problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
208. Comparison of radicals. Which of the two quantities is greater: $\sqrt[8]{8!}$ or $\sqrt[9]{9!} ?$ | 208. In general, $n!<(n+1)^{n}$, since each of the $n$ factors on the left side is less than $(n+1)$. Therefore,
$$
(n!)^{n} n!<(n!)^{n}(n+1)^{n} \quad \text { or } \quad(n!)^{n+1}<[(n+1)!]^{n} .
$$
Taking the $n(n+1)$-th root of both sides of this inequality, we get
$$
(n!)^{\frac{1}{n}}<[(n+1)!]^{\frac{1}{(n+1)}}
... | \sqrt[8]{8!}<\sqrt[9]{9!} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,641 |
209. "Fibonacci Tetrahedron". Find the volume of the tetrahedron whose vertices are located at the points with coordinates $\left(F_{n}, F_{n+1}, F_{n+2}\right), \quad\left(F_{n+3}, F_{n+4}, F_{n+5}\right), \quad\left(F_{n+6}, F_{n+7}, F_{n+8}\right)$ and $\left(F_{n+9}, F_{n+10}, F_{n+11}\right)$, where $F_{i}$ is the... | 209. The Fibonacci sequence satisfies the recurrence relation $F_{n}+F_{n+1}=F_{n+2}$, so any three consecutive Fibonacci numbers satisfy the equation $x+y=z$. Consequently, all four vertices of our tetrahedron lie in the same plane, and its volume is 0.
We can observe that all the reasoning remains valid even if the ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,642 |
210. Regular Octahedron. A regular octahedron with edge $e$ is intersected by a plane parallel to one of its faces. Find the perimeter and area of the resulting section. | 210. If we cut a regular octahedron \(A B C D E F\) along some of its edges, it can then be "unrolled" onto a plane; as a result, we get a parallelogram \(A D D^{\prime} A^{\prime}\), consisting of
 | 213. Let $R$ be the sum of the given vectors. Rotate the entire configuration around the center $O$ by $\frac{2 \pi}{n}$ radians. Then the configuration will coincide with itself, and the vector $R$ will also rotate by $\frac{2 \pi}{n}$ radians and transform into $R^{\prime}$. It is clear that $R=R^{\prime}$, but since... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,646 |
214. Sliding Ellipse. Let two mutually perpendicular lines be given, and let some ellipse move in the plane, constantly touching both lines. Determine what line the center of the ellipse will describe during this motion. | 214. Consider the canonically positioned ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$. The geometric locus of points of intersection of two mutually
perpendicular tangents to this ell... | (^{2}+b^{2})^{\frac{1}{2}}\operatorname{arctg}\frac{^{2}-b^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,647 |
217. Pierced Sphere. The axis of the cylindrical hole drilled in the sphere is 10 cm long and lies on the diameter of this sphere. What is the volume of the remaining part of the sphere? | 217. Let's imagine a sphere of radius $k$, bounded by a thin stretchable membrane that holds a liquid inside the sphere according to the laws of surface tension. Now, let's puncture the sphere along its diameter and insert a cylindrical tube of length $2 k$ that can stretch radially in such a way that it does not lose ... | 523.6\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,650 |
218. To and from work. If a person walks to work and takes transportation back, the total travel time is one and a half hours. If he takes transportation both ways, the entire journey takes him 30 minutes. How much time will the person spend on the road if he walks to and from work? | 218. If a person twice travels the route to work and back using the first method, then in doing so, he will twice travel the distance between home and work by transport and twice walk it, spending a total of 3 hours. Therefore, he can walk to work and back in $3-\frac{1}{2}=2 \frac{1}{2}$ hours. | 2\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,651 |
219. System with an odd base. Show that an integer, written in a system with an odd base, is odd if and only if it contains an odd number of odd digits. | 219. Any number in a base $r$ numeral system can be represented as $a_{0} r^{n}+a_{1} r^{n-1}+\cdots+a_{n} r^{0}$. If in the product of several integers at least one of the factors is even, then the entire product is even; otherwise, it is odd. Furthermore, if $r$ is odd, then $r^{k}$ is also odd; therefore, the parity... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,652 |
221. Rhombic dodecahedra. Show that space can be filled with cells having the shape of rhombic dodecahedra. | 221. Cubic cells can obviously fill space. Consider the part of the cubic lattice shown in the figure. We will leave the central cube untouched, and in each of the
"surrounding" cubes, we w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,654 |
222. Fibonacci Ratio. The Fibonacci sequence $\left\{F_{n}\right\}$ satisfies the recurrence relation $F_{n}=$ $F_{n-1}+F_{n-2}, n \geqslant 3$ and $F_{1}=F_{2}=1$. Show that every fifth term of this sequence is divisible by 5. | 222. Applying this recurrence relation several times, we get
$$
\begin{aligned}
F_{n} & =F_{n-1}+F_{n-2}=F_{n-2}+2 F_{n-3}+F_{n-4}= \\
& =F_{n-3}+F_{n-4}+2 F_{n-4}+2 F_{n-5}+F_{n-4}= \\
& =5 F_{n-4}+3 F_{n-5}
\end{aligned}
$$
Further, $F_{5}=5$, so every fifth term of the Fibonacci sequence is divisible by 5.
[E. M.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,655 |
223. Cryptarithm with multiplication. After performing regular multiplication "in column," a person replaced each even digit with the letter $E$, and each odd digit - with the letter $O^{*}$. As a result, the following expression was obtained:
| $O E E$ E EOEE |
| :--- |
| EOE |
| $O O E E$ |
Restore the given mult... | 223. Since $188 \cdot 8=1504$, the first $O$ must be greater than 1. After multiplying this $O$ by the first $E$ of the multiplier, the result should be $\leqslant 8$; thus, $O=3$, and $E=2$. Numbers of the form $3 E E$, which when multiplied by 2 give $E O E$, can only be $306, 308, 326, 328, 346$, and 348. But if we ... | 348\cdot28=9744 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,656 |
224. Folded Rectangle. Two opposite vertices of a rectangle with sides $x$ and $y$ are overlapped. Find the length of the fold line. | 224. Let the rectangle $ABCD$ have sides $AB=CD=x$, and $AD=BC=y$, where $x \geqslant y$. Fold vertex $C$ onto $A$; denote the fold line as $PQ$. By symmetry, $PC=PA=CQ=AQ$,
148
Problems with a Twist
 \times(m n-1)=0$. Therefore, for $x \neq 0$ and $\neq a+b$
$$
\frac{x-a}{b}+\frac{x-b}{a}=0, \quad \text { that is } \quad(a+b) x=a^{2}+b^{2}
$$
Thus, the third root... | \frac{^2+b^2}{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,659 |
227. Isosceles Triangle. In isosceles triangle \( A B C \), the angle at vertex \( C = 20^{\circ} \). Points \( M \) and \( N \) are chosen on the lateral sides \( A C \) and \( B C \) respectively, such that angle \( A B M = 60^{\circ} \) and angle \( B A N = 50^{\circ} \). Prove, without using trigonometry, that angl... | 227. Since the sum of the interior angles of a triangle is two right angles, or \(180^{\circ}\), \(\angle CBA = \angle CAB = 80^{\circ}\), \(\angle CBM = 20^{\circ}\), and \(\angle BAN = 50^{\circ} = \angle BNA\) (so \(BN = AB\)). Draw \(MR\)
=(-1,0)$. Moreover, it is easy to notice that $x$ cannot take other negative values, since $y^{2}$ is non-negative. Let's write our equation as
$$
\begin{aligned}
& x^{3}+1=(x+1)\left(x^{2}-x+1\right)= \\
& =(x+1)^{2}\left(x-2+\frac{3}{x+1}\right)=y^{2}
\end{aligned}
$$
Further, t... | (x,y)=(-1,0),(0,\1),(2,\3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,662 |
230. "Countries" on a sphere. Suppose that using some triangulation we divided the sphere into "countries". Here, by triangulation we mean such a division of the sphere that each resulting "country" borders (i.e., has a common boundary segment of non-zero length) with exactly three other "countries". A vertex of the gr... | 230. Suppose such a triangulation is possible using graph $G$. If we remove the line connecting two odd vertices (thus forming one "country" bordering four other "countries"), we get a new graph $G^{\prime}$, in which all vertices will be even. Therefore, we can color the "countries" of $G^{\prime}$ in two colors (say,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,663 |
231. Henry's Walk. Henry went for a walk in the countryside sometime between 8 and 9 in the morning, when the hands of his clock were aligned. He arrived at his destination between 2 and 3 in the afternoon; at this time, the hands of his clock were pointing in exactly opposite directions. How long did Henry's walk last... | 231. The walk lasted 6 hours. Suppose we extended the hour hand in the opposite direction. If at the beginning of the walk the hour and minute hands were coincident, then after six hours the hour hand and its extension will simply have swapped places, while the minute hand, having completed exactly six full circles, wi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,664 |
232. Power series. Expand $\frac{1}{(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)}$ into a power series. | 232. We have the following chain of equalities:
\[
\begin{aligned}
& \frac{1}{(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)}=\frac{1-x}{1-x^{16}}= \\
& =(1-x)\left(1+x^{16}+x^{32}+x^{48}+\ldots\right)= \\
& =1-x+x^{16}-x^{17}+x^{32}-x^{33}+\ldots
\end{aligned}
\]
[M. S. Klamkin, M. M., 29, 53 (Sep... | 1-x+x^{16}-x^{17}+x^{32}-x^{33}+\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,665 |
233. Parallel lines in a triangle. Through an internal point $P$ of triangle $A B C$, draw lines parallel to its sides. In this case, each side will be divided into three segments.
Denote t... | 233. Considering the parallelograms and similar triangles depicted in the figure, we can form the following proportions: $a^{\prime}: a=b_{1}: b ; a^{\prime}: a=c_{2}: c ; b^{\prime}: b=c_{1}: c ; b^{\prime}: b=a_{2}: a$; $c^{\prime}: c=b_{2}: b ; c^{\prime}: c=a_{1}: a$. These equalities, together with the identities ... | \frac{^{\}}{}+\frac{b^{\}}{b}+\frac{^{\}}{}=1 | Geometry | proof | Yes | Yes | olympiads | false | 41,666 |
234. Representation of a number as a sum. Represent the number 316 as the sum of two addends, one of which is divisible by 13, and the other by 11. | 234. If we divide 316 by 11, we get a quotient of 28 and a remainder of 8; but $8:(13-11)=4$. Therefore, the two required addends are respectively $4 \cdot 13=52$ and 264. We can also represent 316 as the sum of the numbers $52+11 \cdot 13=195$ and $264-$ $11 \cdot 13=121$. | 195+121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,667 |
## 236. How old is Willie?
- Is this problem your teacher gave you? - I asked Willie. - It looks quite boring.
- No, - Willie replied, - I came up with it myself. - See this polynomial? My age is its root. I mean, if you substitute the number of years I was when I last celebrated my birthday for \( x \), the polynomia... | 236. Willie's friend, relying too much on the "trial and error" method, overlooked one fact that could have been very useful. The fact is that if we take a polynomial \( P(x) \) with integer coefficients and two different integers \( a \) and \( b \), then \( P(a) - P(b) \) will be divisible by \( a - b \). Let's denot... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,669 |
237. Accelerating Particle. A particle, initially at rest, begins to move along a straight line, covers some distance, and then stops; in total, it covers a unit distance and spends a unit of time doing so. Prove that at some moment in time, the acceleration of the particle reaches a value of at least 4 units. We assum... | 237. If we plot the curve of the dependence of $v$ on $t$ in Cartesian coordinates, the area under such a curve (1 square unit) coincides with the area of an isosceles triangle with a base of 1 and a height of 2.
.] | five\2-cent\stamps,\fifty\1-cent\stamps,\\eight\5-cent\stamps | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,671 |
239. Twenty Questions. I want to offer you a variation of the well-known game of "twenty questions," which works as follows. I think of a natural number, and you have to guess it by asking no more than twenty questions, to each of which the answer can only be "yes" or "no." What is the largest number I can think of so ... | 239. If one object is chosen from a finite set of objects, the best way for the questioner to determine it is as follows. At each step, it is necessary to find out whether the given object has a property that is possessed by exactly half of the "suspects." In this case, the number of "suspects" is reduced by exactly ha... | 2^{20} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,672 |
240. Inscribed Spheres. If the faces of a certain hexahedron are equilateral triangles, congruent to the faces of a regular octahedron, then what is the ratio of the radii of the spheres inscribed in these polyhedra?
. Further, we note that \(\frac{V_{t}}{V_{0}} = \frac{1}{4}\) (see the solution to problem 113), where \( V_{0} \) is the volume of the octahedron. From this, denoting the volume of the hexahedron by \( V_{h} \), we get \(\frac{V_{h}}{V_{0}}... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,673 |
242. Sum of squares of binomial coefficients. Find the sum of the squares of the coefficients in the expansion of $(a+b)^{n}$. | 242. If we select $n$ balls at random from $n$ red and $n$ black balls, the number of possible combinations is
$$
\begin{gathered}
C_{2 n}^{n}=C_{n}^{0} \cdot C_{n}^{n}+C_{n}^{1} \cdot C_{n}^{n-1}+C_{n}^{2} \cdot C_{n}^{n-2}+\ldots \\
\ldots+C_{n}^{n-1} \cdot C_{n}^{1}+C_{n}^{n} \cdot C_{n}^{0}
\end{gathered}
$$
But ... | C_{2n}^{n}=\sum_{i=0}^{n}(C_{n}^{i})^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,675 |
243. Christmas Cryptarithm. The Christmas greeting $A M E R R Y X M A S T O A L L^{*}$ is a cryptarithm where each letter represents a specific decimal digit, and each word replaces a square number. Decode this message, given that, in addition, the sum of the digits in each word also matches some square.
Translate the... | 243. Using the table of squares, it can be quickly determined that $A L L$ matches one of the numbers $100,144,400$ or 900, and $T O$ with 36 or 81.
The only four-digit square number whose sum of digits is a square and which has 1, 4, or 9 in the tens place is $7396=X M A S$; therefore, $A L L=$ 900. Since each letter... | 934225739681900 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,676 |
244. Centers of Gravity. Find the centers of gravity of a homogeneous semicircle and a homogeneous semicircular area. | 244. It is evident that the center of gravity of a homogeneous semicircle of radius $r$ lies on the perpendicular to the diameter, erected from the center of the circle, at a distance $y$ from this diameter. If we rotate our semicircle around the diameter, we will obtain a sphere. The first theorem of Pappus states tha... | \frac{2r}{\pi},\frac{4r}{3\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 41,677 |
245. Two Coinciding Triples. None of the three positive numbers $x, y, z$ is less than the minimum of the three positive numbers $a, b, c$. None of the numbers $x, y, z$ is greater than the maximum of the numbers $a, b, c$. It is also known that $x+y+z=a+b+c$ and $x y z=a b c$. Show that the set $\{x, y, z\}$ coincides... | 245. If $x+y+z=a+b+c$ and $x y z=a b c$, then
$$
\begin{gathered}
a b c(a b+b c+c a-x y-y z-z x)= \\
=b c(x-a)(y-a)(z-a)=c a(x-b)(y-b)(z-b)= \\
=a b(x-c)(y-c)(z-c)
\end{gathered}
$$
If none of the last three expressions were zero, then one of them would be positive and the other two negative, so they could not be equ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,678 |
246. Irrational Sum. Prove that the expression $\sum_{n=1}^{\infty} 6^{\frac{2-3 n-n^{2}}{2}}$ is irrational. | 246. The exponents in this expression form a sequence $-1, -4, -8, -13, -19, -26, \ldots$, where the difference between two consecutive terms increases by 1 with each step. If we write our expression in base 6, we get an infinite non-repeating fraction $0,1001000100001000001 .$. , which is clearly an irrational number.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,679 |
247. Six examinees. Suppose that during the exam, six students sit on one bench, with aisles on both sides. They finish the exam in a random order and leave immediately. What is the probability that someone will have to disturb one of the remaining five classmates to get to the aisle? | 247. Let $n$ students be sitting on a bench. Then the probability that the first one of them to finish the exam will not have to disturb anyone when he makes his way to the aisle is obviously $\frac{2}{n}$. Therefore, the probability that someone among 6 students will have to disturb someone among the other 5 is $1-\fr... | \frac{43}{45} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,680 |
248. Construction with a compass. Divide a circle into four equal parts using only a compass. | 248. Let's draw a circle of radius $r$ with center at point $O$. Using the same radius, make marks at points $A, B, C$, and $D$. In this case, the arcs $A B, B C$, and $C D$ will be equal to each other, and the segment $A D$ will coincide with a certain diameter of the given circle. From points $A$ and $D$ as centers, ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,681 |
249. Bonus Fund. In a certain institution, there was a bonus fund. It was planned to distribute it so that each employee of the institution would receive 50 dollars. But it turned out that the last person on the list would receive only 45 dollars. To maintain fairness, it was then decided to give each employee 45 dolla... | 249. In fact, $5 was seized from $95: 5=19$ people; therefore, the original amount of the fund was $20 \cdot 50-5=$ $=995$ dollars. | 995 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,682 |
250. The Court Mathematician's Salary. Once, the court mathematician received his entire annual salary in silver talers, from which he formed nine piles, together constituting a magic square. The king, upon seeing this, was impressed but regretfully noted that in none of the piles was the number of coins a prime number... | 250. As established by L. S. Fryer, the elements of an arbitrary magic square of the third order can actually be represented using three parameters:
$$
\begin{array}{ccc}
(e+x), & (e-x-y) & (e+y) \\
(e-x+y), & e, & (e+x-y) \\
(e-y), & (e+x+y) & (e-x)
\end{array}
$$
It is clear that these numbers can be rearranged so ... | 1350 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,683 |
251. Packing Cylinders. Forty cylinders with a diameter of 1 cm and the same height were tightly packed in a box in 5 rows of 8 cylinders each so that they would not "rattle" during transportation. How many cylinders need to be removed from the box so that, by moving the remaining cylinders and adding the removed cylin... | 251. Only two cylinders need to be removed from the box. Let's renumber the cylinders as shown in the figure. Remove cylinders 6 and 16.
Move cylinders $7-10$ to the right and up (see the figure). Move cylinders 11-15 to the left, cylinders 17-20 up and to the left, and cylinders 21-25 to the left. The remaining cylin... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,684 |
253. Powers of two. Calculate the sum
$$
2+2^{2}+2^{3}+2^{4}+\cdots+2^{n}
$$ | 253. In the binary numeral system, our sum is written using $n$ ones, to which one 0 is appended. If we add two ones to the sum, we get a number that is written in the binary system using one 1, followed by $n+1$ zeros. Therefore, the given sum is equal to $2^{n+1}-2$. | 2^{n+1}-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,686 |
255. Divided Parallelogram. Point $E$ is chosen on side $B C$, and point $F$ is on side $A D$ of parallelogram $A B C D$; $A E$ intersects $B F$ at point $G$, and $E D$ intersects $C F$ at point $H$. Prove that line $G H$ divides the parallelogram into two equal parts. | 255. Pappus' theorem states that if the vertices of a certain hexagon are arranged alternately on two lines, then
the points of intersection of the opposite sides of this hexagon are collin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,688 |
256. Palindromic Squares. A certain square is written in the base-6 numeral system using 5 different non-zero digits. If the digit in the units place is moved from the end of this number to the beginning, the square root of the resulting number matches the square root of the original number, written in reverse order. F... | 256. $12345 \leqslant N^{2} \leqslant 54321$, so $113 \leqslant N \leqslant 221$. If we write the digits of the number $N$ in reverse order, the resulting number will also satisfy these inequalities. The number $N^{2}$ is divisible by 5, so $N$ is also divisible by 5. Let's list all three-digit numbers divisible by 5 a... | 53241 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,689 |
257. Suspicious sum. Let $a$ and $b$ be integers. Can the sum $\frac{a}{b}+\frac{b}{a}$ be an integer? | 257. We can assume without loss of generality that $a$ and $b$ are coprime and $a>0$. If $\frac{a}{b}+\frac{b}{a}=k$, an integer, then the equation $a^{2}+b^{2}=a b k$ would hold, which means $b^{2}=a(b k-a)$. This would imply that $b^{2}$ is divisible by $a$; but this is only possible if $a= \pm b$. Therefore, our sum... | \b | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,690 |
258. Is the square stable? Can a stable square be formed from 18 dominoes $2 \times 1$? We will call a square stable if no straight line (other than the boundaries of the square, of course), formed by the edges of the dominoes, connects opposite sides of this square ${ }^{*}$. | 258. It is impossible to form a solid square using 18 dominoes. Consider a $6 \times 6$ square grid, which is formed by five vertical and five horizontal dividing lines. Each domino covers exactly two cells of this grid. We will cover the entire grid with 18 dominoes. The resulting square will be solid if and only if e... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,691 |
259. Coprime numbers. In which number systems are the numbers 35 and 58 coprime? | 259. Let the greatest common divisor of numbers $x$ and $y$ be denoted by $(x, y)$. Then
$$
(35,58)=(35,23)=(12,23)=(12,11)=(1,11)=1
$$
Therefore, 35 and 58 are coprime in any number system with a base $>8$.
[D. Silverman, M. M., 38, 326 (November 1965).] | 3558coprimeinanysystemwithbase>8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,692 |
260. Five integers. Do there exist 5 consecutive integers such that the sum of the fourth powers of the first four equals the fourth power of the fifth number? | 260. The fourth power of any even number can be represented as $4k$, and that of any odd number as $4k+1$. Consequently, the sum of the fourth powers of any four consecutive numbers has the form $4k+2$ and obviously cannot equal the fourth power of an integer. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,693 |
262. Integer solutions are sought. Find all integer solutions of the equation
$$
y^{2}+y=x^{4}+x^{3}+x^{2}+x
$$ | 262. Multiplying the given equation by 4 and then adding 1 to both sides, we get
$$
4 x^{4}+4 x^{3}+4 x^{2}+4 x+1=(2 y+1)^{2}
$$
For $x=-1$, $y=-1$ or 0; for $x=0$, $y=-1$ or 0; for $x=2$, $y=-6$ or 5; for $x=1$, $y$ will not be an integer.
These are the only 6 integer solutions of the given equation. Indeed, for $x... | -1,-1or0;0,-1or0;2,-6or5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,695 |
263. Composite number. Let $p_{1}$ and $p_{2}$ be two consecutive odd prime numbers, so that $p_{1}+p_{2}=2 q$. Show that $q-$ is a composite number. | 263. The number $q=\frac{\left(p_{1}+p_{2}\right)}{2}$ represents the arithmetic mean of the numbers $p_{1}$ and $p_{2};$ therefore, $p_{1}<q<p_{2}$. But $p_{1}$ and $p_{2}$ are consecutive prime numbers; hence, $q$ is a composite number.
$$
\text { [J. Baum, M. M., 39, } 196 \text { (Mau 1966).] }
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,696 |
264. System of four linear equations. Solve the following system of equations:
$$
\left\{\begin{aligned}
x+7 y+3 v+5 u & =16 \\
8 x+4 y+6 v+2 u & =-16 \\
2 x+6 y+4 v+8 u & =16 \\
5 x+3 y+7 v+u & =-16
\end{aligned}\right.
$$ | 264. If we interchange $x$ and $u$, $y$ and $v$ in equation (1), we will get equation (4), but with the opposite sign on the right-hand side. If we do the same in equation (2), we will get equation (3) with the same caveat. Therefore, $u=-x$ and $v=-y$. Substituting these equalities into (1) and (2), we get
$$
\begin{... | -2,2,v=-2,u=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,697 |
266. When division is exact. For which positive integer $n$ will the quantity $n^{4}+n^{2}$ be divisible by $2 n+1$? | 266.
$$
\begin{aligned}
f(n) & =\frac{n^{4}+n^{2}}{2 n+1}=\frac{n^{2}\left(n^{2}+1\right)}{2 n+1}=\frac{n^{2}}{4}\left[\frac{4 n^{2}+4}{2 n+1}\right]= \\
& =\left(\frac{n}{2}\right)^{2} \cdot\left[2 n-1+\frac{5}{2 n+1}\right]
\end{aligned}
$$
It is obvious that the greatest common divisor of the numbers $n$ and $2n+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,699 |
267. Dubious Equality. The statement «$342=97$» can be made valid by inserting several algebraic signs between the digits, for example $(-3+4) \cdot 2=9-7$. Can the equality be given meaning without inserting any signs? | 267. For such a statement to be true, it is necessary that these two numbers are written in different number systems, that is, $342_{a}=97_{b}$. If $b=10$, then since $3(4)^{2}=48$ and $3(6)^{2}=108, a=5$. Indeed, $3(5)^{2}+4(5)+2=97$.
In the general case, from $3 a^{2}+4 a+2=9 b+7$, it follows that $b=\frac{3 a^{2}+4... | 342_{5}=97_{10} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,700 |
268. Sliced Dodecagon. Cut a regular dodecagon into squares and equilateral triangles.
Let $P_{1}, P_{2}, \ldots, P_{12}$ be the consecutive vertices of a regular dodecagon. What can be said about the intersection of the diagonals $P_{1} P_{9}, P_{2} P_{11}$, and $P_{4} P_{12}$? | 268. Let's describe a circle around the given dodecagon. Then it will become obvious that nine diagonals emanating from any vertex divide the angle at the vertex, which is $150^{\circ}$, into 10 equal angles of $15^{\circ}$ each.
Draw each of the diagonals $P_{1} P_{6}, P_{2} P_{9}, P_{3} P_{8}, P_{4} P_{11}, P_{5} P_... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,701 |
269. No real roots. Show that the equation
$$
1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots+\frac{x^{2 n}}{(2 n)!}=0
$$
has no real roots. | 269. If a real root of the given equation existed, it would be a negative number, say, $(-y)$. But
$$
1-y+\frac{y^{2}}{2!}-\frac{y^{3}}{3!}+\cdots+\frac{y^{2 n}}{(2 n)!}>e^{-y}>0
$$
Therefore, the original equation has no real roots.
[J. Lipman, A. M. M., 67, 379 (April 1960).] | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,702 |
270. Impossible Cube. Prove that in no number system can a three-digit number $\overline{a a a}$ satisfy the relation $\overline{a a \bar{a}}=a^{3}$. | 270. Since $a r^{2}+a r+a=a^{3}, r^{2}+r+1=a^{2}$. Any digit $a$ in any base $r$ number system is always less than the base $r$ of that system. Therefore, the given equality cannot hold for any $a \neq 0$ and $r$.
[Ch. MacCracken, Jr., S. S. M., 52, 241 (March 1952).] | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,703 |
271. Diophantine system * ${ }^{*}$. Prove that the system
$$
\left\{\begin{aligned}
a+b+c & =x+y \\
a^{3}+b^{3}+c^{3} & =x^{3}+y^{3}
\end{aligned}\right.
$$
has infinitely many solutions in integers, with the property that $a, b, c$ form an arithmetic progression. | 271. Let $a=3 d, c=2 b-3 d$. Then $x+y=3 b$ and the second equation can be represented as
$$
(x-y)^{2}=(b-8 d)^{2}-40 d^{2}
$$
It is satisfied, for example, by
$$
x-y=p^{2}-10 q^{2}, \quad b-8 d=p^{2}+10 q^{2}, \quad d=p q
$$
Thus, we find the following two-parameter family of solutions, for which $a, b, c$ form an... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,704 |
272. Calculation of the sum. Let $n$ be a fixed positive number. Set $x_{0}=\frac{1}{n}$ and $x_{j}=\frac{1}{n-1} \sum_{i=0}^{j-1} x_{i}$ for $j=$ $1,2, \ldots, n-1$. Compute the sum $\sum_{j=0}^{n-1} x_{j}$. | 272. We will prove by induction that $\sum_{j=0}^{k} x_{j}=\frac{1}{(n-k)}, k=$ $0,1, \ldots, n-1$. For $k=0$, the given equality is trivially satisfied by definition. Suppose now that $\sum_{j=0}^{k} x_{j}=\frac{1}{(n-k)}$ for some $k, 0 \leqslant k \leqslant n-2$. Then
$$
\sum_{j=0}^{k+1} x_{j}=x_{k+1}+\sum_{j=0}^{k... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,705 |
273. Inequality in a Triangle. Let an acute-angled triangle $A B C$ be given. Choose an arbitrary point $P$ inside it. Show that $P A+P B+P C \geqslant \frac{2}{3} \times$ (perimeter of the triangle formed by the points of tangency of the incircle of $A B C$ with the sides of $A B C$). | 273. It is known that $P A+P B+P C \geqslant 6 r$, where $r$ is the radius of the inscribed circle in triangle $A B C^{\star}$. It is also known that among all triangles inscribed in a certain circle, the maximum perimeter will be for an equilateral triangle. Let $D, E, C$ be the points of tangency of the inscribed cir... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,706 |
274. Packing Squares. Let a set of squares be given, the total area of which is equal to 1. Show that they can be packed inside a square with a side length of $\sqrt{2}$, so that they do not overlap. (For a square with a side length $<\sqrt{2}$, this statement is not true.) | 274. First, we will order all the squares so that their sizes do not increase with the number. Next, we place the first square in the lower left corner of the square $S$ with side $\sqrt{2}$. To the right, we place the next square snugly against it and the lower boundary of $S$, and to the right of it, the next one, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,707 |
275. Maximum Number. Let a set of distinct complex numbers $z_{i}, i=1,2, \ldots, n$, be given, satisfying the inequality
$$
\min _{i \neq j}\left|z_{i}-z_{j}\right| \geqslant \max _{i}\left|z_{i}\right|
$$[^16]
Find the maximum possible $n$ and for this $n$ all sets satisfying the condition of the problem. | 275. Let $\left|z_{m}\right|=\max _{i}\left|z_{i}\right|$. Then on the complex plane, all points $z_{i}$ will be located inside the circle $R$ of radius $\left|z_{m}\right|$ centered at the point $z=0$. Clearly, 6 points $z_{i}$, located on the circumference of $R$ and forming a regular hexagon, together with the point... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 41,708 |
276. One non-multiplicative function. For any natural number $n$, denote by $r_{s}(n)$ the number of solutions to the equation
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{s}^{2}=n
$$
in integers $x_{1}, x_{2}, \ldots, x_{n}$. Let us further set $f_{s}=(2 s)^{-1} r_{s}(n)$. It is known that for $s=1,2,4,8$ the function $f_{s}$ i... | 276. We will show that $f_{s}(2) f_{s}(3) \neq f_{s}(6)$ for $s \neq 1,2,4,8$. Note that if $x_{1}, x_{2}, \ldots, x_{s}$ are integers satisfying the equation $x_{1}^{2}+x_{2}^{2}+\cdots+x_{s}^{2}=2$, then $(s-2)$ of them are 0, and the other two are $\pm 1$. Therefore,
$$
r_{s}(2)=4 C_{s}^{2}=2 s(s-1)
$$
Similarly,
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,709 |
277. Combinatorial problem. Find the number of ordered sets $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $n$ natural numbers, in which
$$
1 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} ; \quad a_{i} \leqslant i \quad(i=1,2, \ldots, n)
$$ | 277. Denoting the number of such sets by $A_{n}$ and setting $A_{0}=1$, we get for $n \geqslant 1$:
$$
A_{n}=\sum_{k=1}^{n} A_{k-1} \cdot A_{n-k} \cdot{ }^{\star}
$$
If $f(x)=\sum_{n=0}^{\infty} A_{n} x^{n}$, then from formula (1) it follows that $f(x)=$ $=1+x \cdot\{f(x)\}^{2}, f(0)=1$, so that
$$
f(x)=\frac{1-\sqr... | A_{n}=\frac{1}{n+1}C_{2n}^{n}=\frac{(2n!)}{n!(n+1)!} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,710 |
278. Another inequality in a triangle. Let the bisectors of the internal angles of triangle $T$ be $\beta_{a}, \beta_{b}, \beta_{c}$, its medians - $m_{a}, m_{b}, m_{c}$, and the radii of its inscribed and circumscribed circles be $r$ and $R$. Prove that
$$
\beta_{a}^{6}+\beta_{b}^{6}+\beta_{c}^{6} \leqslant p^{4}\lef... | 278. First, note that
$$
\begin{gathered}
p^{4}\left(p^{2}-12 r R\right)=p^{4}\left(p^{2}-12 \frac{S}{p} \cdot \frac{a b c}{4 S}\right)=p^{4}\left(p^{2}-\frac{3 a b c}{p}\right)= \\
=p^{3}\left(p^{3}-3 a b c\right)=p^{3}\left[(p-a)^{3}+(p-b)^{3}+(p-c)^{3}\right]
\end{gathered}
$$
where \( S \) is the area of triangle... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,711 |
279. Sequence of composite numbers. Prove that for any natural $n$ there exists a set of $n$ composite numbers forming an arithmetic progression, and such that all these numbers are pairwise coprime. | 279. Note that if $2 \leqslant k \leqslant N$, then $N!+k-$ is a composite number. For any given $n$, choose a prime number $p>n$ and an integer $N \geqslant p+(n-1) n!$. Then the composite integers $N!+p$, $N!+p+n!, \ldots, N!+p+(n-1) n!$ will form an arithmetic progression. Moreover, if $q$ is a common prime divisor ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,712 |
280. What formula? Find the formula for the general term of the sequence $\left\{x_{n}\right\}$, defined recursively:
$$
x_{n}=(n-1)\left(x_{n-1}+x_{n-2}\right), \quad n \geqslant 4, \quad x_{2}=1, \quad x_{3}=2
$$ | 280. Assuming $x_{k}=k!y_{k}$, we get
$$
y_{k}-y_{k-1}=-\frac{1}{k}\left(y_{k-1}-y_{k-2}\right), \quad k \geqslant 4
$$
Using (1) for $k=4,5, \ldots, m$, we get
$$
y_{m}-y_{m-1}=\frac{(-1)^{m}}{m!}, \quad m \geqslant 4
$$
Summing the equalities (2) for $m=4,5, \ldots, n$, we will finally arrive at the answer
$$
x_... | x_{n}=n!\sum_{=2}^{n}\frac{(-1)^{}}{!} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,713 |
281. Points on a segment. Let $0=x_{0}<x_{1}<\cdots<x_{n}=1-$ be points that divide the unit circle into $n$ parts (not necessarily equal). Prove or disprove the following statement: there exists a composite integer $m$ such that in each of the segments $\left[x_{i}, x_{i+1}\right], i=0, \ldots, n-1$ there is at least ... | 281. Let us choose a prime number $p$ such that $\frac{2}{p}<\min \left(x_{i+1}-x_{i}\right)$. Due to this choice, in each of the intervals $\left[x_{i}, x_{i+1}\right]$, there are at least two numbers of the form $\frac{k}{p}$. Let us denote any two such consecutive numbers by $\frac{k}{p}$ and $\frac{(k+1)}{p}$. Sinc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,714 |
282. The condition defining a parabola. Let points $P_{i}(i=0,1, \ldots, n)$ be given on the graph of a nonlinear function $f(x)$. Denote by $m_{i}$ the slope of the line connecting points
$P_{i}$ and $P_{i+1}$ (we assume $P_{n+1}=P_{0}$). Let further $f(x)$ be a function defined for all real $x$ and not identically co... | 282. If $f(x)=a x^{2}+b x+c$ for all $-\infty<x<+\infty$ and $a \neq 0$, then
$$
\begin{aligned}
\sum_{i=0}^{n}(-1)^{i} m_{i} & \left.\left.=\sum_{i=0}^{n}\left\{a\left(x_{i}+x_{i+1}\right)+b\right\}\right)-1\right)^{i}= \\
& = \begin{cases}0 & \text { for odd } n . \\
2 a x_{0}+b & \text { for even } n .\end{cases}
\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,715 |
283. Perfect Square. Find all integers $x, y, z$ for which the value $4^{x}+4^{y}+4^{z}$ is a perfect square. | 283. First, note that $x, y, z$ cannot be negative, as otherwise $4^{x}+4^{y}+4^{z}$ would not be an integer. If $x \leqslant y \leqslant z$, then from the fact that $4^{x}+4^{y}+4^{z}$ is a perfect square, it follows that there exist a positive integer $m$ and a positive odd number $t$ such that
$$
1+4^{y-x}+4^{z-x}=... | {x,y,2y-x-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,716 |
285. The Non-existent Triangle Umbugio. With his characteristic persistence, Professor Euclid Paracelsus Bombast Umbugio is trying to prove the following theorem: if in triangle $ABC$ the orthocenter ${ }^{*} H$, the incenter $I$ of the inscribed circle, and the circumcenter $O$ of the circumscribed circle form an equi... | 285. If triangle $H I O$ were equilateral, the following relations would hold:
$$
\begin{aligned}
& O I^{2}=O H^{2} \\
& O I^{2}=I H^{2}
\end{aligned}
$$
Let $R, \rho, r$ be the radii of the circumcircle, incircle, and the circle inscribed in triangle $H_{1} H_{2} H_{3}$, formed by the points of tangency of the incir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,718 |
286. Generalized Fermat numbers. Let $k$ be a non-zero integer. Show that the numbers $(2 k)^{2^{n}}+1$, where $n=1,2, \ldots$, are pairwise coprime. | 286. The following proof is a simple generalization of Pólya's idea, which can be used to show that there are infinitely many prime numbers. Let $F_{n}=(2 k)^{2^{n}}+1$ and let $m<n$. If $p-$ is a positive integer, then
$$
x^{2 p}-1=(x+1)\left(x^{2 p-1}-x^{2 p-3}-\ldots-x^{2}+x-1\right)
$$
Taking $x=(2 k)^{2^{m}}$ an... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,719 |
287. Sine of cosine. Prove that on the interval ( $0, \frac{1}{2} \pi$ ) there exists a unique pair of numbers $c<d$, such that
$$
\sin \cos c=c, \quad \cos \sin d=d
$$ | 287. Let $\theta(x)=\sin (\cos x)-x, 0 \leqslant x \leqslant \frac{1}{2} \pi$. Then $\theta(0)=$ $=\sin 1>0, \theta\left(\frac{1}{2} \pi\right)=-\frac{1}{2} \pi<0$, we get that $c<d$.
[D. H. Lehmer, AMM, 76, 558 (May 1969).]
SOLUTIONS | proof | Calculus | proof | Yes | Yes | olympiads | false | 41,720 |
288. Zero of a polynomial. Let $f(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}$, where $1 \geqslant a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0$. Furthermore, let $\lambda$ be a complex root of $f$, such that $|\lambda| \geqslant 1$. Prove that $\lambda$ is equal to some root of $1^{\star} .^{* *}$[^17] | 288. The number $\lambda$ is a root of the polynomial $(x-1) f(x)$; therefore
$$
\lambda^{n+1}=\left(1-a_{1}\right) \lambda^{n}+\left(a_{1}-a_{2}\right) \lambda^{n-1}+\ldots+\left(a_{n-1}-a_{n}\right) \lambda+a_{n}
$$
By the triangle inequality
$$
\begin{aligned}
|\lambda|^{n+1} & \leqslant\left(1-a_{1}\right)|\lamb... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,721 |
289. A remarkable number. Find such a number that its fractional part, its integer part, and the number itself form a geometric progression ${ }^{\star}$. | 289. Given the conditions $x(x-[x])=[x]^2$, hence $x=$ $=[x] \frac{(1+\sqrt{5})}{2}, |x| \geqslant |[x]|$ and $x \geqslant 0$.
Further, $[x]+1 \geqslant [x] \frac{1+\sqrt{5}}{2}$ and $[x] \leqslant \frac{2}{\sqrt{5}-1}<2$. Therefore, $[x]=0$ or 1, and $x=0$ or $\frac{1+\sqrt{5}}{2}$.
[D. Silverman, M. M., 43, 56 (Jan... | \frac{1+\sqrt{5}}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,722 |
290. Numbers divisible by 27. Show that the numbers $2^{5 n+1}+5^{n+2}$ are divisible by 27 for $n=0,1,2, \ldots$. | 290. To prove it, it is sufficient to note that
$$
2^{5 n+1}+5^{n+2}=2(27+5)^{n}+5^{n}(27-2)=27 k
$$
[A. Satchiff, M. M., 43, 56 (January 1970).] | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,723 |
291. A number not equal to the sum of squares. Prove that the number $3^{k}$ cannot be represented as the sum of two positive integer squares. | 291. We can assume without loss of generality that $k$ is the smallest integer for which $x^{2}+y^{2}=3^{k}$, and from the problem statement, it follows that $k>0$. Then $x^{2}+y^{2}$ is divisible by 3, which implies that $x$ and $y$ must be divisible by 3. Therefore, $x=3 m$, $y=3 n$ and $(3 m)^{2}+(3 n)^{2}=3^{k}$. B... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,724 |
292. Another inequality in a triangle. Prove that in any triangle \(ABC\), the inequality
$$
\frac{\sin A + \sin B + \sin C}{\sin A \cdot \sin B \cdot \sin C} \geq 4
$$
holds, and equality is achieved only in the case of an equilateral triangle. | 292. I. Let $R$ be the radius of the circumscribed circle. Then the side $a=2 R \sin A$ and so on, and the given inequality will take the form
$$
(a+b+c) R^{2} \geqslant a b c
$$
Since
$$
a+b+c=2 p, \quad a b c=4 R S
$$
where $p$ is the semiperimeter, and $S$ is the area of triangle $ABC$, we obtain from (1)
$$
R ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,725 |
294. An interesting equation. Let the equation $y^{x}=x^{y}$ be given. Answer the following questions:
a. What does the set of positive numbers $x$ represent, for each of which the only positive solution $y$ of the given equation is the trivial solution $y=x$?
b. If we consider those values of $x$ for which there are... | 294. Since $x^{y}=e^{y \ln x}$ for all positive $x$ and $y$, to compare $x^{y}$ with $y^{x}$, we need to compare $y \ln x$ and $x \ln y$. The function $e^{z}$ is strictly increasing. Therefore, $x^{y} > y^{x}$ if and only if $\frac{\ln x}{x} > \frac{\ln y}{y}$. Thus, to answer all questions in this problem, we need to ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,727 | |
295. Trigonometric inequality. Prove that for any set of real numbers $\left\{T_{i}\right\}$, where $i=1,2, \ldots, n$,
$$
\sum_{k, j=1}^{n} \cos \left(T_{k}-T_{j}\right) \geqslant 0
$$ | 295. Let's prove a more general inequality. Take arbitrary real numbers $T_{i}, x_{i}, i=1,2, \ldots, n$. Then
$$
\begin{aligned}
& \sum_{k, j=1}^{n} x_{k}\left[\cos \left(T_{k}-T_{j}\right)\right] x_{j}= \\
& =\sum_{k, j=1}^{n} x_{k}\left\{\cos T_{k} \cos T_{j}+\sin T_{k} \sin T_{j}\right\} x_{j}= \\
& =\left\{\sum_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,728 |
296. Similar Rhombuses. Prove that all rhombuses inscribed in a given rectangle are similar. | 296. I. We will prove a stronger statement: all rhombi "inscribed" in two mutually perpendicular pairs of parallel lines are similar.
First, note that the diagonals of all such rhombi intersect at point \( O \) - the center of the rectangle formed by the given pairs of lines. Let \( A B C D \) be a rhombus, the vertic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,729 |
297. Rational number. Form the number $x=x_{0}, x_{1} x_{2} x_{3} \ldots$ in the decimal system as follows. Let $x_{0}=1$, and $x_{n}$ be the smallest positive remainder obtained by dividing $x_{0}+x_{1}+\cdots+x_{n-1}$ by 9. Show that $x$ is rational. | 297. From the equality $9 n+x_{k+1}=x_{k}+x_{k-1}+\ldots+x_{1}+x_{0}=x_{k}+x_{k}+$ $+9 m$, it easily follows that $x_{k}$ is the remainder of the division of $2^{k-1} x_{0}=2^{k-1}$ by 9.
Let us now prove a more general result. Consider the sequence $\left\{x_{k+1}\right\}$ of remainders of the division of the numbers... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,730 |
298. Can or not? Can the quantities $\sqrt{\sin \theta}$ and $\sqrt{\cos \theta}$ simultaneously take rational values for some $\theta$ in the interval $\left(0, \frac{\pi}{2}\right) ?$ | 298. Let $\sqrt{\sin \theta}=\frac{a}{b}$ and $\sqrt{\cos \theta}=\frac{c}{d}$, where $a, b, c, d$ are positive integers; then
$$
\sin ^{2} \theta+\cos ^{2} \theta=\frac{a^{4}}{b^{4}}+\frac{c^{4}}{d^{4}}
$$
From this, using the identity
$$
\sin ^{2} \theta+\cos ^{2} \theta=1
$$
we obtain
$$
(b d)^{4}=(a d)^{4}+(b ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,731 |
299. All ones. Can a natural number, which in the decimal system is written using $6 k-1$ ones, be prime | 299. A natural number that is written with $6 k-1$ ones in the decimal system can be prime. For example, it is known that $\left(10^{23}-1\right): 9$, which is written with 23 ones, is prime.
Any number that is written with $q=6 k-1$ ones can be represented as $\left(10^{q}-1\right): 9$. Since $6 k-1$ is not divisible... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,732 |
300. Inscribed Polygons. A square and a triangle of equal area are inscribed in a certain semicircle, with one side of the triangle coinciding with the diameter of this semicircle. Show that the center of the circle inscribed in this triangle lies on one of the sides of the given square. | 300. I. Let the diameter of a semicircle with center at point $O$ be $2 R$. The side of the inscribed square is $2 \frac{\sqrt{5} R}{5}$. If $A$ and $B$ are the vertices of the square lying on the given semicircle, then the length of the perpendicular dropped from $A$ to $O B$ is $u=\frac{4 R}{5}$. Our triangle is obvi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,733 |
301. Tetrahedral Inequality. Let $n$ triangles have the same base and the sum of all their lateral sides is known. Prove that the sum of the heights of these triangles is maximal in the case when all triangles are isosceles and equal to each other ${ }^{*}$. | 301. If the base of a triangle and the sum of its two other sides are fixed, the triangle with the greatest height will be isosceles. Similarly, if the base is given and the sum of three segments is fixed, such that each pair can form the missing lateral sides of three triangles with the given base, the sum of the heig... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,734 |
302. Alternating Series. Prove that the alternating series
$$
\sum_{i=1}^{\infty}(-1)^{j+1} \ln \left(1+\frac{1}{i}\right)
$$
converges conditionally, and find its sum. | 302. Let's consider the partial sum of our series:
$$
S_{2 n}=\sum_{i=1}^{2 n}(-1)^{i+1} \ln \frac{i+1}{i}=\ln \left(\frac{\frac{2}{1} \cdot \frac{3}{4} \cdots \frac{2 n}{2 n-1}}{\frac{3}{2} \cdot \frac{5}{4} \cdots \frac{2 n+1}{2 n}}\right)
$$
Denoting by $\sigma_{2 n}$ the expression inside the parentheses under th... | \ln\frac{\pi}{2} | Calculus | proof | Yes | Yes | olympiads | false | 41,735 |
303. The problem of nine points. Let 9 arbitrary points be given in a unit square. Show that among all triangles whose vertices are located at these points, there is at least one whose area does not exceed $\frac{1}{8}$. Generalize this result. | 303. It is easy to show that in any rectangle, any three points define a triangle whose area does not exceed half the area of the given rectangle. (If necessary, we can first compress our rectangle so that the given triangle is inscribed in it. Then we can divide the resulting rectangle into smaller rectangles so that ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,736 |
304. A New Inequality. Show that for arbitrary real numbers $a_{i}>0$ and for any integers $M, P>0$, the following inequality holds:
$$
M \cdot\left(\sum_{i=1}^{M} a_{i}^{P}\right) \leqslant\left(\sum_{i=1}^{M} a_{i}^{P+1}\right)\left(\sum_{i=1}^{M} a_{i}^{-1}\right)
$$ | 304. For any real $\alpha$ and $\beta$, the function
$$
f(x)=\left(\sum_{i=1}^{M} a_{i}^{x-\alpha}\right)\left(\sum_{j=1}^{M} a_{j}^{\beta-x}\right)
$$
is convex, symmetric about the point $x_{0}=\frac{\alpha+\beta}{2}$, and thus increases as $x$ moves away from the point $x_{0}=\frac{\alpha+\beta}{2} \cdot{ }^{\star... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,737 |
305. Isosceles Triangle. Prove that if for the angles \(A, B, C\) of some triangle the following relation holds
\[
\operatorname{tg}(A-B)+\operatorname{tg}(B-C)+\operatorname{tg}(C-A)=0
\]
then this triangle is isosceles. | 305. Denoting $\operatorname{tg} A, \operatorname{tg} B$ and $\operatorname{tg} C$ by $a, b, c$ respectively and applying the formula for the tangent of a difference, we arrive at the relation
$$
\frac{a-b}{1+a b}+\frac{b-c}{1+b c}+\frac{c-a}{1+c a}=0
$$
After elementary transformations, we obtain the equality
$$
(a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,738 |
306. Octagon from two squares. Let there be two concentric squares, the corresponding sides of which are parallel, and the areas are in the ratio $2: 1$. Show that if we draw segments through the vertices of the smaller square, perpendicular to the corresponding diagonals, then these segments together with the sides of... | 306. The difference between the halves of the diagonals is $1-\frac{\sqrt{2}}{2}$, so the lengths of the segments drawn through the vertices of the smaller square are $2-\sqrt{2}$. These segments cut off $\frac{2-\sqrt{2}}{\sqrt{2}}$ from each side of the square at each corner. Therefore, the lengths of the parts remai... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,739 |
307. Divisibility by 30. Prove that for any integer $n$ the expression $\left(n^{5}-n\right)$ is divisible by 30. | 307. An integer and its fifth power always end in the same digit. Therefore, the value of $n^{5}-n$ ends in 0 and, thus, is divisible by 2 and 5. Further, let's factorize it into $(n-1) n(n+1)\left(n^{2}+1\right)$ and note that one of the first three factors must be divisible by 3. Thus, $n^{5}-n$ is divisible by $2 \t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,740 |
308. A simple "game". In the following cryptarithm
| $M A T H$ |
| :--- |
| $M A G S$ |
| $M A T H$ |
| GAME |
the word $G A M E^{*}$ is used instead of some prime number. Find this number. | 308. The letter $M$ must represent one of the elements of the set $\{1,2,3\}$, the letter $A$ one of the elements of the set $\{4,5,9\}$, from which it follows that the letter $G$ coincides with some element of the set $\{4,7,8\} . .^{*}$ Looking at the table of prime numbers, we will find that the only suitable prime ... | 8923 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,741 |
309. The ratio between heights and sides. Let $A B C$ be an acute-angled triangle. Prove that $\frac{1}{2}<\frac{h_{a}+h_{b}+h_{c}}{a+b+c}<$ 1, where $h_{a}, h_{b}, h_{c}$ are the altitudes, and $a, b, c$ are the sides of the given triangle.
* GAME (Eng. - game.) | 309. I. The heights of a triangle satisfy the relations $h_{a}=b \sin C, h_{b}=c \sin A$ and $h_{c}=a \sin B$. Therefore, $h_{a}+h_{b}+h_{c}=$ $b \sin C+c \sin A+a \sin Bb \\
x+y+y+z & >c \\
z+w+w+u & >a
\end{aligned}
$$
By adding all these inequalities, we get $2\left(h_{a}+h_{b}+h_{c}\right)>a+b+c$; hence,
$$
\frac... | \frac{1}{2}<\frac{h_{}+h_{b}+h_{}}{+b+}<1 | Inequalities | proof | Yes | Yes | olympiads | false | 41,742 |
310. Volume of a parallelepiped. Calculate the determinant
$$
\left|\begin{array}{ccc}
a^{2}+b^{2}-c^{2}-d^{2} & 2 b c-2 a d & 2 b d+2 a c \\
2 b c+2 a d & a^{2}-b^{2}+c^{2}-d^{2} & 2 c d-2 a b \\
2 b d-2 a c & 2 c d+2 a b & a^{2}-b^{2}-c^{2}+d^{2}
\end{array}\right|
$$ | 310. I. The determinant is equal to the oriented volume of the parallelepiped constructed on vectors, the projections of which in the Cartesian coordinate system coincide with the rows of this determinant. The square of the length of the first vector (and, by symmetry, the square of the length of each of the other two ... | (^{2}+b^{2}+^{2}+^{2})^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,743 |
311. Trigonometric limit. Prove that the limit
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n-1}\left(\frac{1}{n} \operatorname{tg} \frac{k \pi}{2 n}\right)^{p}
$$
exists for $p>1$ and does not exist for $p \leqslant 1$. | 311. Let
$$
S(n, p)=\sum_{k=1}^{n-1}\left(\frac{1}{n} \tan \frac{k \pi}{2 n}\right)^{p}=\sum_{k=1}^{n-1}\left(\frac{1}{n} \cot \frac{k \pi}{2 n}\right)^{p}
$$
For \(0 < p < 1\), we obtain
$$
\lim _{n \rightarrow \infty} S(n, p)=\sum_{k=1}^{\infty} \lim _{n \rightarrow \infty}\left(\frac{1}{n} \cot \frac{\pi k}{2 n}\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 41,744 |
312. Intersection of Great Circles. Prove that if $n$ distinct great circles on a sphere intersect in more than two points, then they intersect in at least $2 n$ points ($n \geqslant 2$; all possible points common to at least two of the circles are considered). | 312. If at each point no more than two great circles intersect, then the number of intersection points of $n$ such circles is maximal. Each pair of great circles gives 2 intersection points, so the maximum number of these points is $2\left(\frac{n(n-1)}{2}\right)=n(n-1)$.
If we take some great circle and rotate it unt... | 2n | Geometry | proof | Yes | Yes | olympiads | false | 41,745 |
313. Hippie and Geometry. Mr. Hippie, a zealous seeker of truth, dozed off slightly during a lesson. Waking up, he heard the geometry teacher say that by connecting the midpoints of the sides of any quadrilateral, one can obtain a parallelogram. Mr. Hippie, so as not to be outdone by anyone in the art of building hypot... | 313. Let's denote the coordinates of the successive vertices of the given quadrilateral, moving clockwise, as $\left(x_{1} y_{1}\right)$, $\left(x_{2} y_{2}\right)$, $\left(x_{3} y_{3}\right)$, and $\left(x_{4} y_{4}\right)$. Then the points dividing the sides of the quadrilateral in the ratio $r$ will have coordinates... | \frac{1}{24} | Geometry | proof | Yes | Yes | olympiads | false | 41,746 |
314. Points in a Triangle. Let a finite number of points be given inside a certain triangle. Connect these points to each other and to the vertices of the triangle so that the resulting segments do not intersect and divide the entire triangle into smaller triangles. Show that the number of such small triangles is alway... | 314. Let $n$ be the number of small triangles, and $e$ be the number of their sides located inside the original triangle. Out of the $3 n$ sides of all $n$ small triangles, 3 sides belong to the original triangle, and the remaining sides lie inside this triangle, each being counted twice. Therefore, $e=\frac{1}{2}(3 n-... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,747 |
315. Chords in a Sphere. Let three mutually perpendicular chords $A P B, C P D$, and $E P F$ be given in a certain sphere, passing through one point. Determine the radius of the sphere, given that $A P=2 a, B P=2 b, C P=2 c, D P=2 d, E P=2 e$, and $F P=2 f$. | 315. If we choose a rectangular coordinate system, the axes of which are along the given chords, then the center of the sphere will be located at the point with coordinates ( $b-a, d-c, f-e$ ), where we can assume without loss of generality that $b \geqslant a, d \geqslant c$ and $f \geqslant e$. Then we will obtain[^3... | R^{2}=^{2}+b^{2}+^{2}+^{2}+e^{2}+f^{2}-2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,748 |
316. A Glazed Pastry. A pastry made in the shape of a triangular prism with a very small height is covered with a thin layer of glaze on all sides. How should this pastry be divided among 11 friends so that each gets an equal amount of both the pastry and the glaze? | 316. Divide the triangular perimeter into 11 equal parts, and then draw cuts along the lines connecting the center of the inscribed circle with the resulting division points. This problem was first posed by H. S. M. Coxeter for a square. This method also applies to the case of any polygon in which a circle can be inscr... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,749 |
317. Composite values of a polynomial. Show that the polynomial $4 x^{3}+6 x^{2}+4 x+1$ takes values that are composite numbers for all natural $x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 317. To prove this, note that
\[
\begin{gathered}
4 x^{3}+6 x^{2}+4 x+1=(x+1)^{4}-x^{4}= \\
=\left[(x+1)^{2}-x^{2}\right]\left[(x+1)^{2}+x^{2}\right]=(2 x+1)\left[(x+1)^{2}+x^{2}\right]
\end{gathered}
\]
[N. Shaumberger, M. M., 44, 55 (January 1971).] | Combinatorics | MCQ | Yes | Yes | olympiads | false | 41,750 | |
318. A curious product. Prove that
$$
\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \ldots \cos \frac{7 \pi}{15}=\frac{1}{128}
$$ | 318. Let
$$
c=\prod_{i=1}^{7} \cos \frac{i \pi}{15} \quad \text { and } \quad s=\prod_{i=1}^{7} \sin \frac{i \pi}{15}
$$
Then
$$
2^{7} c s=\prod_{i=1}^{7} 2 \cos \frac{i \pi}{15} \cdot \sin \frac{i \pi}{15}=\prod_{i=1}^{7} \sin \frac{2 i \pi}{15}
$$
But
$$
\sin \frac{8 \pi}{15}=\sin \left(\pi-\frac{7 \pi}{15}\righ... | \frac{1}{128} | Algebra | proof | Yes | Yes | olympiads | false | 41,751 |
319. A boy, a girl, and a dog. The boy walks at a speed of 4 km/h, the girl at a speed of 3 km/h, and the dog runs at a speed of 10 km/h. At some point in time, the boy and the girl start moving along the road from the same point and in the same direction, while the dog starts running back and forth between them all th... | 319. We place the dog at an arbitrary point between the positions that the boy and the girl will occupy in an hour, and turn its head in one of the two directions. Then we "rewind the entire film" in reverse, until the boy, the girl, and the dog are all together at the starting point at the initial moment. Thus, the do... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,752 |
320. Divisor. For what values of $N$ does the number $N^{2}-71$ divide by $7 N+55$? | 320. If $\frac{N^{2}-71}{7 N+55}=M$ is an integer, then
$$
N^{2}-7 M N-(55 M+71)=0
$$
When we solve this quadratic equation for $N$, we will find that the expression under the radical must be a perfect square. Since the inequalities hold
$$
\begin{aligned}
& (7 M+15)^{2}=49 M^{2}+210 M+225<49 M^{2}+220 M+284< \\
& \... | N=57orN=-8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,753 |
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