problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
91. Motorboat (I). The smugglers' motorboat has three times the speed of the patrol ship, which is located at a distance of half the way from the boat to the point on the coast that the boat wants to reach. The captain of the boat decides to sail to the goal along two sides of a square. What part of this path will be d... | 91. At the moment in question in the problem, the patrol ship is at point $S$, and the motorboat is at point $M$ (Fig. 165). The motorboat moves along the sides of the square $M N, N W$. The patrol ship will not catch up with the boat on the segment $M N$, as its speed is too low for that. Let $a$ be the length of the ... | \frac{3}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,419 |
92. Motorboat (II). In the previous problem, the captain of the boat decided to sail in such a way as to change course by $90^{\circ}$ once along the way. What route should he choose to definitely avoid the patrol ship and reach his destination as quickly as possible?
## $\Gamma$ L A B A VI
## MATHEMATICAL ADVENTURES... | 92. At the initial moment, the boat is at point $M$, the patrol ship is at point $S$; the boat must reach the shore at point $W$ (Fig. 166). Points $M, S, W$ lie on the same line $M S \equiv S W$. From the text of the problem, it follows that the boat must sail along the sides of the right angle $M N W$, inscribed in a... | \frac{2\sqrt{2}+1}{3}MW | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,420 |
93. A Remarkable Number. Dr. Sharadek has decided to radically change mathematical notation. He considers it a great misunderstanding that there is a number known to children starting school, which, along with its usual notation, has another notation that these children only learn about a few years later, and yet a thi... | 93. Such an amazing number is the number one, which can be written in three ways:
$$
\text { 1, } 100 \% \text { and } 57^{\circ} 17^{\prime} 44^{\prime \prime}, 8 \ldots \text { (1 radian). }
$$ | 1,100,5717^{\}44^{\\},8\ldots | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,421 |
95. Word game. Dr. Sylvester Sharadek announced that he can always guess the word you think of if he is allowed to ask 20 questions, to which the answers should be only "yes" or "no," and if the word is in the dictionary. Do you think he is boasting? | 95. There are no dictionaries containing more than a million words. For example, if a dictionary has 500 pages, we will open it in the middle and find the last word on page 250, let's say, "narcissus." The first question should be: is the word we are guessing located after the word "narcissus" in the dictionary? If the... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,423 |
97. A Strange Society. One person recounted that once he found himself in a society consisting (including himself) of twelve people, and
a) each of them was not acquainted with six other people, but knew all the others;
b) each belonged to some trio of people who knew each other;
c) among those gathered, it was impos... | 97. The required society will be constructed if we seat each of the 12 individuals, who are not acquainted with each other, on a "personal" face of a regular dodecahedron and require that each person gets acquainted with their "neighbors," i.e., the inhabitants of adjacent faces.
Condition e) is satisfied here, as two... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,425 |
100. French cities. Dr. Sharadek, who knew strategy well, was interested in the latest war and in 1940 became acquainted with the map of the French theater of military operations. From this, the following problem probably arose. The distance (as the crow flies, as are all distances in this problem) from Chalon to Vitry... | 100. The five cities of Chalon, Vitry, Chaumont, Sän-Cantän, and Reims form a closed pentagon; one of its sides (the side Chaumont - Sän-Cantän) is the sum of the other four, since $236=86+40+30+80$. This is only possible when the vertices of the pentagon lie on a straight line. The cities are arranged on the line in t... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,428 |
Example 2. Let's determine in which numeral system the following multiplication was performed: $352 \cdot 31=20152$. | Solution. Let $x$ be the base of the numeral system, then the given equality can be written in the form of an equation
$$
\left(3 x^{2}+5 x+2\right)(3 x+1)=2 x^{4}+x^{2}+5 x+2
$$
By performing the multiplication and combining like terms, we get:
$$
2 x^{4}-9 x^{3}-17 x^{2}-6 x=0
$$
It is clear that $x \neq 0$ and t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,430 |
Example 1. The roots of the equation
$$
x^{3}+a x^{2}+b x+c=0
$$
form a geometric progression. What necessary and sufficient condition must the coefficients of the equation satisfy? | Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation. From the condition, it follows that
$$
x_{2}^{2}=x_{1} x_{3}
$$
Multiply both sides of this equation by $x_{2}: x_{2}^{3}=x_{1} x_{2} x_{3}$. Since $x_{1} x_{2} x_{3}=-c$, then
$$
x_{2}^{3}=-c
$$
Moreover, we have:
$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}=-a, \... | ^{3}=b^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,431 |
Example 2. Let's find the sum of the cubes of the roots of the equation
$$
x^{3}+2 x^{2}+x-3=0
$$ | S o l u t i o n. This can be done in various ways, for example, by sequentially calculating the sum of the squares of the roots and then the sum of the cubes of the roots. However, we will use a frequently applied identity in mathematics:
$$
\begin{aligned}
a^{3}+b^{3}+c^{3}-3 a b c= & (a+b+c)\left(a^{2}+b^{2}+c^{2}-a... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,432 |
E x a m p l e 2. We will prove that the equation $x^{2}+4 x-11=8 y$ has no solutions in integers. | Solution. 1st method. Transform the given equation:
$$
x^{2}-11=8 y-4 x
$$
The right-hand side $8 y-4 x$ is even for any integers $x$ and $y$. The left-hand side of the equation, if it has solutions, can only be even when $x^{2}$ is odd, and consequently, $x$ is odd.
Now, rewrite the equation as follows:
$$
x^{2}+4... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,434 |
Example 3. Solve the system of equations in integers:
$$
\left\{\begin{array}{l}
x^{2}-y^{2}-z^{2}=1 \\
y+z-x=3
\end{array}\right.
$$ | Solution. Express $x$ in terms of $y$ and $z$ from the second equation and substitute into the first:
$$
y z-3 y-3 z+4=0
$$
From this,
$$
y=\frac{3 z-4}{z-3}=3+\frac{5}{z-3}
$$
Since $y \in \mathbf{Z}$, then $\frac{5}{z-3} \in \mathbf{Z}$. This is only possible if $z-3= \pm 1$ and $z-3= \pm 5$.
Thus, we get: $z_{1... | {(9;8;4),(-3;-2;2),(9;4;8),(-3;2;-2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,435 |
Example 4. We will prove that the equation
$$
x^{2}+y^{2}+z^{2}=2 x y z
$$
is not solvable in natural numbers. | Solution. We will apply the so-called method of "infinite descent". Assume the opposite.
18
Suppose the given equation is solvable in natural numbers. The right-hand side of the equation is divisible by 2, therefore, its left-hand side must also be divisible by 2. This is only possible if either \(x, y, z\) are even ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,436 |
Example 5. Solve the equation in integers
$$
3^{x}-2^{y}=1
$$ | A solution is. One solution is obvious: $(1 ; 1)$.
It is not difficult to notice that the equation cannot have negative and zero solutions. Therefore, we will look for solutions greater than 1.
Let's rewrite the equation in the following form: $2^{y}=3^{x}-1$.
Since $x \in \mathbf{N}$, the right side of the equation... | (1,1)(2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,437 |
P r o b l e m 6. Let's find a four-digit number that is a perfect square, if its first two digits and last two digits are equal. | Solution. Let $N=\overline{x x y y}$ be the desired number. Write it as a systematic number in base 10 and simplify:
$$
\begin{aligned}
& N=1000 x+100 x+10 y+y \\
& N=1100 x+11 y \\
& N=11(100 x+y)
\end{aligned}
$$
Since (by condition) $N$ is the square of some natural number, the second factor $100 x+y$ must be divi... | 7744=88^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,438 |
Example 7. In a chess tournament, two students from grade VII and several students from grade VIII participated. Each student played one game with each of the other participants. The two seventh-grade students together scored 8 points, and all eighth-grade students scored the same number of points. How many eighth-grad... | Solution. Let $x$ be the number of eighth graders, $y$ be the number of points scored by each of them. The total number of points scored by all participants in the tournament is $x y + 8$. This number is equal to the number of games played.
On the other hand, since there were $x + 2$ participants and each participant ... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,439 | |
Example 1. Solve the equation
$$
\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5}
$$ | Solution. Let $\frac{15 x-7}{5}=t$, where $t \in \boldsymbol{Z}$.
From this,
$$
x=\frac{5 t+7}{15}
$$
Then the equation becomes:
$$
\left[\frac{10 t+39}{40}\right]=t
$$
From the definition of "integer part" it follows that:
$$
0 \leqslant \frac{10 t+39}{40}-t<1
$$
Solving this double inequality, we find that
$$... | x_{1}=\frac{7}{15},x_{2}=\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,440 |
Example 2. Solve the equation
$$
[x-1]=\left[\frac{x+2}{2}\right]
$$ | Solution. 1st method. Let: $[x-1]=t$, $\left[\frac{x+2}{2}\right]=t$
Then
$$
t \leqslant x-1<t+1 \text { and } t \leqslant \frac{x+2}{2}<t+1
$$
or
$$
t+1 \leqslant x<t+2
$$
and
$$
2(t-1) \leqslant x<2 t
$$
The problem has been reduced to finding such integer values of $t$ for which there is a common part of the t... | x\in[3,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,441 |
Example 4. Let's solve the equation
$$
x^{3}-[x]=3
$$ | Solution. Let's transform the given equation
$$
x^{3}=[x]+3
$$
Based on property $1^{0}$ of the "floor" function, we get:
$$
x^{3}=[x+3]
$$
From this, it follows that $x^{3} \in \mathbf{Z}$.
We construct the graphs of the functions $y=x^{3}$ and $y=[x+3]$. A graphical estimate (Fig. 5) shows that the only root lie... | \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,442 |
## Example 2. Solve the equation
$$
f\left(\frac{x+1}{x-2}\right)+2 f\left(\frac{x-2}{x+1}\right)=x
$$ | Solution. Let $\frac{x-2}{x+1}=z$, then $x=\frac{z+2}{1-z}, \quad(z \neq 1, z \neq 0)$ and
$$
f\left(\frac{1}{z}\right)+2 f(z)=\frac{z+2}{1-z}
$$
Replacing $z$ with $\frac{1}{z}$, we get:
$$
f(z)+2 f\left(\frac{1}{z}\right)=\frac{1+2 z}{z-1}
$$
From equations (3) and (4), we find $f(z): f(z)=\frac{4 z+5}{3(1-z)}$
... | f(x)=\frac{4x+5}{3(1-x)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,443 |
Example 3. Solve the equation
$$
2 f(x+y)+f(x-y)=f(x)\left(2 e^{y}+e^{-y}\right)
$$
28 | Solution. Performing the following substitutions:
$$
x=0, y=t ; x=t, y=2 t ; x=t, y=-2 t
$$
we obtain the equations:
$$
\begin{aligned}
& 2 f(t)+f(-t)=a\left(2 e^{t}+e^{-t}\right) \\
& 2 f(3 t)+f(-t)=f(t)\left(2 e^{2 t}+e^{-2 t}\right) \\
& 2 f(-t)+f(3 t)=f(t)\left(2 e^{-2 t}+e^{2 t}\right)
\end{aligned}
$$
where $... | f()=\cdote^{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,444 |
Example 1. Let's find pairs of real numbers $x$ and $y$ that satisfy the equation
$$
x^{2}+4 x \cos x y+4=0
$$ | S o l u t i o n. Several methods can be proposed to solve the given equation.
1st method. Since the equation is quadratic (if we do not consider $x$ under the cosine sign), we can express $x$ in terms of trigonometric functions of the angle $x y$:
$$
x=-2 \cos x y \pm \sqrt{4 \cos ^{2} x y-4}=-2 \cos x y \pm 2 \sqrt{... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,445 |
Example 2. Solve the equation
$$
\frac{|\operatorname{ctg} x y|}{\cos ^{2} x y}-2=\log _{\frac{1}{3}}\left(9 y^{2}-18 y+10\right)
$$ | Solution. The methods considered in Example 1 cannot be used here. Let's try to use the property of boundedness of functions.
On the one hand,
$$
\frac{|\operatorname{ctg} x y|}{\cos ^{2} x y}=\left|\frac{\operatorname{ctg} x y}{\cos ^{2} x y}\right|=\left|\frac{2}{\sin 2 x y}\right| \geqslant 2
$$
and accordingly $... | \frac{\pi}{4}+\frac{\pik}{2},1,k\in\boldsymbol{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,446 |
Example 3. Solve the inequality
$$
2^{y}-2 \cos x+\sqrt{y-x^{2}-1} \leqslant 0
$$ | Solution. Since $y-x^{2}-1 \geqslant 0$, then $y \geqslant x^{2}+1$ and accordingly $2^{y} \geqslant 2$.
Since $\cos x \leqslant 1$, then $-2 \cos x \geqslant-2$. Adding now the inequalities of the same sense:
$$
2^{y} \geqslant 2, -2 \cos x \geqslant-2, \sqrt{y-x^{2}-1} \geqslant 0
$$
we get:
$$
2^{y}-2 \cos x+\sq... | 1,0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,447 |
Example 4. Let's solve the system of equations:
$$
\left\{\begin{array}{l}
\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=2 \sin ^{2} y \\
\sin ^{2} y+\cos ^{2} z=1
\end{array}\right.
$$ | Solution. $\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x \geqslant 2$, since $a+\frac{1}{a} \geqslant 2$ for $a>0$. Since the right-hand side of the first equation $2 \sin ^{2} y \leqslant 2$, the first equation must be satisfied for those values of $x$ and $y$ that satisfy the system:
$$
\left\{\begin{array}{l}
\o... | \frac{\pi}{4}+\frac{\pi}{2}k,\quad\frac{\pi}{2}+\pin,\quad\frac{\pi}{2}+\pi,\quadk,n,\in{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,448 |
Example 5. Let's find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
2^{b x}+(a+1) b y^{2}=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
has at least one solution for any value of $b,(a, b, x, y \in \mathbf{R})$. | S o l u t i o n. Suppose there exists some value $a$ for which the system has at least one solution, for example, for $b=0$.
In this case, the system will take the form:
$$
\left\{\begin{array}{l}
1=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
It is clear that this system is consistent only if $a=1$ or $a=-1$.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,449 |
Example 1. Prove the following relations:
a) $\ln C_{0}(a)=C_{1}(\ln a)$
b) $\left(C_{\alpha}\left(a^{\frac{1}{\alpha}}\right)\right)^{\alpha}=C_{1}(a)$
c) $C_{1}(a b) \leqslant C_{2}(a) C_{2}(b)$
d) $C_{1}(a) C_{1}(b) \geqslant C_{\frac{1}{2}}(a b)$ | Solution. a) $\ln C_{0}(a)=\ln \sqrt[n]{a_{1} a_{2} \ldots a_{n}}=$
$$
=\frac{1}{n}\left(\ln a_{1}+\ln a_{2}+\ldots+\ln a_{n}\right)=C_{1}(\ln a)
$$
b) Let's write the power mean for $a^{\frac{1}{\alpha}}$ :
$$
C_{\alpha}\left(a^{\frac{1}{\alpha}}\right)=\left(\frac{\left(a_{1}^{\frac{1}{\alpha}}\right)^{\alpha}+\le... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,450 |
Example 3. We will prove that if $n>m>0, a_{i}>0, i=$ $=1,2, \ldots, n$, then
$\left(a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m}\right)\left(a_{1}^{n-m}+a_{2}^{n-m}+\ldots+a_{n}^{n-m}\right) \geqslant n^{2} a_{1} a_{2} \ldots a_{n}$. | S o l u t i o n. Using obvious inequalities:
$$
C_{m} \geqslant C_{0}, C_{n-m} \geqslant C_{0}
$$
we obtain:
$$
\begin{gathered}
a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m} \geqslant n \sqrt[n]{\left(a_{1} a_{2} \ldots a_{n}\right)^{m}} \\
a_{1}^{n-m}+a_{2}^{n-m}+\ldots+a_{n}^{n-m} \geqslant n \sqrt[n]{\left(a_{1} a_{2} \l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,451 |
Example 4. Prove the inequality
$$
\left(a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}\right)\left(\frac{b_{1}}{x_{1}}+\frac{b_{2}}{x_{2}}+\ldots+\frac{b_{n}}{x_{n}}\right) \geqslant\left(\sqrt{a_{1} b_{1}}+\ldots+\sqrt{a_{n} b_{n}}\right)^{n}
$$
where all $a_{i}, b_{i}, x_{i}$ are positive real numbers. | S o l u t i o n. In the Cauchy-Bunyakovsky inequality (example 1, v), replace \(a_{i}\) with \(\sqrt{a_{i} x_{i}}, b_{i}\) with \(\sqrt{\frac{b_{i}}{x_{i}}}\). Then the products \(a_{i} b_{i}\) will be replaced by \(\sqrt{a_{i} b_{i}}\):
^{\alpha}
$$
where $\alpha \geqslant 1, \alpha \in \mathbf{R}, h_{1}, h_{2}, h_{3}$ are the lengths of the altitudes of the triangle, $r-$ is the length of the radius of the inscribed circle. | Solution. We notice that
$$
h_{1}+h_{2}+h_{3}=\frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c}=2 S\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
where $S$ is the area of the triangle, and $a, b, c$ are the lengths of its sides.
38
Moreover,
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
s... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,453 |
Example 7. A circle of radius $r$ is inscribed in a triangle, and three circles of radii $r_{1}, r_{2}$, and $r_{3}$ are constructed, each touching two sides of the triangle and the inscribed circle. We will prove that the following inequality holds:
$$
\left(r_{1} r_{2}\right)^{m}+\left(r_{2} r_{3}\right)^{m}+\left(r... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,454 | |
Example 3. Point $M$ lies inside a triangle, $k_{1}, k_{2}, k_{3}$ are the distances from $M$ to the sides of the triangle, $h_{1}, h_{2}, h_{3}$ are the corresponding heights. Find the minimum value of the expression
$$
\left(\frac{k_{1}}{h_{1}}\right)^{\alpha}+\left(\frac{k_{2}}{h_{2}}\right)^{\alpha}+\left(\frac{k_... | S o l u t i o n. We have:
$$
2 S=a k_{1}+b k_{2}+c k_{3}=a h_{1}=b h_{2}=c h_{3}
$$
where $S$ is the area of the triangle.
Divide both sides of the equality $a k_{1}+b k_{2}+c k_{3}=a h_{1}$ by $a h_{1}$:
$$
\frac{k_{1}}{h_{1}}+\frac{b}{a} \cdot \frac{k_{2}}{h_{1}}+\frac{c}{a} \cdot \frac{k_{3}}{h_{1}}=1
$$
Since ... | \frac{1}{3^{\alpha-1}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,455 |
Example 2. Let's prove the inequality
$$
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 8}+\ldots+\frac{1}{n(3 n-1)} \geqslant \frac{1}{n+1}
$$ | S o l u t i o n. The inequality can be proven by the method of mathematical induction (see "Practicum in Algebra"). Let's consider another proof method based on the properties of power means. Applying the inequality $C_{-1} \leqslant C_{1}$ for the numbers $1 \cdot 2, 2 \cdot 5, 3 \cdot 8, \ldots, n(3 n-1)$, we get:
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,456 |
Example 3. Compare the number $a$ with one, if
$$
a=0.99999^{1.00001} \cdot 1.00001^{0.99999} .
$$ | S o l u t i o n. Let's represent the numbers in the given expression as follows:
$$
0.99999=1-\alpha, \quad 1.00001=1+\alpha
$$
where $\alpha=0.00001$. Then
$$
a=(1-\alpha)^{1+\alpha}(1+\alpha)^{1-\alpha}=\left(1-\alpha^{2}\right)\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}
$$
Since $1-\alpha^{2}<1$, then $a<1$
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,457 |
Example 2. We will prove that the number
$$
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}
$$
cannot be an integer for any natural $n$. | Proof. We will bring all the terms to a common denominator. It is clear that the common denominator is an even number.
Among all the fractions of the given sum, consider the one whose denominator includes the number 2 raised to the highest power compared to the other denominators, i.e., the fraction with the denominat... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,458 |
E x a m p l e 1. Let's find the limit of the sum
$$
S_{n}=\frac{3}{4}+\frac{5}{36}+\ldots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$
as $n \rightarrow \infty$. | S o l u t i o n. First, let's try to simplify $S_{n}$. Since the fraction 56
$\frac{2 k+1}{k^{2}(k+1)^{2}}$ can be represented as the difference $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}$, then
$$
\begin{gathered}
\frac{3}{4}=1-\frac{1}{2^{2}} \\
\frac{5}{36}=\frac{1}{2^{2}}-\frac{1}{3^{2}} \\
\frac{2 n+1}{n^{2}(n+1)^{2}}=... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,459 |
Example 2. Let $y \neq-1$. We set,
$$
x_{1}=\frac{y-1}{y+1}, \quad x_{2}=\frac{x_{1}-1}{x_{1}+1}, \quad x_{3}=\frac{x_{2}-1}{x_{2}+1}, \ldots
$$
What is $y$ if $x_{1978}=-\frac{1}{3}$? | Solution. Substituting the value of $x_{1}$ into the second equality, after simplifications we get:
$$
x_{2}=-\frac{1}{y}
$$
Further,
$$
x_{3}=\frac{y+1}{1-y}, \quad x_{4}=y, \quad x_{5}=\frac{y-1}{y+1}=x_{1}
$$
Therefore,
$$
x_{5}=x_{1}, \quad x_{6}=x_{2}, \quad x_{7}=x_{3}, \quad x_{8}=x_{4}, \ldots, \quad x_{19... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,460 |
Example 1. Factor the polynomial
$$
1+x^{5}+x^{10}
$$ | Solution. We have:
$$
\begin{gathered}
x^{10}+x^{5}+1=\frac{\left(x^{10}+x^{5}+1\right)\left(x^{5}-1\right)}{x^{5}-1}=\frac{x^{15}-1}{x^{5}-1}= \\
=\frac{\left(x^{3}-1\right)\left(x^{12}+x^{9}+x^{6}+x^{3}+1\right)}{(x-1)\left(x^{4}+x^{3}+x^{2}+x+1\right)}=\frac{\left(x^{2}+x+1\right)\left(x^{12}+x^{9}+x^{6}+x^{3}+1\ri... | x^{10}+x^{5}+1=(x^{2}+x+1)(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,461 |
Example 2. Let's calculate the sum
$$
a^{2000}+\frac{1}{a^{2000}}
$$
if $a^{2}-a+1=0$. | S o l u t i o n. From the equality $a^{2}-a+1=0$, it follows that:
$$
a-1+\frac{1}{a}=0 \text { or } a+\frac{1}{a}=1
$$
Moreover,
$$
a^{3}+1=(a+1)\left(a^{2}-a+1\right)=0
$$
from which $a^{3}=-1$. Now we have:
$$
\begin{aligned}
& a^{2000}+\frac{1}{a^{2000}}=\left(a^{3}\right)^{666} a^{2}+\frac{1}{\left(a^{3}\righ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,462 |
Example 3. Given three statements:
1) the equation $x+\frac{1}{x}=a$ has no real solutions;
2) the equality $\sqrt{a^{2}-4 a+4}=2-a$ holds;
3) the system $\left\{\begin{array}{l}x+y^{2}=a, \\ x-\sin ^{2} y=-3\end{array}\right.$ has a unique solution.
For which values of $a$ will two of these statements be true, and t... | Solution. Rewrite the second equality as follows: $\sqrt{(a-2)^{2}}=2-a$, or $|a-2|=2-a$.
Therefore, $2-a \geqslant 0$, i.e., $a \leqslant 2$.
Transform the equation $x+\frac{1}{x}=a$ into the form
$$
x^{2}-a x+1=0
$$
For this equation to have no real roots, it is necessary and sufficient that its discriminant $a^{... | -3or\in]-2;2[ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,463 |
Example 2. In each cell of a square table of size $25 \times 25$, one of the numbers +1 or -1 is written arbitrarily. Under each column, the product of all numbers in that column is recorded, and to the right of each row, the product of all numbers in that row is recorded. Prove that the sum of all fifty products canno... | S o l u t i o n. If we multiply all fifty products, then each number in the table will appear twice in the final product: once by row, and once by column. Therefore, the final product is +1.
Since the final product is positive, it can be asserted that an even number of factors equal to -1 are included in its compositi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,464 |
E x a m p l e 5. The numbers from 1 to 50 inclusive are written on a board. It is allowed to erase any two numbers and replace them with one - the absolute value of their difference. (We do not write down the absolute value of zero). After repeating this process multiple times, only one number remains on the board. Can... | Solution. First of all, it is clear that any number remaining on the board is between 0 and 50 (if $0 \leqslant a \leqslant 50$ and $0 \leqslant b \leqslant 50$, then $0 \leqslant|a-b| \leqslant 50$). We will show that the resulting number is necessarily odd. For this, we will track the sum of all numbers written on th... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,465 | |
Example 3. Let there be a set $T$ of right-angled triangles, each of which has a height dropped to the hypotenuse numerically equal to 1. We will prove that from each such triangle, a circle can be cut out whose area will be greater than 0.5. | S o l u t i o n. Obviously, among all the triangles of the given set, we need to find the triangle with the smallest radius of the circle inscribed in it and show that the area of this circle is greater than 0.5.
Let (Fig. 11) $|C D|=1, \widehat{A}=\alpha$ and $|O N|=r$.
Since $|A D|=\operatorname{ctg} \alpha,|D B|=\... | \pi(\sqrt{2}-1)^{2} | Geometry | proof | Yes | Yes | olympiads | false | 35,467 |
Пр и ме р 4. Can a closed self-intersecting broken line be constructed on a plane, each segment of which intersects exactly once, if it consists: a) of 999 segments; b) of 100 segments? (Vertices of the broken line cannot lie on other segments of the broken line.)
Example 4. Can a closed self-intersecting broken line ... | S o l u t i o n. Consider some segment $[A B]$. By the condition, it intersects another segment $[D C]$. Thus, $[A B]$ intersects $[D C]$. Since $[A B]$ and $[C D]$ have already intersected once, no other segments can intersect them according to the condition. Each of the remaining segments must also form a pair of int... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,468 |
Example 5. On a plane, $n$ points are chosen. We will prove that if every triangle formed by three of these points is "empty" (i.e., there are no other points inside it or on its sides), then all points are vertices of some convex $n$-gon. | Solution. Let $M$ be the set of points $A_{i} (i=1,2, \ldots, n)$, satisfying the condition of the problem, i.e., such that any triangle with vertices at three different points $A_{i}, A_{j}, A_{k} \in M$ does not contain other points of this set either inside the triangle or on its boundaries (sides)
$$
\left[A_{i} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,469 |
E x a m p l e 8. We will prove that any convex polygon with an area of one can be enclosed in a parallelogram with an area of two. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,471 | |
3. In a pet store. "I guarantee," said the seller, "that this parrot will repeat every word it hears." The delighted customer bought the wonder-bird, but upon arriving home, discovered that the parrot was "as silent as a fish." Nevertheless, the seller did not lie. How could this be? | 3. The seller told the truth, which means it is true that the parrot repeats every word it hears. The parrot is "as silent as a fish" - therefore, it has not heard a single word. Conclusion: either no word was spoken in its presence, or it is simply "as deaf as a grouse." | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,474 |
5. From the start to the finish, flags are placed at equal distances from each other. An athlete runs the distance from the first flag to the eighth in 8 seconds. How long will it take him to reach the twelfth flag? | 5. Since the athlete runs seven (not eight, as is sometimes thought) intervals between the flags in 8 seconds, he will run 11 intervals in $\frac{8 \cdot 11}{7}=12 \frac{4}{7}$ seconds. | 12\frac{4}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,476 |
6. Points on a Line. If 10 points are placed at equal intervals on a line, they will occupy a segment of length s, and if 100 points are placed, the segment will have a length S. How many times greater is S than s? | 6. Between ten points there are nine intervals, and between a hundred points - ninety-nine. Therefore, $S$ is greater than $s$ by 11 times. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,477 |
7. Stereometric problem. How many faces does a hexagonal pencil have? | 7. It is important to ask: which pencil? If the pencil has not been sharpened yet, then 8, otherwise there may be variations...
Translating the text as requested, while preserving the original line breaks and formatting. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,478 |
8. Correct the error. In the incorrect equation $101=10^{2}-1$, move one digit to make it correct. | 8. Move the digit 2 slightly down: $101=102-1$. | 101=102-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,479 |
9. The Caterpillar's Ascent. The height of the pole is 20 meters. The caterpillar crawls up it, during the day it climbs 5 meters, and at night it slides down 4 meters. How long will it take for the caterpillar to reach the top of the pole? | 9. The usual answer: in 20 days, but... Since the caterpillar climbs 1 meter per day, therefore, in 15 days it will climb 15 meters, and on the sixteenth day it will climb another 5 meters and reach the top of the pole...
 in your hand and pour it into an empty cup. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,481 |
11. Fifty rubles $=$ fifty rubles? It is known that equalities can be multiplied. For example: 2 rub. $=200$ kopecks, and 0.25 rub. $=25$ kopecks. Multiplying these equalities, we get: 0.5 rub. $=5000$ kopecks. What is the error? | 11. You cannot multiply 2 rubles by 0.25 rubles, as there are no square rubles (let alone square kopecks). | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,482 |
12. The Musketeers' Journey. The distance between Athos and Aramis, riding on the road, is 20 leagues. In one hour, Athos travels 4 leagues, and Aramis - 5 leagues. What distance will be between them after an hour? | 12. Well, in which direction was each of the musketeers traveling? The problem statement does not mention this. If they were traveling towards each other, the distance between them would be 11 leagues. In other cases (make a diagram!), the possible answers are: 29 leagues; 19 leagues; 21 leagues. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,483 |
13. To Cut or Not to Cut? A mathematician, finding himself in a small town, decided to get a haircut. There were only two barbershops in the town. Peering into one of the shops, he saw that it was dirty, the barber was dressed sloppily, poorly shaved, and carelessly cut. In the other shop, it was clean, and the owner w... | 13. Can a barber cut his own hair? If not, barbers are forced to cut each other's hair. The first one is cut poorly, and the second one is cut well. Therefore, the second barber cuts poorly, and the first one cuts well. Now it is clear why the mathematician chose the first barber. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,484 |
16.On a circular route, two buses are operating, with an interval of 21 minutes. What would be the interval if three buses were operating on the route? | 16. Since the interval between buses is 21 minutes, the entire route is covered by a bus in 42 minutes. If there are 3 buses, the intervals between them will be 14 minutes. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,487 |
17. Economic forecast. Three chickens laid three eggs in three days. How many eggs will twelve chickens lay in twelve days? | 17. Of course not 12, as sometimes answered (inertial thinking!), but 48. | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,488 |
18. Strange fractions. Can a fraction where the numerator is less than the denominator be equal to a fraction where the numerator is greater than the denominator? | 18. Why not? For example: $\frac{-1}{2}=\frac{1}{-2}$.
9 Carefully read the condition of the problem | \frac{-1}{2}=\frac{1}{-2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,489 |
20. Wine or water? In one glass is wine, in the other is water. A drop of wine is transferred from the first glass to the second, then thoroughly mixed, and a drop of the mixture is transferred back. Which is greater: the amount of wine in the water glass or the amount of water in the wine glass? How would the answer c... | 20.Since the level of liquid in the glasses has not changed, the wine that was poured into the glass of water has been replaced by water. Therefore, there is as much wine in the glass of water as there is water in the glass of wine. The answer will not change if the pouring is done multiple times or if the mixing is no... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,491 |
1. How to cut the chain? A traveler arrived at a hotel. He had no money, only a gold chain consisting of 6 links. For each day of his stay at the hotel, he had to pay with one link of the chain, but the hotel owner warned that he would accept no more than one cut link. How can the traveler cut the chain to stay in the ... | 1. The traveler needs to cut the third ring, so he will have three pieces of chain, 1, 2, and 3 links long. Then on the first day, he will give one link, on the second day two and get one back, and so on. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,492 |
2. How to stay in power? The king decided to dismiss his prime minister but did not want to offend him, and there was no reason to do so. Finally, he came up with the following idea. One day, when the prime minister came to the king, the latter said to him: "I have put two sheets of paper in the portfolio. On one of th... | 2. He must take out one piece of paper and destroy it without reading it. After that, take out the second one and say: "Since it says 'Leave' on this piece of paper, it means that on the one I destroyed was: 'Stay.'" | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,493 |
5. How to toast bread? A frying pan can hold two slices of bread. It takes one minute to toast one slice on one side. Is it possible to toast three slices of bread on both sides in less than 4 minutes? | 5. Yes. Let's say we have three slices of bread: $a A, b B$, and $c C$. The plan for toasting can be conveniently represented by the scheme: $a b ; A c ; B C$. It is not possible to toast the three slices of bread any faster: the three slices of bread have six "sides," and no more than two "sides" can be toasted per mi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,496 |
6. How to defeat Koschei? In the sacred and dense, terrible Murom forest, magical springs of water gush from the ground. Water can be taken from the first nine springs by anyone, but the tenth spring is located in Koschei the Deathless's cave, and no one except Koschei himself can take water from there. The magical wat... | 6. Ivan-the-Prince must drink magic water (from any source) before the duel, then Koshchei, giving him water from the tenth source, will save him from certain death. Koshchei, however, he must give plain water (not magical!) to drink. Koshchei, after drinking plain water and washing it down with water from the tenth so... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,497 |
7. How to Stay Alive? A traveler has been captured by bloodthirsty savages. According to the tribe's laws, every foreigner is asked about the purpose of their visit. If they tell the truth, they will be eaten; if they lie, they will be drowned in the sea. How can the traveler stay alive? | 7. The traveler must say: "I came here for you to drown me." Then, if they want to drown him, it turns out that he told the truth, and for the truth - they eat him. But if they want to eat him, it turns out that he lied, and for lying - they drown him. It is unknown, however, whether the natives will be consistent in t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,498 |
10. How to guess the number? I have thought of one of the numbers: 1, 2, or 3. Can you guess which number I have thought of by asking me just one question, to which I will answer (truthfully!) "yes," "no," or "don't know"? | 10.For example, this: "if I also thought of one of the three numbers 1, 2, or 3, but not the one you thought of, would it be greater than your number?" | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,501 |
14. Where did the dollar go? Three people had lunch, paid 30 dollars (10 dollars each), and left. After some time, the cook noticed that he had overcharged them by 5 dollars and sent the apprentice to return the money. The apprentice gave back 3 dollars (1 dollar to each person), and kept 2 dollars for himself. Three t... | 14. The dollar, in fact, didn't go anywhere. Let's do the calculation. The meal cost 25 dollars, the boy kept 2 dollars for himself, so the meal cost the customers 27 dollars. They were refunded 3 dollars. For the calculation to be correct, you need to add 3 dollars to the 27 dollars, not 2. Everything adds up. | 27+3=30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,505 |
1. Strange report. The school principal reported in his report that there are 3688 students in the school, and boys outnumber girls by 373. But the clever inspector $Р O H O$ immediately realized that there was an error in the report. How did he figure it out? | 1. Yes, a rather strange report. If the number of girls in this school is $x$, then the total number of students is $2 x + 373$, which does not equal 3688, since an odd number cannot be equal to an even number. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,507 |
3. A case at the savings office. Is it possible to exchange 25 rubles using ten bills of 1, 3, and 5 rubles? | 3. It cannot be. And not because such banknotes do not exist. The sum of an even number of odd addends cannot be an odd number. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,509 |
4. Road problem. Can 13 cities be connected by roads so that exactly 5 roads lead out of each city? | 4. Since 5 roads lead out of each city, the total number of roads is 65. Note that in this case, each road $AB$ is counted twice, as leading out of cities $A$ and $B$. Thus, the total number of roads should be even. This leads to a contradiction. Answer: it is impossible. | itisimpossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,510 |
5. Fedya, you are wrong. Fedya wrote on the board the equation: $1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9=20$ (instead of $*$, unknown signs + and - are written on the board in some order). Prove that there is an error in the equation. | 5. Since the expression $1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9$ contains an odd number of odd numbers, the result should be an odd number, therefore, this equality is incorrect. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 35,511 |
7. Points on a Line. Several points are located on a line. Then, between each pair of adjacent points, another point was placed. This process was repeated several times, after which all the marked points were counted. Could the total number of points be 1998? | 7. Note that if at any moment there are $n$ points on the line, then on the next step the number of points becomes the odd number $2 n-1$. Therefore, the number of points on the line will always be odd. Consequently, the number 1998 could not have been obtained in the calculation. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,513 |
8. The Lost Weight. In the set, there were 23 weights with masses of $1 \mathrm{kr,} 2 \mathrm{kr}$, 3 kg, ... 23 kg. Can they be divided into two equal-mass piles if the 21 kg weight was lost? | 8. Note that $(1+23)+(2+22)+\ldots+(11+13)+12-$ is an even number. Therefore, ( $S-21$ ) cannot be divided into two equal-weight piles. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,514 |
9. Without complex calculations. Numbers $a$ and $b$ are integers. It is known that $a+b=1998$. Can the sum $7a+3b$ equal 6799?
15 Even and odd numbers | 9. Since $a$ and $b$ have the same parity, then $7a$ and $3b$ also have the same parity, which means their sum should be even. Since 6799 is an odd number, the problem has no solutions. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,515 |
10. Straight line. Is it possible to draw a straight line so that it intersects all sides (but not the vertices!) of a 2001-gon? | 10. Suppose it is possible, and a straight line intersects all sides (but not vertices) of a certain $2001-$gon.
1 solution. Any polygon divides the plane into two parts. We will "move" along the line. At the beginning of the "movement," we are outside the polygon, when we cross the first side, we will be inside the p... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,516 |
11. An ancient problem. How many kilometers will a traveler cover in 17 days, spending 10 hours a day on this, if he has already traveled 112 kilometers in 29 days, spending 7 hours each day on the road? | 11. Since the traveler has traveled 112 km in 29.7 hours, in $17 \cdot 10$ hours he will travel $\frac{112 \cdot 17 \cdot 10}{29 \cdot 7}=93 \frac{23}{29}$ km (from a French textbook of the $X I X$ century - speeds were not very high back then). | 93\frac{23}{29} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,517 |
12. Buying an album. To buy the album, Masha was short of 2 kopecks, Kolya was short of 34 kopecks, and Fedia was short of 35 kopecks. When they combined their money, it still wasn't enough to buy the album. How much does the album cost? | 12.Since Kolya has one kopek more than Fyodor, he has at least 1 kopek, and 1 kopek is added to Masha's money. But Masha was short of money for the album, so she was given less than 2 kopeks, which means Fyodor has no money at all and is short of the full cost of the album. Answer: the album costs 35 kopeks. | 35 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,518 |
14. What are money for? A banker was walking down the street of a small provincial town when he suddenly saw a $5 bill on the sidewalk. He picked it up, noted the number, and went home for breakfast. Over breakfast, his wife told him that the butcher had sent a bill for $5. Since the banker had no other money on him, h... | 14. After the farmer sold the calf to the butcher, all five participants (the banker, the butcher, the farmer, the trader, and the laundress) found themselves in the same position, namely: each of them owed someone 5 dollars, and was owed exactly the same amount, so the overall balance was zero. The circulation of the ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,520 |
15.Place non-zero integers in the cells of a $4 \times 4$ table so that the sum of the numbers in the corners of each square of size $2 \times 2, 3 \times 3, 4 \times 4$ equals 0. | 15. See the figure:
| -1 | -1 | 1 | 1 |
| :---: | :---: | :---: | :---: |
| 1 | 1 | -1 | -1 |
| -1 | -1 | 1 | 1 |
| 1 | 1 | -1 | -1 |
4 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,521 |
1. Honest Fedya always tells the truth, but once, when he was asked the same question twice in a row, he gave different answers. Could this have happened? | 1. For example, you can ask him twice: «What time is it now?» | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,522 |
2. A strange question. Fedya always tells the truth, while Sasha always lies. They were asked the same question and gave the same answer. Could this be possible? | 2. You can ask: "Are you speaking the truth?" Both a truthful person and a liar should answer "yes!"
Note: The original text's formatting and line breaks have been preserved. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,523 |
4. When was Fedia born? Fedia once said: "The day before yesterday I was 10 years old, and next year I will be 13 years old." Could this have been true? | 4. This could be the case if Fedya was born on December 31, and the conversation took place the next day, that is, January 1. | December31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,525 |
6. On the porch of the house, a boy and a girl are sitting next to each other. Sasha says: "I am a boy." Zhena says: "I am a girl." If at least one of the children is lying, then who is the boy and who is the girl? | 6. From the fact that one of them is lying, it follows that the second one is lying too. Therefore, Sasha is a girl, and Zhenya is a boy. | Sasha | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,527 |
8. Problems with gasoline. Can 50 liters of gasoline be distributed among three tanks so that the first tank has 10 liters more than the second, and after transferring 26 liters from the first tank to the third, the third tank has as much as the second?
19 Was it or wasn't it? | 8. Note that the first and second tanks must contain no less than 26 liters of gasoline. Therefore, it is impossible to distribute 50 liters of gasoline in this way. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,529 |
9. Transport problems. Fedya participates in two math clubs located at opposite ends of Moscow. He has to travel on the same metro line but in opposite directions. Fedya gets on the first train that arrives, regardless of which direction it is coming from. By the end of the year, he found that he had visited the first ... | 9. It could have been worse, for example, if trains in one direction departed at $10.00 ; 10.10 \ldots$, and in the opposite direction at 10.09 ;
21 Was it or wasn't it?
$10.19 \ldots$, then the chances of the schoolchild getting into one of the clubs would be 9 times higher. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,530 |
10. Did it happen or not? This wonderful story is told about one of the greatest German philosophers, I. Kant. One evening he noticed that his wall clock had stopped. To find out the time, he went to visit one of his friends. After staying there for some time, he returned home and set the clock hands correctly. How did... | 10. Probably things happened like this: Kant wound his wall clock and went to visit a friend. Having learned the exact time there, he returned home (not forgetting to take into account the time spent at his friend's). At home, by the clock's indication, he determined how long he had been on the road. He divided this da... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,531 |
11. In Anchuria and Gvayasuèl, the monetary unit is called a dollar, and in both countries, dollars were originally quoted equally. One day, the government of Gvayasuèl decided to equate the Anchurian dollar to ninety Gvayasuèl cents. The next day, a similar rate was introduced for the Gvayasuèl dollar in Anchuria. In ... | 11. The purchasing power of the money used "in the wrong country" is lost, so those who pay "in the wrong country" pay for the beer, in this case, the tavern keeper gives a dollar in change. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,532 |
13. An Arabic Tale. A Flock of Pigeons
$>\Delta \theta \Pi \mathrm{V} \oplus \theta\mathbf{v} \Pi \square \nabla \square \Lambda$
$\oplus \Lambda \nabla \theta \Pi \oplus \mathbf{V V} \varnothing \odot$
ロจ৫ఠ<>VIVOO flew up to a tall tree.
Some of the pigeons perched on the branches, while others settled under the t... | 13. From the condition of the problem, it is clear that the number of pigeons sitting on the branches is two more than those sitting below. Further, it follows from the condition that after one of the pigeons flew up to the branch, the number of pigeons sitting on the branch became twice as many as those sitting on the... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,534 |
14.Strike out a hundred digits from the number $12345678910111213 \ldots 5960 \quad$ so that the resulting number is the largest possible. | 14. The largest possible number should start with the maximum number of nines. We will "move" along the number from left to right, crossing out all digits except 9. We will cross out 27 digits: $12345678910111213141516171819 . . .5960$, 8 digits 19 digits
then 19 digits: $99 \underbrace{20212223242526272829 \ldots .59... | 99999785960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,535 |
1. On a walk. Once, while walking through the land of knights and liars, I met a person who said about himself: “I am a liar.” Who was the person I met? | 1. The one who said "I am a liar" could not have been a liar, for liars never tell the truth. He could not have been a knight either, for knights never lie. He, in fact, was not a native of the country of knights and liars. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,537 |
2. Who is who? Before us are two inhabitants of the country of knights and liars $\mathcal{A}$ and $B . \mathcal{A}$ says: “I am a liar, and $B$ is not a liar.” Who among the islanders $\mathcal{A}$ and $B$ is a knight and who is a liar? | 2. Statement $\mathcal{A}$: "I am a liar, and $B$ is not a liar," is true only if both of its components are true. The statement "I am a liar" cannot be true. Therefore, $\mathcal{A}$ is a liar, and for the entire expression to be a lie, the second part of the statement must also be a lie. Therefore, $B$ is also a liar... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,538 |
3. Expertise is required. Before us are three natives of the country of knights and liars $\mathcal{A}, B$ and $C$. $\mathcal{A}$ says: «We are all liars». $B$ says: «Exactly one of us is a liar». Who is $C$ - a knight or a liar? Can we determine who $B$ is? | 3. $\mathcal{A}$ is a liar, as his statement “We are all liars” cannot be true. From B's statement: “Exactly one of us is a liar,” if it is true, it follows that $B$ and $C$ are both knights. If it is false, then $C$ is the only knight, since $\mathcal{A}$'s statement is false. Therefore, $\mathcal{A}$ is a liar, $C$ i... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,539 |
4. In the government of the country of knights and liars, there are 12 ministers. Some of them are liars, and the rest are knights. Once, at a government meeting, the following opinions were expressed: the first minister said, “There is not a single honest person here,” the second said, “There is no more than one hones... | 4. Note that the number of true statements must match the number of honest people in the government. Further, if a statement from any minister is true, then the statements of each minister who spoke after him are also true. In this case, the only statement that would not lead to a contradiction is: "there are no more t... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,540 |
5. Around a round table, eight people are sitting, each of whom is either a knight or a liar. When asked who their neighbors are, each of them answered: “My neighbors are a liar and a knight.” How many of them were liars? How would the answer change if nine people were sitting at the table? | 5. 6. At the table, there is at least one liar. Indeed, if only knights were sitting at the table, each knight's statement "next to me sits a knight and a liar" would be false, which is impossible. 2. The neighbors
. The presiding officer said that the number of deputies in the State Duma is odd, so he is a liar. If there are no knights among the deputies, then he is even more of a liar! | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,542 |
7. A judicial case. In the country of knights and liars, a crime was committed. Three residents of the country were brought to court: $\mathcal{A}$, $B$, and $C$. When the judge asked a question, $\mathcal{A}$ answered inaudibly. When the judge asked the two remaining individuals again, $B$ said that $\mathcal{A}$ clai... | 7. Obviously, $\mathcal{A}$ said that he is a knight, since $\mathcal{A}$ is a resident of the country. Further, from the condition of the problem, it is clear that $B$ is a knight, and $C$ is a liar. | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,543 |
8. Guide. A tourist arrived in the country of knights and liars. The first local resident he met claimed to be a knight. The tourist was delighted and hired him as a guide. After some time, they met another local resident. The tourist sent the guide to ask him whether he was a knight or a liar. The guide returned and s... | 8. 9. Suppose the guide is a liar, then: a) if the native is a knight, the guide will answer that he is a liar; b) if the native is a liar, he will still say that he is a knight, and the guide will answer that he is a liar. 2. If the guide is a knight, then: a) if the native is a knight, the guide will answer that he i... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,544 |
10. An interesting conversation. Once, among four inhabitants of the country of knights and liars, an interesting conversation took place. $\mathcal{A}$ said: «At least one of us is a liar». $B$ said: «At least two of us are liars». $C$ said: «At least three of us are liars». $D$ said: «There are no liars among us». Bu... | 10.Suppose that $D$ told the truth, then it follows that everyone except him is lying - a contradiction. Therefore, he is a liar. (This also follows from the condition of the problem.) If the statement of $C$ is true, then the statements of the first two are also true - a contradiction. Therefore, $C$ is also a liar, a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,545 |
11. In the bookbinding workshop. A piece fell out of the book, the first page of which is numbered 328, and the number of the last page is written with the same digits but in a different order. How many pages are in the missing piece? | 11. Since the last page must have a higher number of a different parity than the initial one, its number is 823. Answer: 496 pages fell out.
${ }^{1}$ Notation: $p-p$ king, $l-l$ liar.
27 In a country of knights and liars | 496 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,546 |
12.From Wittenberg to Göttingen, two students are walking. The first one walked 7 miles daily. The second one walked 1 mile on the first day, 2 miles on the second day, 3 miles on the third day, and so on, walking 1 mile more each day than the previous day. When will the second catch up with the first? | 12. On the first day, the second student walked 6 miles less than the first, on the second day 5 miles less, and so on. By the seventh day, the students will walk the same distance. After that, on the eighth day, the second student will walk 1 mile more, on the ninth day 2 miles more, and so on. They will meet at the e... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,547 |
14. Insert parentheses in the equation: $1: 2: 3: 4: 5: 6: 7: 8: 9: 10=7$, to make it true. | 14. Answer: $1: 2: 3: 4: 5:(6: 7: 8: 9: 10)=7$. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,549 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.