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 |
|---|---|---|---|---|---|---|---|---|---|
92. A phone call with Xinjiang. One person expected to be given a phone call with Xinjiang. Bored, he started writing down the number $0.12345 \ldots$, where the $n$-th digit in the decimal expansion is $n$. Being meticulous, the person quickly performed the necessary calculations and recorded the number in standard fo... | 92. Since
\[
\begin{aligned}
S & =0.1+0.02+0.003+0.0004+\ldots \\
0.1 S & =\quad 0.01+0.002+0.0003+\ldots \\
0.9 S & =0.1+0.01+0.001+0.0001+\cdots= \\
& =\frac{0.1}{1-0.1}=\frac{1}{9}
\end{aligned}
\]
[^27]it follows that, \( S=\frac{10}{81}=0.123456790123456790 \ldots \) - this is a repeating decimal, which does not ... | S=\frac{10}{81}=0.123456790123456790\ldots | Number Theory | proof | Yes | Yes | olympiads | false | 41,525 |
94. A system of five linear equations. Solve the following system of equations:
$$
\left\{\begin{array}{l}
x+y+z+u=5 \\
y+z+u+v=1 \\
z+u+v+x=2 \\
u+v+x+y=0 \\
v+x+y+z=4
\end{array}\right.
$$ | 94. By adding all five equations and dividing the result by 4, we get
$$
x+y+z+u+v=3
$$
Now it only remains to subtract each of the original equations from this equation in turn to obtain the desired solution $v=-2$, $x=2, y=1, z=3, u=-1$. | v=-2,2,1,3,u=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,527 |
95. Almost universal theorem. Formulate a theorem that is valid for any integer $n$, except for $n=5, 17$ and 257. | 95. The theorem states that $(n-5)(n-17)(n-257) \neq 0$.
[L. Moser, M. M., 25, 49 (September 1951).] | (n-5)(n-17)(n-257)\neq0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,528 |
96. Trisection of an Angle. In triangle $ABC$, segments $BD$ and $BE$ divide angle $B$ into three equal parts, and $CD$ and $CE$ divide angle $C$ into three equal parts. $E$ is the point closer to side $BC$. Prove that angle $BDE$ is equal to angle $EDC$. | 96. Since $E$ is the point of intersection of two of the angle bisectors of triangle $BCD$, segment $DE$ is the third angle bisector.
[K. F. Pinsk, M. M., 34, 182 (January 1961).] | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,529 |
97. The Fibonacci sequence. Consider the Fibonacci sequence $1,1,2,3,5,13, \ldots$, whose terms $F_{n}$ (Fibonacci numbers) satisfy the relation $F_{n+2}=F_{n}+F_{n+1}$, $F_{1}=F_{2}=1$. Now consider the sequence of digits in the units place of the Fibonacci numbers. Will this sequence be cyclic, that is, can it be obt... | 97. Yes, it is possible. Each member of the desired sequence $11235831 \ldots$ can be obtained by adding the previous two members and taking the digit in the units place of this sum.
In this sequence, two odd and one even members alternate. There are $5 \cdot 5=25$ ordered pairs composed of odd digits. Therefore, afte... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,530 |
98. Two cubic equations. Let $a, b, c$ be the roots of the equation $x^{3} + q x + r = 0$. Write the equation whose roots are the numbers
$$
\frac{b+c}{a^{2}}, \quad \frac{c+a}{b^{2}}, \quad \frac{a+b}{c^{2}}
$$ | 98. Since the coefficient of $x^{2}$ in the original equation is 0, it follows that $a+b+c=0$. Therefore, $b+c=-a, c+a=-b$ and $a+b=-c$. In other words, we need to find the equation whose roots are $-\frac{1}{a},-\frac{1}{b}$ and $-\frac{1}{c}$ - the reciprocals of the roots of the original equation, taken with the opp... | rx^{3}-^{2}-1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,531 |
99. Cryptarithm-product. The product of three consecutive even numbers is $87_{* * * * *} 8$. Find these numbers and fill in the blanks in the given product. | 99. None of the digits in the units place of the factors can be zero. $4 \cdot 6 \cdot 8=192$, and $2 \cdot 4 \cdot 6=48$. $(87)^{\frac{1}{3}}=4.4 \ldots$, and $(88)^{\frac{1}{3}}=4.4 \ldots$. Therefore, the product $442 \cdot 444 \cdot 446=87526608$
$[M$. M., 37, 360 (November 1964).] | 442\cdot444\cdot446=87526608 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,532 |
102. Deck of Cards. The cards in the deck are sequentially numbered from 1 to 7, and then thoroughly shuffled. Five cards are randomly drawn sequentially from the deck. What is the probability that the numbers on these cards will be in increasing order? | 102. We are only interested in the order of the drawn cards. There are 5! permutations of five numbers in total; therefore, the desired probability is $\frac{1}{120}$.
[ Дж. Р. Ziegler, M. M., 23, 278 (May 1950).]
Note: The reference at the end is kept in its original form as it appears to be a citation in a non-Engl... | \frac{1}{120} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,535 |
103. Midperpendicular. Prove that the perpendicular erected from the midpoint of the segment connecting the bases of two altitudes of a triangle bisects the third side of this triangle into two equal parts. | 103. A side of the triangle is the diameter of the circle passing through the feet of the altitudes dropped on the other two sides of this triangle. The segment connecting these two feet is a chord of this circle. Therefore, the perpendicular erected from the midpoint of our segment will pass through the center of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,536 |
104. Divisibility condition. For which integer $a$ does the polynomial $x^{13}+x+90$ divide by $x^{2}-x+a ?$ | Let $f(x)=x^{2}-x+a, g(x)=x^{13}+x+90$. Then $f(0)=$ $a, f(1)=a, g(0)=90, g(1)=92$. Therefore, the greatest common divisor of 90 and 92, which is 2, must divide $a$. Further, $f(-1)=$ $a+2, g(-1)=88$; hence $a$ is neither 1 nor $-2 ; f(-2)=a+6$, $g(-2)=-8104$, so $a \neq-1$. Therefore,
$$
\begin{gathered}
a=2^{\star} ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,537 |
105. The Farmer's Task. A certain farmer must buy 100 heads of cattle for 100 dollars. If each calf costs 10 dollars, each lamb 3, and each piglet 0.5 dollars, then how many calves, lambs, and piglets will the farmer buy? | 105. The average cost of one head of cattle is 1 dollar. The cost of each calf differs from the average by +9 dollars, each lamb by +2 dollars, and each piglet by $-\frac{1}{2}$ dollar. Therefore, for each calf, the farmer must buy 18 piglets, and for each lamb, 4 piglets. Consequently, since $5(1+18)+(1+4)=100$, he mu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,538 |
106. Prime factorization. Find the prime divisors of the number 1000027. | 106. $1000027=(100)^{3}+(3)^{3}=(100+3) \cdot(10000-300+9)=$ $103 \cdot 9709=103 \cdot 7 \cdot 1387=103 \cdot 7 \cdot(1460-73)=103 \cdot 7 \cdot 73 \cdot 19$. | 103\cdot7\cdot73\cdot19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,539 |
108. Unique Square. What square is equal to the product of four consecutive odd numbers? | 108. If $n(n+2)(n+4)(n+6)=m^{2}$, then $\left(n^{2}+6 n+4\right)^{2}=m^{2}+16$. However, among the squares, only 0 and 9 have the form $a^{2}-16$; and since $m^{2}$ is odd, the sought square is $9=(-3)(-1)(1)(3)$.
[D. L. Silverman, M. M., 38, 60 (January 1965).] | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,541 |
109. Strict Inequality. Show that $n^{n}>1 \cdot 3 \cdot 5 \cdot$ $7 \cdot \ldots \cdot(2 n-1)$. | 109. The arithmetic mean of a set of numbers, not all of which are equal, is greater than their geometric mean; therefore,
$$
n=\frac{n^{2}}{n}=\frac{1+3+5+7+\cdots+(2 n-1)}{n}>[1 \cdot 3 \cdot 5 \cdot 7 \ldots(2 n-1)]^{\frac{1}{n}}
$$
Therefore, \( n^{n}>1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots(2 n-1) \).
[F. E. Néme... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 41,542 |
110. Sum of Cosines. Calculate the sum
$$
\cos 5^{\circ}+\cos 77^{\circ}+\cos 149^{\circ}+\cos 221^{\circ}+\cos 293^{\circ}
$$ | 110. Project the sides of an arbitrary polygon onto a line lying in the plane of this polygon. Then the sum of such projections, taken with the appropriate sign, is zero. Now let's take a regular pentagon with a unit side and note that its exterior angle is $72^{\circ}$. The terms
$$
$[M . M ., \mathbf{2 6}, 286($ May 1953)].$ | 9(91) | Number Theory | proof | Yes | Yes | olympiads | false | 41,544 |
112. Triangular numbers in the nonary system. Show that each member of the infinite sequence $1,11,111$, $1111, \ldots$ represents a triangular number*, written in the nonary (base 9) numeral system. | 112. Triangular numbers have the form $\frac{n(n+1)}{2}$, so 1 is a triangular number in any base. We will proceed by induction. It is easy to see that each term of the original sequence can be obtained by multiplying the previous term by the base of the given system and then adding 1. If we operate in base 9 and some ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,545 |
113. The ratio of the volumes of two polyhedra. For a regular tetrahedron and an octahedron with equal edge lengths, find the ratio of their volumes without calculating the volume of each of the given polyhedra.
^{3}\) of the volume of the larger tetrahedron (since their edges are in the ratio of \(1:2\)). The volume of four such smaller... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,546 | |
115. Simple multiplication. Simplify, as much as possible, the fraction $\frac{116690151}{427863887}$. | 115. The sum of the digits of the numerator is 30, so it is divisible by 3. The difference between the sums of the alternating digits of the denominator is $32-21=11$; hence, the denominator is divisible by 11. Factoring out these factors, we get
$$
\frac{116690151}{427863887}=\frac{38896717 \cdot 3}{38896717 \cdot 11... | \frac{3}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,548 |
116. Sum of Sines. Prove that in any triangle, the sum of the sines of all its angles does not exceed $\frac{3 \sqrt{3}}{2}$, and equality is achieved only in the case of an equilateral triangle. | 116. Suppose that the maximum, which obviously exists*, is achieved for two unequal angles $A$ and $B$. Then the sum of the sines is
$$
\begin{gathered}
\sin A+\sin B+\sin \left[180^{\circ}-(A+B)\right]= \\
=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin (A+B)
\end{gathered}
$$
But this expres... | \frac{3\sqrt{3}}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 41,549 |
117. Two Ferries. Two ferries travel between two opposite banks of a river at constant speeds. Upon reaching a bank, each immediately starts moving in the opposite direction. The ferries departed from opposite banks simultaneously, met for the first time 700 meters from one of the banks, continued to their respective b... | 117. Ferry $A$ departs from the shore, travels 700 m, and meets ferry $B$. By this time, they have covered a total distance equal to the width of the river. $A$ continues to the opposite shore, turns around, and after turning, travels another 400 m, meeting $B$ again. By this time, they have covered a total distance eq... | 1700 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,550 |
118. At lunch. Albert and Bertha Jones have five children: Christina, Daniel, Elizabeth, Frederick, and Grace. The father decided that during a certain cycle of lunches, all family members would sit at a round table in a new arrangement each day, and over the entire cycle, each family member should sit next to every ot... | 118. There are $6: 2=3$ different pairs between which Albert must sit; therefore, the cycle consists of three dinners. Let's denote each family member by the initial letter of their name, arrange the entire family in order, and, breaking the circle to the left of Albert, write down all the initials in a row. Then we ta... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,551 |
120. Три водителя. Три водителя грузовиков зашли в придорожное кафе. Один водитель купил четыре сандвича, чашку кофе и десять пончиков на общую сумму 1 доллар 69 центов. Второй водитель купил три сандвича, чашку кофе и семь пончиков за 1 доллар 26 центов. Сколько заплатил третий водитель за сандвич, чашку кофе и пончик... | 120. Используя данные относительно покупок первых двух водителей, можно составить два уравнения:
$$
\begin{aligned}
4 s+c+10 d & =169 \\
3 s+c+7 d & =126
\end{aligned}
$$
где $s, c$ и $d$ обозначают соответственно стоимость в центах одного сандвича, одной чашки кофе и одного пончика ${ }^{*}$. Умножим первое уравнени... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,553 |
122. Without expanding the brackets. Solve the following equation without expanding the brackets:
$$
(12 x-1)(6 x-1)(4 x-1)(3 x-1)=5 \text{. }
$$ | 122. Substituting $x=\frac{y}{12}$ into the original equation, we get
$$
(y-1)(y-2)(y-3)(y-4)=120=2 \cdot 3 \cdot 4 \cdot 5
$$
Any integer root of this equation will turn its left side into the product of four consecutive integers. $y=-1$ and $y=6$ are two such roots.
Now let's use the known relationships between th... | -\frac{1}{12},\frac{1}{2},\frac{5+i\sqrt{39}}{24},\frac{5-i\sqrt{39}}{24} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,555 |
123. Problem with prime numbers. The product of a certain three-digit number by a certain two-digit number has the form
| $p p p$ |
| ---: |
| $p p$ |
| $p p p p$ |
| $p p p p$ |
| $p p p p p$ |
Here the letter $p$ can represent any prime digit other than one (not necessarily the same one). Restore all the numbers an... | 123. We need to find a three-digit and a one-digit number, the product of which is a four-digit number, and only the digits $2,3,5$ or 7 are allowed. There are only four possible solutions: $3 \cdot 775=2325, 5 \cdot 555=2775$, $5 \cdot 755=3775$ and $7 \cdot 325=2275$.
Since none of the three-digit numbers in these f... | 25575 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,556 |
124. Intersection of Great Circles. It is known that on a sphere, $n$ great circles intersect in $n(n-1)$ points in general. How should the numbers $1,2, \ldots, n(n-1)$ be placed in these points so that the sums of the numbers located on each circle are equal to each other. (Recall that the great circles lie in planes... | 124. Place the number $b$ at some point, and at the point located at the opposite end of the corresponding diameter, place the "complementary" number, which is $n(n-1)+1-b$. Continue
this ... | [n(n-1)+1](n-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,557 |
125. Dear club. Ten people decided to found a club. If there were 5 more people, each would contribute 100 dollars less. How much money did each member contribute? | 125. If the original number of participants increases by $50 \%$, then each person's share is only $\frac{2}{3}$ of the previous individual contribution. Thus, 100 dollars constitute $\frac{1}{3}$ of the original contribution, and each club member paid 300 dollars.
$[$ [ M. M., 32, 229 (March 1959).] | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,558 |
126. Binomial coefficients. Find the largest value of $y$ such that in the expansion of some binomial, $y$ consecutive coefficients are in the ratio $1: 2: 3: \ldots: y$. Determine this expansion and list the corresponding coefficients. | 126. For the condition of the problem to be satisfied at $y=3$, for some $n$ and $k$, the equalities $2 C_{n}^{k}=C_{n}^{k+1}$ and $3 C_{n}^{k}=C_{n}^{k+2}$ must hold. After simplification, we get $n=3 k+2$ and $3(k+1)(k+2)=$ $(n-k)(n-k-1)$. Solving this system of equations and discarding the negative solution, we get ... | 3,n=14,k=4,coefficients:1001,2002,3003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,559 |
127. A number that is divisible by 8640. Show that for any integer $x$ the number
$$
x^{9}-6 x^{7}+9 x^{5}-4 x^{3}
$$
is divisible by 8640. | 127. The product of $n$ consecutive integers is divisible by $n$. Moreover, the product of four consecutive integers is divisible by $2^{3}$. Further,
$$
\begin{aligned}
N & =x^{9}-6 x^{7}+9 x^{5}-4 x^{3}= \\
& =[(x-2)(x-1) x][(x-1) x(x+1)][x(x+1)(x+2)]= \\
& =[(x-2)(x-1) x(x+1)(x+2)][(x-1)x][x(x+1)]= \\
& =[(x-2)(x-1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,560 |
129. Infinite Product. Calculate the following infinite product: $3^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 27^{\frac{1}{27}} \ldots$. | 129. Let $N=3^{\frac{1}{3}} \cdot 3^{\frac{2}{9}} \cdot 3^{\frac{3}{27}} \ldots=3^{\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots+\frac{n}{3^n}+\ldots}=3^{M}$. Then
$$
\frac{M}{3}=\frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+\cdots+\frac{n-1}{3^{n}}+\cdots
$$
Therefore,
$$
\left(1-\frac{1}{3}\right) M=\frac{1}{3}+\f... | \sqrt[4]{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,562 |
130. Impossible Square. Prove that for no positive integer $n$ the number $n^{4}+2 n^{3}+2 n^{2}+2 n+1$ can be a perfect square. | 130. We have $\left(n^{2}+n\right)^{2}=n^{4}+2 n^{3}+n^{2}<n^{4}+2 n^{3}+2 n^{2}+n+1<$ $n^{4}+2 n^{3}+3 n^{2}+2 n+1=\left(n^{2}+n+1\right)^{2}$. Therefore, our number is strictly between two consecutive squares. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,563 |
132. Determinants of nine digits. Nine positive digits can be arranged in the form of a third-order determinant in 9! ways. Find the sum of all such determinants. | 132. If we interchange two adjacent rows of a determinant, it will change its sign. If we permute the rows of a certain third-order determinant in all possible ways, we will obtain three positive and three negative determinants, equal to each other in absolute value. Consequently, our 9! determinants break down into $\... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,565 |
133. Six common points. Find six common points of the curves with equations
$$
2 x^{2}+3 x y-2 y^{2}-6 x+3 y=0
$$
and
$$
3 x^{2}+7 x y+2 y^{2}-7 x+y-6=0
$$ | 133. If two second-order curves have more than $2 \cdot 2=4$ common points, then their equations must be degenerate and have a common divisor. The two given equations can be written as
\[
\begin{gathered}
(x+2 y-3)(2 x-y)=0 \\
(x+2 y-3)(3 x+y+2)=0
\end{gathered}
\]
Therefore, all points on the line $x+2 y-3=0$ belong... | (-1,2),(1,1),(0,\frac{3}{2}),(3,0),(4,-\frac{1}{2}),(5,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,566 |
134. Shuffled Deck. Prove that if in a shuffled standard deck of cards, among the top 26 cards there are more red cards than there are black cards among the bottom 26 cards, then in this deck there are at least 3 cards of the same color in a row. | 134. The number of red cards in the upper half of the deck necessarily matches the number of black cards in its lower half. Therefore, the initial premise is always false, and then according to the laws of logic, any statement follows from it.
$$
\text { [L. Moser, M. M., 26, } 167 \text { (January 1953).] }
$$
$[\ll... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 41,567 |
135. Equal Angles. In an acute-angled triangle $A B C$, drop the altitude $A H$. Choosing an arbitrary point $D$ on $A H$, draw the line $B D$ until it intersects side $A C$ at point $E$. Then draw the line $C D$ until it intersects side $A B$ at point $F$. Prove that angle $A H E$ is equal to angle $A H F$.
=5
$$
Considering the first equation, we find from here the desired value $k=4$. For this $k$ we have $x=\frac{1}{3}, y=\frac{2}{3}$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,569 |
137. Diophantine equation. Prove that for any positive integer $a$ the equation $x^{2}-y^{2}=a^{3}$ is solvable in integers. | 137. Let $x+y=a^{2}, x-y=a$, from which $x=\frac{a(a+1)}{2}$ and $y=\frac{a(a-1)}{2}$. Since for any integer $a$ the numerator of each of these fractions is the product of an even and an odd number, the $x$ and $y$ defined in this way are integers and satisfy the original equation.
[L. E. B., A. M. M., 61, 548 (Octobe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,570 |
138. Dividing one triangle into two similar ones. Show that any given triangle can be divided into four parts by straight cuts, from which it is then possible to form two triangles similar to the given one. | 138. Let's choose points \(D, E\), and \(F\) on the sides \(AB, BC\), and \(CA\) of the given triangle \(ABC\) respectively, such that \(AD: AB = CE: CB = AF: FC = 1: 5\). Then, take a point \(G\) on side \(AC\) such that \(AG = 2AF\), and mark the midpoint \(H\) on \(DE\). Now, we need to cut the triangle along the li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,571 |
139. Moving Digits. A certain number, which is written using fewer than 30 digits, starts (if we move from left to right) with the digits 15, that is, it has the form $15 \ldots$. If we multiply it by 5, the result can be obtained simply by moving these two digits to the right end; in the end, we get a number of the fo... | 139. Let $f$ be a proper fraction, in the decimal periodic expansion of which one period coincides exactly with the initial number $15 \ldots$ By the conditions of this problem, $5 f=$ $0, \ldots 15 \ldots 15 \ldots$, and $100 f=15, \ldots 15 \ldots 15$, from which $95 f=15$ and $f=\frac{3}{19}$. Expanding $\frac{3}{19... | 157894736842105263 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,572 |
140. Vertices of a tetrahedron. Show that any (not necessarily regular) tetrahedron has at least one vertex where all dihedral angles are acute. | 140. If a plane angle at one of the vertices of a tetrahedron is right or obtuse, then the sum of all plane angles at this vertex is greater than $\pi$ radians*. If at least one of the angles at each vertex of the tetrahedron were right or obtuse, then the sum of all plane angles of the tetrahedron would exceed $4 \pi$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,573 |
141. Happy Prisoners. A certain jailer, carrying out a partial amnesty, proceeded as follows. First, he opened all the cells. Then he locked every second cell (in this prison all cells were arranged in a single row). On the third stage, he turned the key in every third cell (with each turn of the key, an open cell is l... | 141. Let the key in cell number $q$ be turned $t$ times. Then $t$ is equal to the number of divisors of $q$. Thus, if $q=p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \ldots p_{k}^{a_{k}}$, where $p_{i}$ are distinct prime numbers, then $t=$ $\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)$. Further, if at leas... | the\lucky\ones\were\in\cells\whose\\were\perfect\ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,574 |
143. Mental arithmetic. Multiply 96 by 104. | 143. Applying the identity $(a-b)(a+b)=a^{2}-b^{2}$, we get: (96) $(104)=(100-4)(100+4)=10000-16=9984$. | 9984 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,576 |
144. Unique Triad. Prove that there exists only one set of three distinct positive integers that have no common divisor greater than 1 and have the property that the sum of any two of these numbers is divisible by the third number. | 144. It is required to find three different integers satisfying the system of equations
$$
\left\{\begin{array}{l}
x+y=m z \\
y+z=n x \\
z+x=p y
\end{array}\right.
$$
where $m, n, p$ are positive integers. The condition that this homogeneous system has a non-zero solution can be written as
$$
\left|\begin{array}{ccc... | 1,2,3 | Number Theory | proof | Yes | Yes | olympiads | false | 41,577 |
145. The Partitioned Corner. One of the corners of a rectangular room is partitioned off using two identical screens, each 4 meters long. How should the screens be placed to maximize the area of the partitioned section?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 145. The fenced area has the shape of a quadrilateral, two sides of which, emanating from the vertex opposite to[^29]
the right angle, are equal to each other. From four quadrilaterals cong... | 8(\sqrt{2}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,578 |
146. Coprime numbers. Let $a, b, c$ be integers that have no common divisor other than 1, and let $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Show that the numbers $(a+b),(a-c)$ and $(b-c)$ are perfect squares. | 146. If $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then $a+b=\frac{a b}{c}$. Since $a$ and $b$ are integers, $c$ can be factored into a product, say, $c=q r$, where one factor divides $a$, and the other divides $b$, so $a=m q, b=p r$. Therefore, $m q+p r=m q p r: q r=m p$. Since all three numbers $a, b, c$ have no common di... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,579 |
147. Diophantine system. Solve the following system of equations in positive integers:
$$
\left\{\begin{aligned}
a^{3}-b^{3}-c^{3} & =3 a b c \\
a^{2} & =2(b+c)
\end{aligned}\right.
$$ | 147. The first equation of the system can be written as
$$
(a-b-c)\left[a^{2}+(b-c)^{2}+a b+b c+c a\right]=0
$$
Since the second factor cannot be zero for positive \(a, b, c\), we obtain
$$
a=b+c=\frac{a^{2}}{2}
$$
Therefore, the only solution to our system in positive integers is \(a=2, b=c=1\).
[E. W. Marchand, ... | =2,b==1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,580 |
148. Bimedians of a tetrahedron. Prove that in a regular tetrahedron, the lines connecting the midpoints of opposite edges intersect each other at right angles.
 | 148. A regular tetrahedron can be inscribed in a cube, with the opposite edges of the tetrahedron coinciding with the non-parallel diagonals of the opposite faces of the cube. Therefore, the midpoints of the edges of the tetrahedron coincide with the centers of the corresponding faces of the cube. This means that the s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,581 |
149. Scattered set. In a printing house, a multiplication line of the form $\overline{a b c} \cdot \overline{b c a} \cdot \overline{c a b}^{*}$ was being set; but the set scattered, and the digits of the product got mixed up. As a result, the product was printed as 2342355286. It is known that $a>b>c$ and that the unit... | 149. Let $n=a b c$, and $N$ be the desired product. If $c=1$, then the largest possible $n$, namely 981, would yield a product $N=159080922$, which is too small. Therefore, $982 \leqslant n \leqslant$ 987. The sum of the digits of $N$ is $35 \equiv 2(\bmod 3)$; hence, $n$ and the numbers obtained by permuting the digit... | 328245326 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,582 |
151. Three out of six. Prove that if you randomly select 6 people, then either among them there will be three, each of whom is acquainted with the other two, or - three, none of whom are acquainted with the others. | 151. Here we assume that the acquaintance of two people is mutual, that is, if one person is acquainted with another, then the second is also acquainted with the first. Identifying each of the 6 people with a vertex of an octahedron, we reduce the problem to an equivalent one: if each edge and each diagonal of the octa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 41,584 |
152. Simplification. Simplify the expression
$$
\frac{(4+\sqrt{15})^{\frac{3}{2}}+(4-\sqrt{15})^{\frac{3}{2}}}{(6+\sqrt{35})^{\frac{3}{2}}-(6-\sqrt{35})^{\frac{3}{2}}}
$$ | 152. Multiplying the numerator and the denominator of our fraction by $2 \frac{3}{2}$, we get
$$
\begin{aligned}
\frac{(8+2 \sqrt{15})^{\frac{3}{2}}+(5-2 \sqrt{15}+3)^{\frac{3}{2}}}{(12+2 \sqrt{35})^{\frac{3}{2}}-(7-2 \sqrt{35}+5)^{\frac{3}{2}}} & = \\
\frac{\left(\sqrt{5}+\sqrt{3}^{3}+(\sqrt{5}-\sqrt{3})^{3}\right.}{... | \frac{7}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,585 |
153. The product of three prime numbers. A certain number is the product of three prime factors, the sum of the squares of which is 2331. There are 7560 numbers (including 1) less than this number and coprime with it. The sum of all divisors of this number (including 1 and the number itself) is 10560. Find this number. | 153. Let $N=p q r$. Then $p^{2}+q^{2}+r^{2}=2331$; hence, each of these prime numbers is less than $(2331)^{\frac{1}{2}}<49$, and all prime numbers are odd.
The sum of all divisors of the number $N$ is $(1+p) \cdot(1+q) \cdot(1+r)=$ $=10560=11 \cdot 960$. The only multiple of 11 not exceeding 49 and greater than some ... | 8987 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,586 |
154. Representation of a rational number as a sum. Prove that any positive rational number can be represented as a finite sum of different terms of the harmonic series $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n}, \ldots$[^6] | 154. Let $\frac{a}{b}$ be a rational number. Then it can be represented as the sum of $a$ repeated terms of the harmonic series
$$
\frac{a}{b}=\frac{1}{b}+\frac{1}{b}+\ldots \frac{1}{b}
$$
Leave the first term unchanged, and transform the remaining ($a-1$) terms using the identity
$$
\frac{1}{n}=\frac{1}{n+1}+\frac{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,587 |
155. Condition of isosceles. Prove that if in a certain triangle the perpendiculars erected from the bases of the bisectors of the internal angles of this triangle intersect at one point, then the given triangle is isosceles.
 and \( y \) be positive numbers. Which of the fractions is greater:
$$
\frac{x^{2}+y^{2}}{x+y} \quad \text { or } \frac{x^{2}-y^{2}}{x-y} ?
$$ | 158. To answer the question of the problem, it is enough to notice that
$$
\frac{x^{2}-y^{2}}{x-y}=x+y>x+y-\frac{2 x y}{x+y}=\frac{x^{2}+y^{2}}{x+y}
$$ | \frac{x^{2}-y^{2}}{x-y}>\frac{x^{2}+y^{2}}{x+y} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,591 |
159. Curvilinear Triangle. Find the radius of the circle inscribed in the curvilinear triangle, where two sides coincide with the legs of a given right triangle $A B C$, and the third side is a semicircle constructed on the hypotenuse $A B$ as the diameter, and located outside the triangle $A B C$. | 159. Let us introduce a rectangular coordinate system, with axes directed along the legs of triangle $ABC$. In this coordinate system, the coordinates of the center of the inscribed circle and the center of the given semicircle are respectively $(r, r)$ and $\left(\frac{a}{2}, \frac{b}{2}\right)$, where $r$ is the radi... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,592 | |
160. Pandiagonal Heterosquare. We define a pandiagonal heterosquare as an arrangement of the first $n^{2}$ positive integers in a square, such that no two sums of numbers along any row, column, or diagonal (straight or broken) are the same*. Does there exist an $n$ for which these $4 n$ sums coincide with consecutive i... | 160. If we sum all the numbers along each row, column, or diagonal, then each number will be counted four times. Therefore, the total sum of all such sums is \(4\left[\frac{n^{2}\left(n^{2}+1\right)}{2}\right]=2 n^{2}\left(n^{2}+1\right)\). If there existed some \(k\) such that our \(4n\) sums represented consecutive n... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,593 |
161. Multiple of $2^{m+1}$. Prove that the smallest integer greater than $(\sqrt{3}+1)^{2 m}$ is divisible by $2^{m+1}$. | 161. Consider the expression $I=(\sqrt{3}+1)^{2 m}+(\sqrt{3}-1)^{2 m}$, which is clearly an integer. Since $(\sqrt{3}-1)^{2 m}$ is less than 1, $I$ coincides with the smallest integer greater than $(\sqrt{3}+1)^{2 m}$. Further,
\[
\begin{aligned}
I & =(4+2 \sqrt{3})^{m}+(4-2 \sqrt{3})^{m}= \\
& =2^{m}\left[(2+\sqrt{3}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,594 |
162. In a nonagon. Let $A B$ and $B C$ be two adjacent sides of a regular nonagon inscribed in a circle with center $O$. Let, further, $M$ be the midpoint of $A B$, and $N$ be the midpoint of the radius perpendicular to $B C$. Show that the angle $O M N=30^{\circ}$.
 of the fence and two straight lines connecting the ends of this section with the center of the square. The areas of all such triangles are equal to each other, and their total height is $\frac{1}{2}$ o... | 16000\mathrm{}=16\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,598 |
166. Triangular numbers from odd squares. Prove that every odd square ends in the octal system (that is, in the base 8 system) with 1; and if this one is discarded, the remaining part will represent some triangular number. | 166. Note that $(2 n+1)^{2}=4 n^{2}+4 n+1=4 n(n+1)+1=8 k+1$, since among two consecutive integers one is necessarily even. Discarding 1 is the same as dividing $4 n(n+1)$ by the base 8. In doing so, we precisely obtain the triangular number $\frac{n(n+1)}{2}$.
[G. U. Milne, N. M. M., 10, 313 (May 1936).] | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,599 |
167. Inscribed Circles. Which of the two triangles has a larger inscribed circle: the triangle with sides 17, 25, and 26, or the triangle with sides 17, 25, and 28? | 167. Answer: none. The radius of the inscribed circle of a triangle with sides $a, b, c$ is calculated by the formula
$$
r=\frac{S}{p}=\left[\frac{(p-a)(p-b)(p-c)}{p}\right]^{\frac{1}{2}}, \quad \text { where } \quad 2 p=a+b+c
$$
From this, for each of our triangles, the radius value will be 6.
A rare example of "ob... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,600 |
168. Inscribed Dodecagon. For a convex dodecagon inscribed in a circle, the lengths of some six sides are $\sqrt{2}$, and the length of each of the remaining six sides is $\sqrt{24}$. What is the radius of the circle?
^{2}=(\sqrt{2})^{2}+(\sqrt{24})^{2}-2 \sqrt{2} \cdot... | \sqrt{38} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,601 |
169. An Impressive Diophantine Equation. Find at least one solution to the equation $a^{3}+b^{4}=c^{5}$ in positive integers. | 169. Since $2^{24}+2^{24}=2^{25}$, we get $\left(2^{8}\right)^{3}+\left(2^{6}\right)^{4}=$ $\left(2^{5}\right)^{5} ;$ from which $a=256, b=64, c=32$.
[Leo Moser, M. M., 26, 53 (September 1954).] | =256,b=64,=32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,602 |
170. Antifreeze. The radiator of a car with a capacity of 21 quarts* is filled with an 18% alcohol solution. How many quarts of this solution need to be drained from the radiator so that, by replacing them with the same amount of 90% solution, a 42% alcohol solution is obtained in the end. | 170. The percentage of alcohol in the old solution differs from the percentage of alcohol in the new (or mixed) solution by $-24 \%$, and the percentage of alcohol in the solution added to the radiator differs from the percentage of alcohol in the new solution by $+48 \%$. Therefore, for each quart of $90 \%$ solution ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,603 |
171. Maximum - minimum without analysis. Find the maximum and minimum values of the expression $\frac{x^{2}-2 x+2}{2 x-2}$, without using methods of mathematical analysis. | 171. Let's perform the transformations:
$$
f(x)=\frac{x^{2}-2 x+2}{2 x-2}=\frac{1}{2}\left[x-1+\frac{1}{x-1}\right]=\frac{x^{2}}{2(x-1)}-1
$$
The sum of two reciprocals takes the smallest absolute value when these numbers are equal to $\pm 1$. Therefore, $f(x)$ takes the smallest absolute value
= \\
=\left(\operatorname{tg} 117^{\circ}+\operatorname{tg} 118^{\circ}\right):\left(1-\operatorname{tg} 117^{\circ} \cdot \operatorname{tg... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,605 |
175. A divisor of its palindrome. In what base does 792 divide 297?
Note: The original problem statement is in Russian, but the provided translation is in English as requested. | 175. In any number system with base $B \geqslant 10$ the following inequalities hold:
$$
2(297)<2(300)=600<792<800=4(200)<4(297)
$$
Therefore, $792=3(297)$, so $7 B^{2}+9 B+2=3\left(2 B^{2}+9 B+7\right)$, or $B^{2}-18 B-19=0$. Discarding the negative root of this quadratic equation, we get $B=19$.
[D. L. Silverman, ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,608 |
176. "Square" Dad. Having once entered into a legal marriage in California, my neighbor has now reached an age that is a perfect square. The product of the digits representing the father's age equals the age of his wife. The age of their daughter is equal to the sum of the digits in the father's age, and the age of the... | 176. The fact that the marriage was legal reduces the number of possible options to three*
| 64 | 10 | | 49 | 13 | 36 | 9 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 24 | 6 | | 36 | 9 | 18 | 9 |
If we now compare the ages of the children with the ages of the parents, we will find a single possible option... | 49,36,13,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,609 |
177. Tetrahedron - through a straw. Take a flexible thin-walled cylinder of diameter $d$, for example, a cocktail straw. What is the edge $e$ of the maximum regular tetrahedron that can be pushed through this straw? | 177. If we form a parallelogram from four equilateral triangles of side $e$ placed adjacent to each other, then in this parallelogram, the length of any segment drawn parallel to its longer side will be equal to $2 e$. From this, it follows that if we form a regular tetrahedron from this parallelogram and then draw any... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,610 |
178. Multiple of $(a-1)^{2}$. Show that $a^{n+1}-n(a-1)-a$ is divisible by $(a-1)^{2}$ for any positive integer $n$.[^9] | 178.
$$
\begin{gathered}
f(a, n)=a^{n+1}-n(a-1)-a= \\
=a\left(a^{n}-1\right)-n(a-1)= \\
=(a-1)\left[a\left(a^{n-1}+a^{n-2}+\cdots+1\right)-n\right] .
\end{gathered}
$$
Since the polynomial in the square brackets turns to zero when $a=1$, this polynomial is divisible by $(a-1)$. Therefore, $f(a, n)$ is divisible by $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,611 |
179. Determinant of Pascal's Triangle. Pascal arranged the binomial coefficients in the following table (Pascal's Triangle) $)^{*}$
| 1 | 1 | 1 | 1 | 1 | 1 | $\ldots$ |
| ---: | ---: | ---: | ---: | ---: | ---: | :--- |
| 1 | 2 | 3 | 4 | 5 | 6 | $\cdots$ |
| 1 | 3 | 6 | 10 | 15 | 21 | $\cdots$ |
| 1 | 4 | 10 | 20 | 35... | 179. The law according to which our table is composed is that each element is equal to the sum of two other elements, one of which is directly above the given element, and the other is to the left of the given element. Applying to our determinant of the $n$-th order the operation of subtracting columns (column) $_{i} -... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 41,612 |
180. Rational coordinates. Prove that for any rational number $x$ there exists at least one rational number $y$, such that the pair $(x, y)$ will satisfy the equation
$$
2 x^{3}+2 y^{3}-3 x^{2}-3 y^{2}+1=0
$$ | 180. Our third-order curve is the union of an ellipse and a straight line, which becomes obvious if we write the original equation in the form
$$
(x+y-1)\left(2 x^{2}-2 x y+2 y^{2}-x-y-1\right)=0
$$
Taking on the line $x+y-1=0$ any point with a rational coordinate $x$, we find that the corresponding coordinate $y=1-x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 41,613 |
181. Closed construction. Take an arbitrary point \( P \) on the side \( BC \) of some triangle \( ABC \) and then mark a point \( Q \) on the side \( AB \) such that \( BQ = BP \). Next, on \( CA \), mark a point \( R \) such that \( AR = AQ \); on \( BC \), mark a point \( P' \) such that \( CP' = CR \); on \( AB \),... | 181. Let $P_{0}, Q_{0}$ and $R_{0}$ be the points where the incircle touches the sides of triangle $ABC$; then $P P_{0}=Q Q_{0}=$
$=R R_{0}=P^{\prime} P_{0}=Q^{\prime} Q_{0}=R^{\prime} R_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,614 |
182. Grouped Numbers. Consecutive odd numbers are grouped as follows: $1 ;(3,5) ;(7,9,11) ;(13$, $15,17,19) ; \ldots$. Find the sum of the numbers in the $n$-th group.[^10] | 182. The $n$-th group contains $n$ integers; therefore, the number of integers in all groups from the 1st to the $n$-th inclusive is $\frac{n(n+1)}{2}$, and the number of integers in all groups up to the $(n-1)$-th inclusive is $\frac{(n-1) n}{2}$. These two sets form arithmetic progressions with a common difference of... | n^3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,615 |
183. Sixteen-point sphere. Can the radius of the sixteen-point sphere of some tetrahedron be half the radius of the sphere circumscribed around this tetrahedron? (A sixteen-point sphere is called a sphere that passes through the centers of the circles circumscribed around the faces of the given tetrahedron.) | 183. Let $R$ be the radius of the circumscribed sphere, and $r$ be the radius of the sixteen-point sphere. If the given tetrahedron is regular, then $r=\frac{R}{3}$; if, however, it has three right dihedral angles at some vertex, then $r=\infty$ (since in this case all the centers of the circumscribed circles lie in th... | \frac{R}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,616 |
184. Intersecting Circles. The three points of intersection of three circles passing through one point, not coinciding with this point, lie on one straight line. Prove that the centers of these circles and their common point of intersection lie on some new circle. | 184. Let the circles with centers at $O_{1}, O_{2}, O_{3}$ pass through a common point $A$. Furthermore, circles $\left(O_{1}\right)$ and $\left(O_{2}\right)$ intersect at point $B$; $\left(O_{1}\right)$ and $\left(O_{3}\right)$ intersect at point $C$; and circles $\left(O_{2}\right)$ and $\left(O_{3}\right)$ intersect... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,617 |
185. Golf Tournaments. Professional golfers decided to hold a tournament among 16 members of their club. In each match, four players participate, and over the course of the tournament, each player must be in the same foursome with each of the other players exactly once. How should the participants be distributed into f... | 185. Let's take the following four permutations of four letters: $P_{1}(x y z w), P_{2}(z w x y), P_{3}(w z y x)$, and $P_{4}(y x w z)$.
Each player meets each of the other 15 players exactly once; therefore, the tournament must consist of $\frac{15}{3}=5$ rounds. Arrange the initials of the 16 participants in a $4 \t... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,618 |
186. Square triangular numbers. Show that there are infinitely many numbers, each of which is both triangular and square. | 186. If $T[n]=\frac{n(n+1)}{2}$ is the $n$-th triangular number, which is also a square, then the number $T[4 n(n+1)]=4 T[n](2 n+1)^{2}$ is also a perfect square. Since the first triangular number 1 obviously has the required property, there are infinitely many square triangular numbers.
[O. U. Sylvester, A. M. M., 69... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,619 |
187. System with three unknowns. Solve the following system of equations:
$$
\left\{\begin{aligned}
x+y+z & =6 \\
x y+y z+z x & =11 \\
x y z & =6
\end{aligned}\right.
$$ | 187. Recalling the connection between the roots and coefficients of a third-order equation with one unknown, we notice that $x, y, z$ coincide with the roots of the cubic equation.
$$
a^{3}-6 a^{2}+11 a-6=0
$$
Further,
$$
\begin{gathered}
(a-1)(a-2)(a-3)=0 \\
a=1,2,3
\end{gathered}
$$
Since the original system is s... | (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,620 |
189. Baseball Competitions. The wins and losses of the National League teams of the United States as of July 14, 1965, are given below.
| | Wins | Losses | | Wins | Losses |
| :--- | :---: | :---: | :--- | :---: | :---: |
| Chicago | 41 | 46 | New York | 29 | 56 |
| Cincinnati | 49 | 36 | Philadelphia | 45 | 39 |
| ... | 189. We will first compare each team with a certain hypothetical team, whose wins constitute $50 \%$, and the difference $B-\Pi$ is, obviously, zero. If our team has $B > \Pi$, then its wins exceed $50 \%$ and, by definition, we will classify it in the upper class. If our team has $B \leq \Pi_{C}$ or if $B_{A}=B_{C}$, ... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 41,622 |
191. Intersection of Diagonals. Find the number of intersection points of the diagonals of a convex $n$-gon. | 191. Consider a polygon with $n$ sides, where $n \geqslant 4$. Any four vertices of this polygon define some
quadrilateral, the diagonals of which coincide with two intersecting diagonals ... | C_{n}^{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,624 |
192. Vanishing Triples. Let the sum of the elements in each of the two given numerical triples be zero. Show that the sums of the cubes of the elements of these triples are in the same ratio as the products of their elements.
Translate the above text into English, please retain the original text's line breaks and form... | 192. Given that $a+b+c=0$ and $d+e+f=0$. Therefore,
$$
\begin{gathered}
(a+b)^{3}=(-c)^{3} \\
a^{3}+b^{3}+c^{3}=-3 a b(a+b)=3 a b c
\end{gathered}
$$
Similarly,
$$
d^{3}+e^{3}+f^{3}=-3 d e(d+e)=3 d e f
$$
Finally,
$$
\frac{a^{3}+b^{3}+c^{3}}{d^{3}+e^{3}+f^{3}}=\frac{a b c}{d e f}
$$
[Aaron Bachman, S. S. M., 38, ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,625 |
193. Condition for Factorization. Let $a$ and $b$ be two integers, each of which is coprime with 3 and their sum $a+b$ equals $3 k$. Show that the polynomial $x^{a}+x^{b}+1$ can be factored ${ }^{*}$. | 193. Let $f(x)=x^{a}+x^{b}+1=x^{a}+x^{3 k} \cdot x^{-a}+1$. The cubic roots of 1 are $1, \omega=\frac{-1+i \sqrt{3}}{2}$, and $\omega^{2}$. Since $a$ is coprime with 3, either $\omega^{a}=\omega$ and $\omega^{-a}=\omega^{2}$, or $\omega^{a}=\omega^{2}$ and $\omega^{-a}=\omega$. In either case,
$$
f(\omega)=f\left(\ome... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,626 |
194. Parallel Resistors. If we connect two conductors in parallel, with resistances equal to $x$ and $y$ respectively, the resistance $z$ of such a segment can be found from the relation
$$
\frac{1}{z}=\frac{1}{x}+\frac{1}{y} ; \quad x, \quad y, \quad z>0
$$
Find positive integers $x, y, z$ that satisfy this relation... | 194. Since $x, y, z$ are positive, $x>z$ and $y>z$. Let $x=$ $z+u$, and $y=z+v ; u, v>0$. The equation $\frac{1}{z}=\frac{1}{x}+\frac{1}{y}$ reduces to $z^{2}=u v$. Therefore, for any $z$, we need only factor $z^{2}$ into the product of two positive integers $w$ and $v$.
[Marion L. Gains, N. M. M., 19, 100 (November 1... | z^2=uv | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,627 |
195. Curve of minimum length. What is the curve of minimum length that divides an equilateral triangle into two equal-area parts[^11]
 | 195. Let's fix one vertex of the triangle and reflect it along with the desired curve several times relative to the sides emanating from this vertex. In doing so, we obtain a regular hexagon and a closed curve that divides this hexagon into two equal areas, as shown in the figure. Since the length of the curve is minim... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,628 |
196. Interesting square. Find a nine-digit integer of the form $\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$, which represents the product of the squares of four different prime numbers, and
$$
\overline{b_{1} b_{2} b_{3}}=2 \overline{\left(a_{1} a_{2} a_{3}\right)} \quad\left(a_{1} \neq 0\right)
... | 196. Note that
$$
\begin{gathered}
N^{2}=\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}=\overline{a_{1} a_{2} a_{3}} \cdot 1002001= \\
=\overline{a_{1} a_{2} a_{3}} \cdot 7^{2} \cdot 11^{2} \cdot 13^{2}
\end{gathered}
$$
Therefore, the number $\overline{a_{1} a_{2} a_{3}}$ is the square of a prime n... | 289578289or361722361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,629 |
197. Test Series. Professor Tester conducts a series of tests, based on which he assigns the test subject an average score. After answering the last test, John realized that if he had scored 97 points on this last test, his average score would have been 90. On the other hand, if he had scored only 73 points on the last... | 197. If the difference in points obtained in one test, equal to $97-73=24$ points, causes a change in the average score by $90-87=3$ points, then the series contains a total of $\frac{24}{3}=8$ tests. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,630 |
198. Coincidence of points in a quadrilateral. Show that in any quadrilateral, the midpoint of the segment connecting the midpoints of the diagonals coincides with the point of intersection of the segments connecting the midpoints of the opposite sides of the given quadrilateral. | 198. The midpoints of the sides of an arbitrary quadrilateral $ABCD$ serve as the vertices of a certain parallelogram $MQNP$, and the diagonals of the parallelogram $PQ$ and $MN$, intersecting at point $O$, are bisected. Now consider the quadrilateral $ACDB$ with sides $AC, CD, DB$, and $BA$. The points $M, N$ and $X, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,631 |
199. Irreducible Fractions. Let $n$ be a positive integer greater than 2. Show that among the fractions $\frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \ldots, \frac{n-1}{n}$, there is an even number of irreducible fractions. | 199. If $\frac{k}{n}$ is irreducible, then $1-\frac{k}{n}=\frac{n-k}{n}$ also represents an irreducible fraction. Therefore, irreducible fractions occur in the given sequence in pairs and, consequently, their number is always even.
[N. Anning, M. M., 27, 284 (May 1954).] | proof | Number Theory | proof | Yes | Yes | olympiads | false | 41,632 |
200. Tea Service. A trader has displayed a silver tea service on the shelf with the cost price of each item and its retail price in dollars:
| Sugar Bowl | HKHC | 6.72 | Creamer | HCKH | 6.00 |
| :--- | :---: | ---: | :--- | ---: | ---: |
| Tray | AMSL | 50.16 | Teapot | SIAB | 91.08 |
| Tongs | $N B L T$ | 1.72 | Spo... | 200. Since the cost price of any item is less than its retail price, we can, by studying the prices of the sugar bowl and the creamer, conclude that $H=0$. Comparing these prices further, we find
$$
\overline{K 0 C}: \overline{C K 0}=672: 600=28: 25 .
$$
From this, it follows that $K 0$ is a multiple of 25, meaning t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,633 |
201. Segments defining an equilateral triangle. Three segments of lengths 3, 4, and 5 cm connect an internal point \( P \) of an equilateral triangle to its vertices. What is the length of the side of this triangle?
$. Since it is always true that
$$
\left(N_{1} d+r\right)-\left(N_{2} d+r\right)=d\left(N_{1}-N_{2}\right)
$$
the desired number is 23, and the corre... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,635 |
203. Nine digits. Find a nine-digit number, all digits of which are different and do not contain zero, and the square root of which has the form $\overline{a b a b c}$, where $\overline{a b}=c^{3}$. | 203. The digit $c$ can only coincide with 3 or 4. However, the number $(64644)^{2}$ is ten-digit. Therefore, the only solution to our problem will be $(27273)^{2}=743816529$.
[N. Farnum, S. S. M., 63, 603 (October 1963).]
This solution will be unique even if we drop the restriction $a b=c^{3}$. Moreover, $27273=3(909... | 743816529 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,636 |
204. Cutting Congruent Triangles. Consider two congruent triangles on a plane that can be obtained from each other by a mirror reflection. Into what parts should these triangles be cut so that each of them can be transformed into the other by simply moving the obtained parts on the plane (without reflections)? | 204. One of the two given triangles does not need to be cut at all, while the other should be cut into several isosceles triangles. If the original triangle is a right triangle,
^{\frac{1}{2}}+\left(1-\frac{1}{x}\right)^{\frac{1... | 205. Let $b=\left(x-\frac{1}{x}\right)^{\frac{1}{2}}$ and $a=\left(1-\frac{1}{x}\right)^{\frac{1}{2}}$; then
$$
x=a+b
$$
and since $n \neq 0$
$$
b-a=\frac{b^{2}-a^{2}}{b+a}=\frac{x-1}{x}=1-\frac{1}{x}
$$
Adding (1) and (2), we get
$$
2 b=x-\frac{1}{x}+1=b^{2}+1
$$
from which $b=1$.
Therefore,
$$
x-\frac{1}{x}=1... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,638 |
206. Characteristics of men. In a certain group of men, $70 \%$ have brown eyes; $70 \%$ have dark hair, $85 \%$ are taller than 5 feet 8 inches, and $90 \%$ weigh more than 140 pounds. What percentage of men definitely have all four of these characteristics? | 206. Let's take 100 men and record each of the four characteristics for each man who possesses the characteristic. As a result, we will have $70+75+85+90=320$ records. Since there are 320 records for 100 men, we can observe that with the most even distribution of characteristics among the men, each man would have at le... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,639 |
207. Region of Constant Width. Let $A$ and $B$ be internal points of some region of constant width 1. Show that there exists a path from $A$ to $B$ that has a common point with the boundary $b$ of the given region and whose length $\leqslant 1$.
$(A C + C B) + (B D + D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 41,640 |
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