English or French. It is known that 5 people in the first group and 5 in the second group study both languages. The number of students studying French in the first group is three times less than in the second group. The ... | 4. Let the number of people studying French in the first group be $x$, then in the second group it is $3x$. Let the number of people studying English in the second group be $y$, then in the first group it is $4y$. The total number of people in the first group is $x+4y-5$, and in the second group it is $3x+y-5$. By equa... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. In 8 "Y" grade, there are quite a few underachievers, but Vovochka performs the worst of all. The pedagogical council decided that either Vovochka must correct his failing grades by the end of the quarter, or he will be expelled. If Vovochka corrects his grades, then 24% of the class will be underachievers, but if h... | # Solution
Let there be $n$ students in the class now. According to the condition, $0.24 n = 0.25(n-1)$, which means $0.01 n = 0.25$. Therefore, $n = 25$. One person makes up $4 \%$ of 25, so there are now $24 + 4 = 28 \%$ of underachievers.
Answer: $28 \%$. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The heights of an acute-angled triangle $ABC$, drawn from points $B$ and $C$, were extended to intersect the circumscribed circle at points $B_{1}$ and $C_{1}$. It turned out that the segment $B_{1} C_{1}$ passes through the center of the circumscribed circle. Find the angle $BAC$. | # Solution
Since $B_{1} C_{1}$ is the diameter of the circle, $\angle B_{1} B C_{1}=\angle B_{1} C C_{1}=90^{\circ}$, therefore, $B C_{1} \| A C$ and $C B_{1} \| A B$ (see figure). Since $B C_{1} \| A C$, then $\angle C_{1} B A=\angle A=\alpha$. Similarly, $\angle B_{1} C A=\angle A=\alpha$. The degree measure of the ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.5
For a triangle, one of whose angles is equal to $120^{\circ}$, it is known that it can be cut into two isosceles triangles. What can the other two angles of the original triangle be?
## Number of points 7 | Answer:
$40^{\circ}$ and $20^{\circ}$ or $45^{\circ}$ and $15^{\circ}$.
## Solution
Let in triangle $ABC$, $\angle B=120^{\circ}$. The cut specified in the problem must pass through one of the vertices of the triangle (otherwise, two triangles will not be formed upon division). It can pass through vertex $B$ or anot... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out of the plane? | Solution. Consider 100 nodes - the intersection points of lines from the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second directions. The second sector - nodes lying on the second lines (excluding points lying in the first sector) and so... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest natural $k$ does the number 2016 $\cdot$ 20162016 $\cdot$ 201620162016 $\cdot$... 20162016...2016( $k$ factors) divide without remainder by $3^{67}$? | Solution.
We will show that in the case of 27 factors, the product is divisible by $3^{67}$, but not by $3^{68}$.
$2016 \cdot 20162016 \cdot \ldots \cdot \underbrace{20162016 . .2016}_{27 \text { times }}=$
$=2016^{27} \cdot(1 \cdot 1001 \cdot 1001001 \cdot \ldots \cdot \underbrace{1001001 \ldots 1001}_{27 \text { t... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" f... | # Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Calculate:
$$
\left(\frac{1+2}{3}+\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{2017+2018}{2019}\right)+\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{673}\right)
$$ | Answer: 1346.
Solution. We have
$$
\begin{aligned}
& \left(\frac{1+2}{3}+\frac{4+5}{6}+\frac{7+8}{9}+\ldots+\frac{2017+2018}{2019}\right)+\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{673}\right)= \\
= & \left(\frac{(3-2)+(3-1)}{3}+\frac{(6-2)+(6-1)}{6}+\frac{(9-2)+(9-1)}{9}+\ldots\right. \\
& \left.\ldots+\frac{(2... | 1346 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Square $A B C D$ is inscribed in circle $\omega$. On the smaller arc $C D$ of circle $\omega$, an arbitrary point $M$ is chosen. Inside the square, points $K$ and $L$ are marked such that $K L M D$ is a square. Find $\angle A K D$. | Answer: $135^{\circ}$.
Solution.
We will prove the equality of triangles $D A K$ and $D C M$. Let's check that the conditions of the first criterion for the equality of triangles are met. Se... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Given a positive number $a$. It is known that the equation $x^{3}+1=a x$ has exactly two positive roots, and the ratio of the larger to the smaller one is 2018. The equation $x^{3}+1=a x^{2}$ also has exactly two positive roots. Prove that the ratio of the larger to the smaller one is also 2018. | Solution. Let the positive roots of the equation $x^{3}+1=a x$ be denoted by $x_{1}$ and $x_{2}$ $\left(0<x_{1}<x_{2}, x_{2}: x_{1}=2018\right)$. Substitute them into the equation and divide the two resulting equations by $x_{1}^{3}$ and $x_{2}^{3}$:
\[
\begin{aligned}
& x_{1}^{3}+1=a x_{1} \Longleftrightarrow 1+\left... | 2018 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 4.2. Petya took half of the candies from the box and put them in two pockets. Deciding that he took too many, Petya took out 6 candies from each pocket and put them back into the box. By how many more candies did the box have than Petya's pockets? | Answer: 24.
Solution. Let $x$ be the number of candies in one of Petya's pockets before he returned some of the candies to the box. At that moment, there were $2 x$ candies in the box.
After Petya took out 6 candies from each pocket, he had
$$
(x-6)+(x-6)=2 x-12 \text{ candies }
$$
and there were $2 x+12$ candies i... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. On the meadow, there were 12 cows. The shepherds brought a flock of sheep. There were more sheep than the cows' ears, but fewer than the cows' legs. How many sheep were there if there were 12 times more sheep than shepherds? | Answer: 36 sheep.
Solution. Since there are 12 times more sheep than shepherds, the number of sheep is divisible by 12. In addition, the number of sheep is more than $12 \cdot 2=24$ (the number of ears of cows) and less than $12 \cdot 4=$ 48 (the number of legs of cows). The only number divisible by 12 and between 24 ... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. From sticks of the same length, a row of 700 octagons was laid out as shown in the figure. How many sticks were used in total?
Answer, option 1. 7001.
Answer, option 2. 3501.
Answer,... | Solution option 1. In the first octagon, there are 8 sticks. Constructing each subsequent octagon adds 7 sticks. In total, it will be $8+7 \cdot 999=7001$ sticks. | 7001 | Geometry | MCQ | Yes | Yes | olympiads | false |
2.4. The standard 12-hour clock face is arranged as shown in the figure. At noon, both the hour and minute hands were vertical, and now the hour hand is exactly pointing to the "18 minutes" mark.
 divisions. Subtracting 60 divisions from 204 as many times as possible (considering that the min... | 24 | Geometry | MCQ | Yes | Yes | olympiads | false |
6.4. To be freed from Ivan the Fisherman's net, the Golden Fish gives him a reward: either 4 pearls and 1 diamond; or 2 emeralds and 2 diamonds; or 2 pearls, 1 emerald, and 1 diamond. After several releases of the Golden Fish, Ivan the Fisherman received 1000 pearls, 800 emeralds, and some number of diamonds. How many ... | Solution 1. Let there be $x$ actions of the first type, $y$ actions of the second type, and $z$ actions of the third type. From the problem statement, we get two equations $4 x+2 z=2000, 2 y+z=800$, and we need to find $x+y+z$. By adding the first equation to twice the second, we get $4 x+4 y+4 z=3600$, from which $x+y... | 900 | Number Theory | MCQ | Yes | Yes | olympiads | false |
8.4. In the Go section, there are 55 boys of different ratings. The boys decided to play a tournament, each with each other in one game. To make it more interesting, some of the boys were allowed to use a computer's help exactly once during the tournament. If in a game, one of the boys uses a computer's help and the ot... | Solution Option 1. Let's number the players in descending order of their strength from first to fiftieth.
We will assume that each won game brings the player one point. Consider two players who ultimately scored more points than the two strongest participants. It is clear that the number of one of them is not less tha... | 45 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
6. On 52 cards, the numbers from 1 to 52 are written. It is considered that 1 is older than 52, and in all other pairs, the card with the larger number is older. The cards are lying on the table in a random order face down. In one question, Petya can find out which card is older in any pair. Can Petya guarantee to find... | 6. Answer: it can. We will divide the cards into 13 quartets. In each quartet, we will make the following comparisons: by dividing the quartet into pairs, we will determine the higher cards in the pairs (let $\mathrm{a}>\mathrm{b}$ and $c>d$), then we will determine the higher of the two higher cards (let $a>c$). Card ... | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Inside the square $A B C D$, points $K$ and $M$ are marked (point $M$ is inside triangle $A B D$, point $K$ is inside $B M C$) such that triangles $B A M$ and $D K M$ are equal $(A M=K M, B M=M D, A B=K D)$. Find $\angle K C M$, if $\angle A M B=100^{\circ}$ | Answer: $35^{\circ}$.
Solution. Note that triangles $A B M$ and $A M D$ are also equal by three sides. Thus, point $M$ lies on the diagonal $A C$ of the square, meaning $\angle M C D=\angle M A D=\angle M A B=45^{\circ}($ Fig. 1$)$.
In addition, from the equality of triangles $B A M$ and $D K M$, it follows that $\an... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. Every month Ivan pays a fixed amount from his salary for a mortgage, and the remaining part of the salary is spent on current expenses. In December, Ivan paid $40 \%$ of his salary for the mortgage. In January, Ivan's salary increased by $9 \%$. By what percentage did the amount spent on current expenses increase ... | Answer: 15
Solution. Let Ivan's December salary be $100 r$. Then Ivan paid $40 r$ for the mortgage, and in December, he spent $60 r$ on current expenses. In January, Ivan's salary was $109 r$, so he spent $109 r - 40 r = 69 r$ on current expenses. Thus, the amount spent on current expenses increased by $9 r$, which is... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. In the product
$$
24^{a} \cdot 25^{b} \cdot 26^{c} \cdot 27^{d} \cdot 28^{e} \cdot 29^{f} \cdot 30^{g}
$$
the seven exponents $a, b, c, d, e, f, g$ were replaced by the numbers $1, 2, 3, 5, 8, 10, 11$ in some order. Find the maximum number of zeros that the decimal representation of this product can end with. | Answer: 32
Solution. The prime factor 5 enters the factorization of this product with multiplicity $2 b+g$, so there are no more than $2 b+g$ zeros in this product. Since $b \leqslant 11$ (the largest of the given exponents) and $b+g \leqslant 11+10$ (the sum of the two largest of the given exponents), then $2 b+g \le... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. The geometric progression $b_{1}, b_{2}, \ldots$ is such that $b_{25}=2 \operatorname{tg} \alpha, b_{31}=2 \sin \alpha$ for some acute angle $\alpha$. Find the number $n$ for which $b_{n}=\sin 2 \alpha$.
# | # Answer: 37
Solution. Let $q$ be the common ratio of our progression. Using the fact that $b_{31}=b_{25} q^{6}$ and $\sin 2 \alpha=2 \sin \alpha \cos \alpha$. We have $q^{6}=\frac{b_{31}}{b_{25}}=\frac{2 \sin \alpha}{2 \tan \alpha}=\cos \alpha$.
On the other hand, $\frac{\sin 2 \alpha}{2 \sin \alpha}=\cos \alpha$, f... | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Given a rectangular parallelepiped $2 \times 3 \times 2 \sqrt{3}$. What is the smallest value that the sum of the distances from an arbitrary point in space to all eight of its vertices can take?
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $ABC$, where $AB = AC$ and $\angle ABC = 53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $AK$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $AC$;
- $KM = AB$
- angle $MAK$ is the maximum possible.
How many degrees does angle $BAM... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
 on the circle. Add two more marked points to them, forming a decagon $A_{1} A_{2} \ldots A_{5} B_{1} B_{2} \ldots B_{5}$. Then the segments $A_{1} B_{1}, A_{2} B_{2}, \ldots, A_{5} B_{5}$ intersect each other, and among them, there are t... | 143 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. On a shelf, there are 12 books. In how many ways can 5 of them be chosen if books standing next to each other cannot be chosen? Justify your answer. | Solution: Let's place one more, the thirteenth book, on the right side of the shelf. We will mentally glue each selected book to the book to its right. This is possible since we do not select the 13th book and do not select adjacent books. This will result in 5 glued two-volume sets and 3 separate books. Therefore, the... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 5. The factory paints cubes in 6 colors (each face in its own color, the set of colors is fixed). How many varieties of cubes can be produced? | Solution: Suppose the procedure of painting the cube goes as follows: an unpainted cube is placed in a machine in a certain fixed position, and then its faces are painted in a certain order: bottom, top, right, left, front, back. First, let's calculate how many ways such a painting can be done. The bottom face can be p... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. Option 1.
Find the value of the expression $101+102-103-104+105+106-107-108+\ldots-619$? | Answer: 100.
Solution. Subtract and add the number 620 at the end of the expression. Then the numbers from 101 to 620 will be divided into $(620-100): 4=130$ quartets. And in each quartet, the sum of the numbers will be equal to -4. Then the value of the expression will be equal to $(101+102-103-104)+(105+106-107-108)... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
On a sheet, two rectangles are drawn. It is known that both the length and the width of the second rectangle are 3 cm larger than the length and width of the first rectangle, and the area of the second rectangle is 48 cm ${ }^{2}$ larger than the area of the first rectangle. Find the perimeter of the sec... | Answer: 38.
Solution: Let the first rectangle have a length of $a$ cm and a width of $b$ cm. Then the length of the second rectangle is $a+3$ cm, and the width is $b+3$ cm. From the condition, it follows that $(a+3)(b+3)-a b=48$, then $3(a+b)=39, a+b=13$ and $(a+3)+(b+3)=19$. Therefore, the perimeter is 38 cm.
## Var... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1 Sasha wrote the number 765476547654 on the board. He wants to erase several digits so that the remaining digits form the largest possible number divisible by 9. What is this number? | Answer: 7654765464.
Solution: The sum of the digits of the original number is $3 \cdot(7+6+5+4)=66$. According to the divisibility rule for 9, the sum of the erased digits must give a remainder of 3 when divided by 9. It is impossible to select digits with a sum of 3. Digits with a sum of $3+9=12$ can be chosen, and w... | 7654765464 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1
In the garden, there are 26 trees - apple trees and pears. It turned out that among any 18 trees, there is at least one apple tree, and among any 10 trees, there is at least one pear. How many pears are there in the garden? | Answer: 17.
Solution: Since among 18 trees there is at least one apple tree, there are no more than 17 pears. And since among any 10 trees there is at least one pear, there are no more than 9 apple trees. There are 26 trees in total, so there are 9 apple trees and 17 pears.
## Variant 2
In the garden, there are 26 t... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
Petya wrote down all positive numbers that divide some natural number $N$. It turned out that the sum of the two largest written numbers is 3333. Find all such $N$. If there are several numbers, write their sum in the answer. | Answer: 2222.
Solution: Note that one of the listed numbers will be equal to $N$. Since the sum of the two largest listed numbers is odd, these numbers have different parity. This means that the number 2 is a divisor of $N$, so the second number is $N / 2$. According to the condition, $N + N / 2 = 3333$. Therefore, $N... | 2222 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.1. Circular weights weigh 200 grams, square ones weigh 300 grams, and triangular ones weigh 150 grams. 12 weights were placed on the pan balance as shown in the figure. Which pan is heavier and by how many grams?
![](https://cdn.mathpix.com/cropped/2024_05_06_c60160c61c72db74c711g-1.jpg?height=577&width=493&... | Answer: The right side is 250 grams heavier.
Solution. The weight of the left pan in grams is
$$
1 \cdot 300+2 \cdot 150+3 \cdot 200=1200
$$
The weight of the right pan in grams is
$$
1 \cdot 150+2 \cdot 200+3 \cdot 300=1450
$$
Thus, the right pan is 250 grams heavier than the left pan. | 250 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.2. In class 4A, each child has no fewer than 11 classmates and no fewer than 13 female classmates. What is the smallest number of children that can study in this class? | Answer: 26.
Solution. It is not difficult to verify that a class consisting of 12 boys and 14 girls satisfies the condition of the problem. Now let's prove that it cannot be less.
It is clear that there are both boys and girls in the class. Each boy in the class has at least 11 classmates, so there must be at least 1... | 26 | Combinatorics | 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 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 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?
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 8.6. 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 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 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 located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles of $... | 16 | Logic and Puzzles | 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 11.3. 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. Factorize the number $n$ into prime factors. For a number to be a square, it is necessary that all prime numbers in this factorization 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 look at the power of two th... | 1800 | Number Theory | 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 |
10.2 It is known that $10 \%$ of people own no less than $90 \%$ of all the money in the world. For what minimum percentage of all people can it be guaranteed that these people own $95 \%$ of all the money? | Solution. Let $100 S$ be the total amount of money in the world (regardless of the currency), and the total number of people be $100 N$ (where $N$ can be non-integer, which is not important). Then at least $90 S$ of the money is owned by $10 N$ people (let's call this group oligarchs). The remaining $90 N$ people (let'... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the island of knights and liars (liars always lie, knights always tell the truth), each resident supports exactly one football team. In a survey, all residents of the island participated. To the question "Do you support 'Rostov'?", 40% of the residents answered "Yes". To a similar question about 'Zенit', 30% answ... | Solution. Let $x \%$ of the island's inhabitants be liars. Then $(100-x) \%$ are knights. Since each knight answered affirmatively to exactly one of the questions, and each liar to three, we have $(100-x)+3 x=40+30+50$, from which $x=10$. Since none of the island's inhabitants said they support CSKA, all liars support ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Let quadrilateral $A B C D$ be inscribed. The rays $A B$ and $D C$ intersect at point K. It turns out that points $B, D$, and the midpoints $M$ and $N$ of segments $A C$ and $K C$ lie on the same circle. What values can the angle $A D C$ take?
 An apprentice chef took two buckets of unpeeled potatoes and cleaned everything in an hour. In the process, $25 \%$ of the potatoes went into peels. How long did it take for him to accumulate exactly one bucket of peeled potatoes? | Solution. Since a quarter of the potatoes went into peels, then in one hour, three quarters of two buckets of cleaned potatoes were obtained. This means that in one hour, the trainee cook got $\frac{3}{2}$ buckets, and exactly one bucket in 40 minutes.
Answer. 40 minutes.
## Comments on Evaluation.
Correct answer on... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. There are more than 20 but fewer than 30 people in the class, and everyone has a different birthday. Petya said: "There are twice as many people older than me as there are younger than me in the class." Katya said: "There are three times fewer people older than me as there are younger than me in the class." How many... | Answer: 25.
Solution: From Petya's words, it is clear that without him, the number of students in the class is a multiple of 3 (2 parts + 1 part). That is, together with him, there can be 22, 25, or 28 people in the class. Similarly, from Katya's words, it is clear that without her, the number of students in the class... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Three three-digit numbers, in the notation of which all digits except zero participate, add up to 1665. In each number, the first digit was swapped with the last digit. This resulted in three new three-digit numbers. What is the sum of the new numbers | Solution: The sum of the last digits of the three original numbers is 5, 15, or 25. However, 5 and 25 are excluded, as they cannot be represented as the sum of three different digits (from 1 to 9), so 15 remains. Therefore, the sum of the middle digits is also 15, and the sum of the first digits is 15. It then becomes ... | 1665 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Petya cut an $8 \times 8$ square along the cell boundaries into parts of equal perimeter. It turned out that not all parts were equal. What is the maximum number of parts he could have obtained?
# | # Answer: 21.
Solution: Let $S$ be the smallest area among the figures obtained by Petya.
If $S=1$, then this figure is a $1 \times 1$ square and its perimeter is 4. Then all of Petya's figures have a perimeter of 4, meaning they are all $1 \times 1$ squares. But this means that all figures are equal, which contradic... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. Find the largest natural number, all digits of which are different, and the product of these digits is a square of a natural number. | Answer: 986431. Solution: Obviously, among these digits there is no zero. Further, we have $1 \cdot 2 \cdot \ldots \cdot 9=2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, we need to remove the digits 5 and 7, and also need to make the odd power of two even. This means we need to remove the digit 2: obviously, we... | 986431 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the pop ensemble "Sunshine," everyone plays either the violin or the double bass. The average age of those who play the violin is 22 years, and those who play the double bass is 45 years. Igor changed his instrument and started playing the violin instead of the double bass. As a result, the average age of those w... | Solution. Let $x$ be the number of people who play the contrabass, excluding Igor, and $y$ be the number of people who play the violin (also excluding Igor).
From the condition, it follows that the total age of those who play the contrabass was initially equal to: $(x+1) \cdot 45$, and when Igor changed the instrument... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. 3.1. Find the number of four-digit numbers composed of the digits $1,2,3,4,5,6,7$ (each digit is used no more than once) that are divisible by 15. | Answer: 36.
Solution.
For a number to be divisible by 15, it must be divisible by 3 and by 5. Therefore, according to the divisibility rule for 5, the last digit can only be 5. The sum of the digits of the number must be divisible by 3. For this, the sum of the first three digits must give a remainder of 1 when divid... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. 5.1. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7. | Answer: 60718293.
## Solution.
We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7.
We see that the ... | 60718293 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. 6.1. The figure "lame rook" can move to an adjacent cell in one move. On a $20 \times 20$ board, crosses were placed in all cells that the "lame rook" can reach from the top-left corner in exactly 10 moves. How many cells were marked with a cross? | Answer: 36.
## Solution.
Let's set up a coordinate system so that the top-left cell has coordinates $(0,0)$. In one move, the sum of the rook's coordinates changes by 1. In 10 moves, the sum of the coordinates cannot exceed 10, and it will become an even number.
. It turned out that the distance between $a$ and $b$ is half the distance between $b$ and $c$. Find $M$.
, $ab+bc+ac=-84$ (2), $abc=M$ (3). According to the condition, $2(b-a)=c-b$ or $3b=c+2a$. Considering (1), $c... | 160 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. 8.1. How many increasing arithmetic progressions of 22 different natural numbers exist, in which all numbers are no greater than 1000? | Answer: 23312.
## Solution.
Consider the 22nd term of each such progression, it will have the form $a_{22}=a_{1}+21d$. This means that $a_{1}$ and $a_{22}$ will have the same remainders when divided by 21. Each pair of numbers not exceeding 1000, giving the same remainders when divided by 21, defines one of the requi... | 23312 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Indicate the smallest number ending in 37 with the sum of its digits equal to 37 and divisible by 37. | Answer: 99937.
Solution. The number is the smaller, the fewer digits are required to write it. Two digits - these are the last two digits, their sum is 10. Therefore, the sum of the other digits is 27, and there are at least three of them, since the largest digit is 9. Thus, the number 99937 satisfies three of the fou... | 99937 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Cyclists Petya, Vlad, and Timur simultaneously started a warm-up race on a circular cycling track. Their speeds are 27 km/h, 30 km/h, and 32 km/h, respectively. After what shortest time will they all be at the same point on the track again? (The length of the cycling track is 400 meters.) | Answer: 24 min.
Solution. Vlad rides 3 km/h faster than Petya. Therefore, he will overtake him by one lap (i.e., ride 400 m more) in $\frac{0.4}{3}$ h $=8$ min. Timur rides 2 km/h faster than Vlad. Therefore, he will overtake him by one lap in $\frac{0.4}{2}$ h $=12$ min. The least common multiple of the numbers 8 and... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Karlson counts 1000 buns baked by Fräulein Bock: «one, two, three, ..., nine hundred ninety eight, nine hundred ninety nine, thousand». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer: 2611.
Solution. One word will be required to pronounce 37 numbers: $1,2,3,4,5,6,7$, $8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200,300,400$, $500,600,700,800,900,1000$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required f... | 2611 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. A semicircle with diameter $A B$ and center at point $O$ is divided by points $C$ and $D$ into three parts such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$... | Answer: $20^{\circ}$.
Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of inte... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.6. At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote in "VK" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each of the brunettes ... | Solution: According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times but... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false |
5. Private Petrov took a bucket of unpeeled potatoes and cleaned them in 1 hour. In the process, $25 \%$ of the potatoes went into peels. How long did it take for him to have half a bucket of peeled potatoes? | Answer. In 40 minutes.
Solution. Since a quarter of the potatoes went into peels, Petrov received three quarters of a bucket of peeled potatoes in 1 hour. This means that a quarter of a bucket of peeled potatoes Petrov received in 20 minutes, and half a bucket - in 40 minutes.
## Grading Criteria.
- Correct solution... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The figure shown on the right is made of matches (the side of a small square is one match). The area of the entire shaded figure is 300 square centimeters. Find the total length of all the matches used.
. What did the total weight of the tomatoes become? | Solution. The weight of the dry matter in tomatoes is $1 \%$ of 1 ton - 10 kg. After drying, the weight of the dry matter does not change and constitutes $5 \%(100 \%-$ $(99 \%-4 \%))$ of the total weight of the tomatoes. In other words, the total weight is 200 kg.
Remark. Perhaps, the student can be given 2 - 3 point... | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.2. (7 points)
The master makes a whole number of parts in one hour, more than 5, and the apprentice makes 2 parts less. The master completes the order in a whole number of hours, and two apprentices together - one hour faster. How many parts does the order consist of? | Answer: 24.
Solution: Let $x$ be the number of parts in the order; $y$ (parts per hour) be the master's productivity $(y>5)$; $y-2$ (parts per hour) be the apprentice's productivity; $2 y$ - 4 (parts per hour) be the productivity of two apprentices.
Then $\frac{x}{y}$ is the time for the master to complete the order,... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.3. (7 points)
The base of the pyramid with vertex $P$ is a quadrilateral $A B C D$, where the sum of angles $A$ and $D$ is five times smaller than the sum of angles $B$ and $C$. Find the angle between the planes of the faces $P A B$ and $P C D$, if both are perpendicular to the base. | Answer: $60^{\circ}$.
Solution: Let the planes $A P B$ and $C P D$ intersect along the line $P X$ (point $X$ lies in the plane $A B C$, see the figure).
Since $(A P B) \perp(A B C)$ and $(C ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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