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Task 3. For a nine-digit number, the following operation is allowed: any digit of the number can be replaced by the last digit of the sum of the digits of this number. Is it possible to obtain the number 123456789 from the number 133355555 using such operations? | Answer: No, it cannot.
Solution. The sum of nine odd digits is an odd number, and its last digit will also be odd. Therefore, even digits cannot appear in such a process.
## Criteria
4 6. A complete and justified solution is provided.
In the absence of a correct solution, the highest applicable criterion from those... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,255 |
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 | 14,257 |
Problem 6. Given the quadratic trinomials $x^{2}+a x+b, x^{2}+c x+d$, and $x^{2}+e x+f$. It turns out that any two of them have a common root, but all three do not have a common root. Prove that exactly two of the following three inequalities are satisfied:
$$
\begin{aligned}
& \frac{a^{2}+c^{2}-e^{2}}{4}>b+d-f \\
& \... | Solution. Let the roots of the first quadratic trinomial be $x_{1}, x_{2}$; the second $x_{1}, x_{3}$; and the third $-x_{2}, x_{3}$. Then (by Vieta's theorem) we can say that
$$
\begin{gathered}
a=-\left(x_{1}+x_{2}\right) ; \quad c=-\left(x_{1}+x_{3}\right) ; \quad e=-\left(x_{2}+x_{3}\right) \\
b=x_{1} x_{2} ; \qua... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,258 |
1.1. For what least natural value of \( b \) does the equation
$$
x^{2}+b x+25=0
$$
have at least one root? | Answer: 10
Solution. The equation has at least one root if and only if the discriminant $D=b^{2}-4 \cdot 25=$ $=b^{2}-100$ is greater than or equal to 0. For positive $b$, this condition is equivalent to the condition $b \geqslant 10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,259 |
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 | 14,260 |
3.1. It is known that the area of the shaded region of the figure is $\frac{32}{\pi}$, and the radius of the smaller circle is 3 times smaller than the radius of the larger circle. What is the length of the smaller circle?
^{2}=9\pi R^{2}$. Therefore, the area of the shaded part is $S_{2}-S_{1}=8\pi R^{2}$. We have $8\pi R^... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,261 |
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 | 14,262 |
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 | 14,264 |
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?
. Is the number 2017 a member of this sequence? Justify your answer. | 2. Since $25, 41, 65$ are terms of an arithmetic progression, then $25=a_{1}+k d ; 41=a_{1}+n d ; 65=a_{1}+m d$, where $k, n, m$ are natural numbers. From these three equations, it follows that $16=(n-k) d, 24=(m-n) d$. From these two equations, we get:
$8=(m-2 n+k) d$. Since $2017=1+2016$, and $2016=252 \cdot 8=3 \cd... | Yes,itis | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,268 |
3. In an isosceles trapezoid, the bases are 9 cm and 21 cm, and the height is 8 cm. Find the radius of the circumscribed circle around the trapezoid. | 3. Let's introduce a rectangular coordinate system as follows: the x-axis is directed to the right along the larger side of the trapezoid, point $O$ is chosen at the midpoint of the larger side, and the y-axis is directed perpendicular to the x-axis (see the figure).
 x_{0}+(c-d)=0$, from which we obtain: $(b-a) x_{0}=d-c$. Since by condition $b>a, d>c$, then $b-a>0$ and $d-c>0$. Therefore, $x_{0}>0$. But a positive number c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,270 |
5. The game starts with the number 2. On a turn, it is allowed to add to the current number any natural number less than it. The player who reaches the number 1000 wins. Who will win with correct play? | 5. Let's call a winning position the one from which, by making a move, we win; and a losing position the one from which we lose. Obviously, all positions from which we can move to a winning position are losing (if we move to such a position, the opponent can move to a winning position on the next move and win). Convers... | thefirst | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,271 |
1. In the cells of a $3 \times 3$ square, letters are written (see the figure). Is it possible to arrange them so that any two letters that were initially a knight's move apart end up in cells that are a king's move apart? | Answer: yes.
| $a$ | $b$ | $c$ |
| :---: | :---: | :---: |
| $d$ | $e$ | $f$ |
| $g$ | $h$ | $i$ |
| $a$ | $h$ | $c$ |
| :---: | :---: | :---: |
| $f$ | $e$ | $d$ |
| $g$ | $b$ | $i$ |
Solution. See fig. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,273 |
2. The quadratic trinomial $a x^{2}+2 b x+c$ has two distinct roots, while the quadratic trinomial $a^{2} x^{2}+$ $2 b^{2} x+c^{2}$ has no roots. Prove that the roots of the first trinomial have different signs. | Solution. Let $x_{1}, x_{2}, D_{1}$ be the roots and the simplified discriminant of the first quadratic, and $D_{2}$ the simplified discriminant of the second quadratic.
By Vieta's theorem, $x_{1} x_{2}=c / a$, so it is sufficient to prove the inequality $a c < 0$. The second quadratic has no roots, so $D_{2}=b^{4}-a^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,274 |
3. A circle passes through vertex $\mathrm{B}$ of triangle $\mathrm{ABC}$, intersects sides $\mathrm{AB}$ and $\mathrm{BC}$ at points X and Y respectively, and is tangent to side AC at its midpoint M. It is known that AX $=$ XM. Prove that CY $=$ YM. | Solution. From the isosceles triangle AXM: $\square \mathrm{XAM}=\square \mathrm{AMX} . \square \mathrm{AMX}=\square \mathrm{XBM}$ (the angle between a chord and a tangent is equal to the inscribed angle subtended by this chord). Then in triangle $\mathrm{AMB} \square \mathrm{BAM}=\square \mathrm{ABM}$, so $\mathrm{MC}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,275 |
4. Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the water. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have caught on the three days - Monday... | Answer: 50.
Solution. Estimation. On Tuesday and Thursday, the old man caught no more fish than on Monday and Wednesday, so over the specified three days, he caught no less than half of 100, which is no less than 50 fish.
Example. If the old man caught 25 fish each of the first four days and caught nothing on Friday,... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,276 |
5. Given $n+1$ pairwise distinct natural numbers, all less than $2n$ ($n>1$). Prove that among them, there exist three numbers such that the sum of two of them equals the third. | Solution. Consider the largest number among the given ones. Let it be x<2n. We will show that there are two numbers with a sum of x. Suppose there are no such numbers.
If x is odd. Then from each pair ( $1, x-1),(2, x-2), \ldots,((x-1) / 2,(x-1) / 2+1)$ no more than one number is taken, and in total no more than ( $x-... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,277 |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | 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 | 14,280 |
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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,281 |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
 in the squares so that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,285 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to the line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on the line $A L$. Carefully finding the intersection point of thes... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,286 |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,288 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,289 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,290 |
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 | 14,291 |
Problem 9.7. Through points $A(0 ; 14)$ and $B(0 ; 4)$, two parallel lines are drawn. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,293 |
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 | 14,294 |
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 $B C=9$, $C D=30$.
. Petya wrote down the abscissas of all six points of intersection and multiplied them. Prove that the result does not depend on the choice of the number $k$. | 10.1. A line parallel to the line $y=k x$ has the equation $y=k x+b$ The abscissas of its intersection points with the hyperbola are both roots of the equation $\frac{k}{x}=k x+b, \quad$ equivalent to the equation $k x^{2}+b x-k=0$. The product of the roots of this equation is -1. Multiplying three such products, we ge... | -1 | Algebra | proof | Yes | Yes | olympiads | false | 14,296 |
10.2. Two circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Tangents are drawn from point $M$ to these circles, and the segments from $M$ to the points of tangency are equal. Prove that point $M$ lies on the line $A B$. | 10.2. Let the points of tangency be denoted as $K$ and $L$, as shown in the figure. By the problem's condition, $M K = M L$. Suppose that point $M$ does not lie on the line $A B$. Draw the line $A M$. Let it intersect the circle $\omega_{1}$ again at point $C$, and the circle $\omega_{2}$ again at point $D$. By the pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,297 |
10.3. All points of a circle are colored in two colors. Prove that there are infinitely many isosceles triangles inscribed in this circle, the vertices of which are colored in the same color. | 10.3. Consider five points on a circle, located at the vertices of an inscribed regular pentagon. Since the points can be colored in two colors, three of them are colored in one color. But any three vertices of a regular pentagon form the vertices of an isosceles triangle. And since there are infinitely many regular pe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,298 |
10.4. Two athletes are swimming along the pool lane at constant but different speeds. Meeting for the first time at point $C$ at some moment, they turn around, each swims to their end of the lane, immediately turn around and swim back, meet for the second time at point $D$, and so on. Where will their 20th meeting occu... | # 10.4. Answer: At point $D$.
Let's use a graphical representation. Along the x-axis, we plot time, and along the y-axis, we plot distance. On the y-axis, we mark the edges of the pool as points $A$ and $B$, and the meeting points as $C$ and $D$. The movement of one swimmer is represented by a solid line, and the othe... | D | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,299 |
10.5. In a round-robin volleyball tournament, $n$ teams participated. Prove that, based on the results of the tournament, it is possible to compile a list of teams such that each team has won against the team immediately following it in the list. | 10.5. We will prove the statement by induction on the number of teams $n$. For $n=2$, the statement is true. Suppose it is true for $n=k$ and we will prove it for $n=k+1$. According to the inductive hypothesis, $n$ teams can be arranged in a list that satisfies the condition of the problem: $B_{1}, B_{2}, \ldots, B_{n}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,300 |
1. Robinson Crusoe replenishes his supply of drinking water from a source every second day, gathers fruits every third day, and goes hunting every fifth day. Today, September 13, is a tough day for Robinson: he has to do all three of these tasks. When will Robinson have his next tough day? | Answer: October 13.
## Solution.
We will count how many days have passed since the "heavy" day. If this number is divisible by 2, then Robinson needs to replenish his water supply. If it is divisible by 3, then he needs to replenish his fruit supply. If it is divisible by 5, then he needs to go hunting. If he does al... | October13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,301 |
2. The plane departed from Perm on September 28 at noon and arrived in Kirov at 11:00 AM (all departure and arrival times mentioned in the problem are local). At 7:00 PM the same day, the plane departed from Kirov to Yakutsk and arrived there at 7:00 AM. Three hours later, it departed from Yakutsk to Perm and returned ... | # Solution.
The plane was absent in Perm for 23 hours. Out of these, it was stationed in Kirov for 8 hours (from 11 to 19) and for 3 hours in Yakutsk. In total, out of these 23 hours, it was stationed $8+3=11$ (hours), i.e., the plane was in the air for $23-11=12$ (hours).
## Grading Criteria.
- Correct solution -7 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,302 |
3. On a glade, 25 gnomes gathered. It is known that 1) every gnome who put on a hat also put on shoes; 2) 12 gnomes came without a hat; 3) 5 gnomes came barefoot. Which gnomes are more and by how many: those who came in shoes but without a hat, or those who put on a hat? | Answer. There are 6 more gnomes who put on a cap.
## Solution.
From condition 2, it follows that $25-12=13$ gnomes came in a cap.
From condition 1, we get that exactly 13 gnomes came both in a cap and in shoes.
From condition 3, it follows that a total of $25-5=20$ gnomes came in shoes.
Thus, $20-13=7$ gnomes came... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,303 |
4. The difference of the squares of two numbers is 6, and if each of these numbers is decreased by 2, then the difference of their squares becomes equal to 18. What is the sum of these numbers? | Answer: -2.
## Solution.
Given:
\[
\begin{aligned}
& a^{2}-b^{2}=6 \\
& (a-2)^{2}-(b-2)^{2}=18
\end{aligned}
\]
There are different ways to proceed.
## Method 1.
\((a-2)^{2}-(b-2)^{2}=a^{2}-4a+4-b^{2}+4b-4=a^{2}-b^{2}-4(a-b)\). Since from the first condition \(a^{2}-b^{2}=6\), we get \(6-4(a-b)=18\). Hence, \(a-b... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,304 |
5. In an isosceles triangle $ABC$ with base $AB$, a point $D$ is chosen on side $CB$ such that $CD = AC - AB$. Point $M$ is the midpoint of $AD$. Prove that angle $BMC$ is obtuse. | # Solution.
Since $C D=A C-A B=B C-A B$, we obtain that $D B=A B$, which means that triangle $A B D$ is isosceles. Then its median $B M$ is also an altitude, i.e., angle $B M D$ is a right angle. Therefore, $\angle B M C=\angle B M D+\angle D M C=90^{\circ}+\angle D M C>90^{\circ}$ is obtuse.
, from which an equilateral triangle needs to be formed. Can this be done if a) p $=100$; b) p = 99? (The sticks cannot be broken, all p sticks must be used.) | 7.5. There are n sticks of lengths 1, 2, 3, ..., n (cm), from which an equilateral triangle needs to be formed. Can this be done if a) n = 100; b) n = 99? (The sticks cannot be broken, and all sticks must be used.)
Answer: a) no; b) yes. Hint: Clearly, the total length of all sticks must be divisible by 3 in order to ... | )no;b)yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,311 |
# Task 11.1
Does there exist a function $f(x)$, defined for all real numbers, such that $f(\sin x) + f(\cos x) = \sin x$?
Points: 7 | Answer:
Does not exist
Solution
Suppose such a function exists.
Then $f(\sin 0) + f(\cos 0) = \sin 0$, that is, $f(0) + f(1) = 0$.
But $f(\sin \pi / 2) + f(\cos \pi / 2) = \sin \pi / 2$, that is, $f(0) + f(1) = 1$. Contradiction. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,312 |
# Problem 11.2
In each cell of a $2021 \cdot 2021$ square table, one of the numbers 1 or -1 is written arbitrarily. Below each column, the product of all numbers in that column is written. To the right of each row, the product of all numbers in that row is written. Prove that the sum of the 4042 written products canno... | Solution
Let's find the product of all 2021 numbers written under each column and all 2021 numbers written to the right of the rows. Since in this product each of the numbers in the square table appears twice, the product of these 4042 products, each of which has 2021 factors, will be positive, i.e., equal to 1. Since... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,313 |
# Task 11.3
In the vertices of a heptadecagon (17-sided polygon), different integers were written (one in each vertex). Then all the numbers were simultaneously replaced with new ones: each was replaced by the difference of the two following it in a clockwise direction (subtracting the next from the adjacent one). Cou... | # Answer:
It could not.
Solution. Let the original numbers in the vertices of the heptadecagon be: $a_{1}, a_{2}$, ..., $a_{17}$ (numbered clockwise). Then, after the specified replacement, the numbers in the vertices will be: $a_{2}-a_{3}, a_{3}-a_{4}, \ldots, a_{16}-a_{17}, a_{17}-a_{1}, a_{1}-a_{2}$.
Notice that ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,314 |
# Task 11.4
Two circles intersect at points $A$ and $B$. It turns out that the radii $O A$ and $O B$ of the first circle are tangents to the second circle. A line is drawn through point $A$ and intersects the circles again at points $M$ and $N$. Prove that $M B \perp N B$. Number of points 7
# | # Solution
Let $\angle B M N=\alpha, \angle B N M=\beta$ (see fig. a, b).
## First Method.
See fig. a. Note that $\angle O A B=\angle O B A=\alpha$ (by the theorem on the angle between a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,315 |
# Problem 11.5
Prove that for any natural $n$
$$
\frac{1}{3+1^{2}}+\frac{1}{3+2^{2}}+\frac{1}{3+3^{2}}+\ldots+\frac{1}{3+n^{2}}<\frac{4}{5}
$$
## Number of points 7
# | # Solution
To prove the inequality, we will strengthen it:
$\frac{1}{3+1^{2}}+\frac{1}{3+2^{2}}+\frac{1}{3+3^{2}}+\ldots+\frac{1}{3+n^{2}} \leq \frac{4}{5}-\frac{2}{2 n+1}$
Now we can apply the method of mathematical induction.
Let's check for $n=1$
$\frac{1}{4} \leq \frac{4}{5}-\frac{2}{3}$ - false
Let's check f... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,316 |
9.4. Let the mass of the diamond be x, its cost is kx ${}^{2}$, where k is the proportionality coefficient. Let a piece with mass p break off from the diamond, then the cost of the broken part is kp ${}^{2}$, and the cost of the remaining part is k(x-p) ${}^{2}$. According to the problem:
$\kappa(x-p)^{2}=(1-0.36) \ka... | Answer: one fifth broke off | \frac{p}{x}=0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,317 |
1.1. At the stadium, through the checkpoints numbered 1, 2, 3, and 4, an equal number of fans entered. They then passed through several more checkpoints, moving along the arrows, and at the fork, the fans were evenly distributed. At checkpoint 7, 15 people passed the control.
Solution. If each compartment in the carriage has 6 seats, then compartment number 5 contains seats 25 to 30. Your seat can be any of these six. In the current carriage, where each compartment has 4 seats, seats 25, 26, 27, and 28 will be in compartment number 7, and seats 29 and 30 will be i... | 7,8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,319 |
3.1. A gardener grows white and red flowers: peonies and roses. In his garden, there are 301 stems, among which 135 are roses and 182 are red flowers. What is the smallest number of red peonies that can grow in the gardener's garden? | Answer: 47
Solution. Estimation: The number of red peonies is the difference between the number of red flowers and the number of red roses. The number of red roses is no more than the total number of roses, i.e., no more than 135. Therefore, the number of red peonies is no less than $182-135=47$.
Let's provide an exa... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,320 |
4.1. Mom baked four raisin buns for breakfast for her two sons. $V$ In the first three buns, she put 7, 7, 23 raisins, and some more in the fourth. It turned out that the boys ate an equal number of raisins and did not divide any bun into parts. How many raisins could Mom have put in the fourth bun? List all the option... | Answer: $9,23,37$ (All answers)
Solution. If the boys ate the buns equally, then the fourth bun can only have 23 raisins. If one of them ate three buns, and the other one, then the one who ate one bun ate the one with the most raisins, which is either the one with 23 raisins or the fourth one. In the first case, the f... | 9,23,37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,321 |
6.1. In February of a non-leap year, Kirill and Vova decided to eat ice cream according to certain rules. If the date was even and the day of the week was Wednesday or Thursday, they each ate seven portions of ice cream. If the day of the week was Monday or Tuesday and the date was odd, they each ate three portions of ... | # Answer: 110
Solution. February has 28 days, which is exactly 4 Mondays, 4 Tuesdays, and so on. Moreover, if Monday falls on an even date in one week, the following Monday will fall on an odd date, and vice versa. This means that the same day of the week falls on an even date twice and an odd date twice, regardless o... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,323 |
9.1. On a circle, 1000 points are marked, each colored in one of $k$ colors. It turns out that among any five pairwise intersecting segments, whose endpoints are 10 different marked points, there will be at least three segments, each of which has endpoints of different colors. For what smallest $k$ is this possible?
(... | Answer. For $k=143$.
Solution. Suppose there are 8 points of one color (say, red) 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 | 14,326 |
9.2. Let $n$ be a natural number. An integer $a>2$ is called $n$-decomposable if $a^{n}-2^{n}$ is divisible by each number of the form $a^{d}+2^{d}$, where $d$ is a natural divisor of $n$, different from $n$. Find all composite natural numbers $n$ for which there exists an $n$-decomposable number.
(S. Kudrya) | Answer. $n=2^{m}$ for $m>1$.
Solution. For $n=2^{m}$, any natural number $a>2$ is $n$-decomposable due to the equality
$$
a^{2^{m}}-2^{2^{m}}=(a-2)\left(a^{1}+2^{1}\right)\left(a^{2}+2^{2}\right) \ldots\left(a^{2^{m-1}}+2^{2^{m-1}}\right)
$$
Indeed, among the factors on the right side, there are all numbers of the f... | 2^{}for>1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,327 |
9.3. On a line, $n+1$ distinct segments are marked; one of the points on the line belongs to all these segments. Prove that among the marked segments, one can choose distinct segments $I$ and $J$, intersecting in a segment of length at least $\frac{n-1}{n} d$, where $d$ is the length of segment $I$.
(I. Bogdanov, V. U... | The first solution. Let's introduce coordinates on our line. Let the given segments be $I_{0}=\left[a_{0} ; b_{0}\right], I_{1}=\left[a_{1} ; b_{1}\right], \ldots$, $I_{n}=\left[a_{n} ; b_{n}\right] ;$ we choose the numbering of the segments such that $a_{0} \leqslant a_{1} \leqslant \ldots \leqslant a_{n}$. If $b_{k} ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,328 |
9.1. Given two quadratic trinomials $f(x)$ and $g(x)$, it is known that the trinomials $f(x)$, $g(x)$, and $f(x)+g(x)$ each have two roots. It turns out that the difference between the roots of the trinomial $f(x)$ is equal to the difference between the roots of the trinomial $g(x)$. Prove that the difference between t... | The first solution. Note that the difference of the roots of a reduced quadratic trinomial $x^{2}+b x+c$ is equal to the square root of its discriminant, that is, $\sqrt{b^{2}-4 c}$.
Let the two given trinomials be $f(x)=x^{2}+b_{1} x+c_{1}$ and $g(x)=x^{2}+b_{2} x+c_{2}$. According to the condition, they have a commo... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,329 |
9.2. Initially, 250 letters are written in a row - 125 letters A and 125 letters B in some order. Then, in one operation, you can take any segment of consecutive letters, where there are an equal number of letters A and B, and reverse the order of the letters in this segment, swapping all A's to B's and all B's to A's ... | Answer: No.
First solution. Let's number the positions in the string from left to right with numbers from 1 to 250. Suppose in the initial string, $x$ letters A are in odd positions (i.e., positions with odd numbers). We will show that this quantity does not change in the resulting strings.
Indeed, suppose for some o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,330 |
9.3. Each natural number greater than 1000 has been colored either red or blue. It turned out that the product of any two different red numbers is blue. Can it happen that no two blue numbers differ by 1? (S. Berlov) | Answer: It cannot.
First solution. Suppose it is possible.
Lemma. If the number $n$ is blue, then $n^{2}$ is red.
Proof. Since $n$ is blue, the numbers $n-1$ and $n+1$ are red, otherwise there would be two blue numbers differing by 1. Therefore, the number $n^{2}-1=(n-1)(n+1)$ is blue. Hence, $n^{2}$ is red.
Obviou... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,331 |
9.4. Point $X$ lies strictly inside the circumcircle of triangle $A B C$. Let $I_{B}$ and $I_{C}$ be the centers of the excircles of this triangle, touching sides $A C$ and $A B$ respectively. Prove that $X I_{B} \cdot X I_{C} > X B \cdot X C$.
(D. Brodsky) | Solution. Let $\Gamma$ be the circle with diameter $I_{B} I_{C}$. Since $C I_{C} \perp C I_{B}$ and $B I_{C} \perp B I_{B}$, points $B$ and $C$ lie on $\Gamma$ (see Fig. 1).
Let $I$ be the incenter of triangle $A B C$. If point $X$ lies inside angle $B I C$, then angles $X B I_{C}$ and $X C I_{B}$ are obtuse, so $X I_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,332 |
9.1. On the board, the numbers $\sqrt{2}$ and $\sqrt{5}$ are written. It is allowed to add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that it is possible to write the number 1 on the board. | Solution: The simplest way is to provide a sequence of numbers that will lead to the number 1. For example, the following sequence works:
$$
\begin{gathered}
\sqrt{2}+\sqrt{5}, \quad 2 \sqrt{2}+\sqrt{5}, \quad 3 \sqrt{2}+\sqrt{5}, \quad \sqrt{2}-\sqrt{5}, \quad 2 \sqrt{2}-\sqrt{5}, \quad 3 \sqrt{2}-\sqrt{5} \\
(\sqrt{... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 14,333 |
9.2. Do there exist
a) such different natural numbers $a, b, c$, and $d$ that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$;
b) such different natural numbers $a, b, c, d$, and $e$ that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}+\frac{1}{e}$? Justify your answer. | Solution: a) For example, $a=12, b=3, c=6$ and $d=4\left(\frac{1}{12}+\frac{1}{3}=\frac{5}{12}=\frac{1}{6}+\frac{1}{4}\right)$ or $a=56, b=7, c=28$ and $d=8\left(\frac{1}{56}+\frac{1}{7}=\frac{9}{56}=\frac{1}{28}+\frac{1}{8}\right)$.
b) For example, $a=24, b=3, c=12, d=8, e=6\left(\frac{1}{24}+\frac{1}{3}=\frac{3}{8}=... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,334 |
9.3. On a straight road between a dog in a kennel and a cat, a kilogram of sausages was placed, and the animals simultaneously rushed towards them. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and finished eating at the same time. It is known tha... | # Solution:
Method 1. While the dog was eating its half of the sausages, the cat ate half as much, which is a quarter of all the sausages. The remaining sausages (in the amount of $1 / 2 - 1 / 4 = 1 / 4$) the cat ate while the dog had not yet reached the sausages. During this time, the cat could have run a quarter of ... | )thesausageswereplacedclosertothedog;b)thecat'pathis1.4timeslongerthanthedog'path | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,335 |
9.4. On the side $AC$ of triangle $ABC$, a point $D$ is taken. Prove that the incircles of triangles $ABD$ and $CBD$ cannot divide the segment $BD$ into 3 equal parts. | Solution: Assume the opposite. Suppose such a configuration is possible, and one circle touches segments $A B A D$ and $B D$ at points $K, E$ and $F$ respectively, while the second circle touches segments $C B C D$ and $B D$ at points $G, H$ and $I$ respectively. Without loss of generality, assume that point $I$ lies b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,336 |
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 | 14,337 |
9.6. Let \(a\) and \(b\) be positive numbers. Find the minimum value of the fraction \(\frac{(a+b)(a+2)(b+2)}{16ab}\). Justify your answer. | Solution: By the inequality between the arithmetic mean and the geometric mean of two numbers, the following three inequalities hold:
$$
a+b \geqslant 2 \sqrt{a b}, \quad a+2 \geqslant 2 \sqrt{2 a}, \quad b+2 \geqslant 2 \sqrt{b}
$$
Multiplying the left and right sides of these three inequalities, we get
$$
(a+b)(a+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,338 |
1. Olga traveled from the city to her cottage by taxi with a transfer in the village. From the city to the village, she took "Taxi-A" and paid 6 rubles per km for the ride. From the village to the cottage, she took "Taxi-B" and paid 4 rubles per km for the ride. The next day, Olga traveled back, and both taxi companies... | Solution. No, it couldn't. The difference in the cost of a trip on different days for "Taxi-A" is 3 rubles per km, and for "Taxi-B" it is 6 rubles per km. Both differences are divisible by 3, so regardless of the length of the route, the change in cost must be a number divisible by 3. Therefore, it is impossible to pay... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,339 |
2. Prove the inequality for any real $a, b$
$$
a^{2}+4 b^{2}+4 b-4 a+5 \geq 0
$$
For which $a, b$ does the equality hold? | Solution.
$$
\begin{gathered}
a^{2}+4 b^{2}+4 b-4 a+5 \geq 0 \Leftrightarrow \\
a^{2}-4 a+4+4 b^{2}+4 b+1 \geq 0 \Leftrightarrow \\
(a-2)^{2}+(2 b+1)^{2} \geq 0
\end{gathered}
$$
The last inequality is obtained as the sum of two valid inequalities $(a-2)^{2} \geq 0$, $(2 b+1)^{2} \geq 0$.
Equality is achieved when $... | =2,b=-\frac{1}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 14,340 |
3. On a circle, 2021 points are marked and these points are connected by segments to form a convex inscribed 2021-gon, which is divided into triangles by its diagonals. In this process, no two diagonals intersect at interior points (the only common points of different diagonals are the vertices). Prove that among the r... | Solution. We are given an inscribed polygon divided into triangles. All the resulting triangles are inscribed in the same circle. The center of the circle can lie inside only one triangle, or on a diagonal, or outside the polygon. As is known, a triangle is acute if and only if the center of the circumscribed circle li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,341 |
4. Given an isosceles trapezoid. Two circles are drawn: the first circle touches the lateral sides at the vertices of one base of the trapezoid, the second circle touches the lateral sides at the vertices of the other base of the trapezoid. A diagonal of the trapezoid intersects each of the circles, respectively, formi... | Solution. There are two possible arrangements of the circles. The first arrangement. Let the first circle cut off a chord $BK$ from the diagonal $BD$, and the second circle cut off a chord $MD$ from the diagonal $BD$. The chords $BK$ and $MD$ do not intersect. We need to prove that $BK = MD$. Let $BK = x, KM = y, MD = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,342 |
5. The equation $x^{3}+p x^{2}+q x+r=0$, where $p, q, r$ are integers, has three distinct integer roots $x_{1}, x_{2}, x_{3}$. Prove that the numbers $q$ and $r$ have no common divisors if and only if the numbers in each of the pairs $\left(x_{1}, x_{2}\right),\left(x_{2}, x_{3}\right),\left(x_{3}, x_{1}\right)$ have n... | Solution. Since the equation has three distinct roots $x_{1}, x_{2}, x_{3}$, the left side can be factored as
$$
x^{3}+p x^{2}+q x+r=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)
$$
Expand the brackets on the right
$$
x^{3}+p x^{2}+q x+r=x^{3}-\left(x_{1}+x_{2}+x_{3}\right) x^{2}+\left(x_{1} x_{2}+x_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,343 |
1. Restore an example of dividing two numbers if it is known that the quotient is five times smaller than the dividend and seven times larger than the divisor.
# | # Solution
Since the quotient is five times smaller than the dividend, the divisor is 5. Since the quotient is seven times larger than the divisor, it is equal to $5 \cdot 7=35$.
Answer $175: 5=35$. | 175:5=35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,345 |
2. We will call a number beautiful if it reads the same from right to left as it does from left to right. For example, the number 12321 is beautiful. Find all beautiful five-digit numbers that use only the digits 1 and 0.
# | # Solution
Note that the leading digit cannot be 0. Therefore, the first and last positions must definitely be the digit 1. It is not difficult to calculate that the problem has four solutions - two solutions with zero in the middle and two solutions with one in the middle (students are not required to prove this).
A... | 10001,10101,11011,11111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,346 |
3. In one country, knights and liars live. Knights always tell the truth, while liars always lie. Knights always carry a sword with them, while liars do not. Two knights and two liars gathered together and looked at each other. Who among them could have said the following phrases: 1) "All of us are knights." 2) "Among ... | # Solution
1) The knight could not say this, as it is false, so this statement could only have been made by any of the liars. 2) This could be said by any knight, as it is true for him. It could also be said by any liar, as it is false for him. 3) No one could have said this statement, as it is false for any knight an... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,347 |
4. A third of the kids stayed in the summer camp, while the rest went on an excursion. Those who stayed in the camp ate a quarter of the prepared food for lunch, and those who returned from the excursion in the evening received portions one and a half times larger than those given for lunch. How much food was left for ... | # Solution
Two-thirds of the children went on an excursion, which is twice as many as those who remained in the camp. If they had received the same portions as they were given for lunch, they would have eaten twice as much as was eaten for lunch, that is, half of the prepared food. Since in reality their portions were... | nothingwasleft | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,348 |
2. Find all real solutions of the equation $x^{2}+2 x \sin (x y)+1=0$. | Solution If we consider the given equation as a quadratic equation in terms of $\mathrm{x}$, then its discriminant will be equal to $4 \sin ^{2}(\mathrm{X}) - 1$. The discriminant must be non-negative, so $\sin ^{2}(\mathrm{X}) \geq 1$, i.e., $\sin (\mathrm{X})= \pm 1$. Let $\sin (\mathrm{x})=1$. Then we get the equati... | \1,-\pi/2+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,350 |
3. Let's call white numbers those of the form $\sqrt{a+b \sqrt{2}}$, where $\mathbf{a}$ and $\mathbf{b}$ are integers, not equal to zero. Similarly, let's call black numbers those of the form $\sqrt{c+d \sqrt{7}}$, where $c$ and $d$ are integers, not equal to zero. Can a black number equal the sum of several white numb... | Solution We will look for a black number equal to the sum of two white ones. Let's take conjugate white numbers:
$\sqrt{A+B \sqrt{2}}+\sqrt{A-B \sqrt{2}}=\sqrt{C+D \sqrt{7}}$
Squaring both sides of the equation, we get:
$$
2 A+2 \sqrt{A^{2}-2 B^{2}}=C+D \sqrt{7}
$$
Thus, it is sufficient to choose such numbers $A$ ... | \sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}=\sqrt{6+2\sqrt{7}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,351 |
# 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 | 14,352 |
1. All cells of a $2 \times 24$ board are filled with ones and zeros. Each one has exactly one neighbor that is a one, and each zero has exactly two neighbors that are zeros. Provide an example of such an arrangement (neighbors are considered to be signs in cells sharing a side). | Solution. Note that if the columns on the board alternate between pairs of crosses and $2 \times 2$ squares with zeros, then all conditions are met. Repeat the $2 \times 3$ group 8 times: in the left column, a pair of crosses, the rest - zeros.
Criteria. Full solution - 7 points. If the example is fully drawn and ther... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,353 |
2. Matvey decided to start eating properly and every day he ate one bun less and one pear more than the previous day. In total, during the time of proper nutrition, he ate 264 buns and 187 pears. How many days was Matvey on a proper diet? | Answer: 11 days.
Solution: If we "reverse" the sequence of the number of buns, while leaving the pears unchanged, the total number of buns and pears will not change, and the difference between the number of buns and pears eaten each day will become constant. Since $264-187=77=7 \cdot 11$, the correct diet lasted eithe... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,354 |
3. Once, 20 animals gathered in the jungle, 11 of them were tigers, who always tell the truth, and the remaining 9 were monkeys, who always lie. Around a stump, $n \geq 3$ animals sat, each of them stating, "Exactly one of my two neighbors is a monkey." For which $n$ is this possible? (It is necessary to find all possi... | Answer. $n=3,4,5,6,7,8,9,12,15$.
Solution. First, there is a case when only monkeys sit behind the stump: in this case, all $n \leq 9$ are suitable.
Now let there be at least one tiger behind the stump. Let this tiger be $A$, its neighbors be tiger $B$ and monkey $C$. Then on the other side of $B$ sits a monkey, and ... | 3,4,5,6,7,8,9,12,15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,355 |
4. Dima and Vasya are playing the following game. Vasya has a water tap and three pots with capacities of 3, 5, and 7 liters. Vasya needs to measure out 1 liter of water using these pots. However, there is a small catch - after every two transfers Vasya makes (filling a pot counts as a transfer, pouring out any amount ... | Answer: Yes, he will.
Solution. Let's denote Vasya's moves with the letter V, and Dima's moves with the letter D. See the diagram below.
Criteria. Full solution - 7 points. Reasoning of the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,356 |
5. There is one slip with the digit 5, two slips with the digit 3, and a hundred slips with the digit 2.
a) In how many ways can a ten-digit number be formed from them such that the product of the digits ends in 0?
b) All such numbers were written in ascending order. What number is in the 455th position? | Answer. a) 460 ways; b) 5322222322.
Solution. a) Among the selected cards, there will definitely be cards with a two, so the product of the digits is even. It ends in 0 if and only if the card with a five is also among the selected ones. The number of cards with a three can be 0, 1, or 2. | 460;5322222322 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,357 |
11.1. Does there exist an eight-digit number without zero digits, which when divided by its first digit gives a remainder of 1, when divided by its second digit gives a remainder of 2, ..., and when divided by its eighth digit gives a remainder of 8? | Answer. Does not exist.
Solution. Suppose such a number exists. Since the remainders when dividing by 8 digits are different numbers from 1 to 8, the digits of the eight-digit number are different.
Furthermore, if a number gives a remainder of 8 when divided by a digit, then this digit is 9. Therefore, the last digit... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,358 |
11.2. The equation $(x+a)(x+b)=9$ has a root $a+b$. Prove that $a b \leqslant 1$. | Solution. Substituting the root $x=a+b$ into the equation, we obtain the equality $(a+b+a)(a+b+b)=(2a+b)(2b+a)=9$. Then $9=5ab+2(a^2+b^2) \geqslant 5ab+2 \cdot 2ab=9ab$, from which $ab \leqslant 1$. (We used the inequality $a^2+b^2 \geqslant 2ab$, which is equivalent to $(a-b)^2 \geqslant 0$.)
Comment. The solution ap... | \leqslant1 | Algebra | proof | Yes | Yes | olympiads | false | 14,359 |
11.3. Workers were laying a floor of size $n \times n(10<n<20)$ using two types of tiles: $2 \times 2$ and $5 \times 1$. It turned out that they managed to completely cover the floor such that the same number of tiles of each type was used. For which $n$ could this have happened? (Cutting tiles or overlapping them is n... | Answer: $12, 15, 18$.
Solution. Let the workers use $x$ tiles of each type. Then the area occupied by the tiles is $4x + 5x = 9x = n^2$. Therefore, $n^2$ must be divisible by 9, which means $n$ must be divisible by 3. Thus, the possible values for $n$ are $n=12, n=15$, and $n=18$.
We will show how to tile the require... | 12,15,18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,360 |
11.4. The midpoint of the edge $S A$ of the triangular pyramid $S A B C$ is equidistant from all vertices of the pyramid. Let $S H$ be the height of the pyramid. Prove that $B A^{2}+B H^{2}=C A^{2}+C H^{2}$. | The first solution. Let $M$ be the midpoint of edge $S A$. Since $M A = M S = M C$, in triangle $A S C$, the median $M C$ is twice the length of side $A S$ to which it is drawn. Therefore, triangle $A S C$ is a right triangle with hypotenuse $A S$. Similarly, triangle $A S B$ is a right triangle with hypotenuse $A S$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,361 |
11.5. Do there exist natural numbers $a$ and $b$, greater than a thousand, such that for any $c$, which is a perfect square, the three numbers $a, b$, and $c$ cannot be the lengths of the sides of a triangle? | Answer. They exist.
Solution. Let, for example, $a=10000^{2}+10000, b=1001$. Suppose there exists $c=d^{2}$ such that the numbers $a, b$, and $c$ are the lengths of the sides of some triangle. Then the triangle inequalities $a+b>c, b+c>a$, $a+c>b$ must hold. Consider the first two of these: $a+b>c$ and $c>a-b$. Note t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,362 |
# 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 | 14,363 |
# 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 | 14,364 |
# 3. Option 1.
Along a road, 10 lampposts were placed at equal distances, and the distance between the outermost posts was $k$ meters. Along another road, 100 lampposts were placed at the same distances, and the distance between the outermost posts was $m$ meters. Find the ratio $m: k$. | Answer: 11.
Solution: Let the distance between adjacent posts be $x$ meters. Then, in the first case, the distance between the outermost posts is $(10-1) x=9 x=k$ meters. And in the second case, $(100-1) x=99 x=m$ meters. Therefore, the desired ratio is $m: k=(99 x):(9 x)=11$. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,365 |
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