problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
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
$$
\sum_{j=0}^{k+1} x_{j}=x_{k+1}+\sum_{j=0}^{k... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
275. Maximum Number. Let a set of distinct complex numbers $z_{i}, i=1,2, \ldots, n$, be given, satisfying the inequality
$$
\min _{i \neq j}\left|z_{i}-z_{j}\right| \geqslant \max _{i}\left|z_{i}\right|
$$[^16]
Find the maximum possible $n$ and for this $n$ all sets satisfying the condition of the problem. | 275. Let $\left|z_{m}\right|=\max _{i}\left|z_{i}\right|$. Then on the complex plane, all points $z_{i}$ will be located inside the circle $R$ of radius $\left|z_{m}\right|$ centered at the point $z=0$. Clearly, 6 points $z_{i}$, located on the circumference of $R$ and forming a regular hexagon, together with the point... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
308. A simple "game". In the following cryptarithm
| $M A T H$ |
| :--- |
| $M A G S$ |
| $M A T H$ |
| GAME |
the word $G A M E^{*}$ is used instead of some prime number. Find this number. | 308. The letter $M$ must represent one of the elements of the set $\{1,2,3\}$, the letter $A$ one of the elements of the set $\{4,5,9\}$, from which it follows that the letter $G$ coincides with some element of the set $\{4,7,8\} . .^{*}$ Looking at the table of prime numbers, we will find that the only suitable prime ... | 8923 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
322. Limit of a Sum. Calculate the value of the expression
$$
\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{k^{2}+3 k+1}{(k+2)!}
$$ | 322.
$$
\begin{gathered}
\sum_{k=0}^{n} \frac{k^{2}+3 k+1}{(k+2)!}=\sum_{k=0}^{n}\left(\frac{1}{k!}-\frac{1}{(k+2)!}\right)= \\
=\left(\frac{1}{0!}+\frac{1}{2!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+2)!}\right)= \\
=\frac{1}{0!}+\frac{1}{1!}-\frac{1}{(n+1)!}-\frac{1}{(n+2... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
342. Simple calculation. Let $\left(x+\frac{1}{x}\right)^{2}=3$; determine what $x^{3}+\frac{1}{x^{3}}$ is equal to. | 342.
$$
\begin{aligned}
x^{3} & +\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)= \\
& =\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=0
\end{aligned}
$$
[M. Demos, M. M., 45, 102 (February 1972).] | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
373. Cutting a Cube. In a mathematical journal, the following question and answer appeared.
Question: "A carpenter wants to cut a wooden cube with a side of 3 inches into 27 cubes with a side of 1 inch. He can easily do this by making 6 cuts while keeping the pieces together so they don't fall apart. What is the minim... | 373. After the first cut, the cube splits into 2 parts. The larger of these (consisting of 17 one-inch cubes) contains one central cube, for four faces of which another cut is required. After the last of these is made, there will remain at least two one-inch cubes undivided - regardless of any rearrangement of the piec... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
388. Three Napkins. One lady made 3 round napkins with radii of 2, 3, and 10 inches, respectively. She placed them on a round table so that each napkin touched the other two and the edge of the table. What is the radius of the table top? | 388. If the centers of napkins with radii of 2, 3, and 10 inches are located at points \( C, A \), and \( B \) respectively, then these points form the vertices of a right triangle with sides 5, 12, and 13 inches. Completing the figure \( ABC \) to a rectangle, we denote its fourth vertex as \( O \). From point \( O \)... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ex. 1. Find the area of a right triangle if the length of the shortest median is 13, and one of the legs is 10. | Ex. 1. 120.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ex. 19. In triangle $ABC$, points $M$ and $N$ are chosen on sides $AB$ and $AC$. It is known that $\angle ABC=70^{\circ}, \angle ACB=50^{\circ}, \angle ABN=20^{\circ}$, $\angle ACM=10^{\circ}$. Find $\angle NMC$.
(y+x)=(a-c)(a+c) \Rightarrow 7 b=(a-c)(a+c)$. Sinc... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ex. 129. In an integer-sided triangle, two sides are equal to 10. Find the third side, given that the radius of the inscribed circle is an integer. | Ex. 129. Answer: 12. Solution. Let the third side be denoted by $a$, and the angle subtending it by $\alpha$. Then $\sin \alpha=\frac{10+10+a}{10 \cdot 10} \cdot r$. At the same time, $1 \leq a \leq 19 \Rightarrow r \leq 4$. If $\sin \alpha=1$, then $x=5, r=4$, but a triangle with such data does not exist. According to... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 7. At the Dentist's Office
In the waiting room, two women and ten men are waiting for their turn. At their disposal are eight copies of the latest magazine and four copies of the morning newspaper. In how many ways can they distribute the newspapers and magazines among themselves, if both women insist on reading th... | 7. If both women will be reading a newspaper, then the corresponding number of distribution options for newspapers and magazines will be equal to the number of ways in which the remaining two copies of the newspaper can be distributed among ten men, i.e.
$$
C_{10}^{2}=\frac{10!}{2!8!}=45
$$
If the two women will be r... | 255 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 10. Auto Rally
This was an extremely dangerous auto rally. It started with a small and very narrow bridge, where one out of five cars fell into the water. Then came a terrible sharp turn, where three out of ten cars ended up in the ditch. Further along the way, there was such a dark and winding tunnel that one out ... | 10. For a car to successfully complete an auto rally, it must not get into an accident on the bridge (which happens in 4 out of 5 cases), nor on a curve (7 out of 10), nor in a tunnel (9 out of 10), nor, finally, on a sandy road (3 out of 5). Thus, we get
$$
\frac{4 \cdot 7 \cdot 9 \cdot 3}{5 \cdot 10 \cdot 10 \cdot 5... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 11. How Many Are We?
To help you answer this question, let's say that the probability that at least two of us have the same birthday is less than $1 / 2$, but this would be false if one more person joined us. | 11. The probability that two people do not share the same birthday is $364 / 365$. For three people, the probability that no two of them share the same birthday is (364/365)$\cdot$(363/365), and so on. In the case of $n$ people, the same probability is
$$
(364 / 365) \cdot(363 / 365) \cdot \ldots \cdot(365-n+1) / 365
... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 12. Convoy of Ships
In 1943, in one American port, there were four escort ships, seven cargo ships, and three aircraft carriers. From these ships, a transport convoy was formed to deliver food and ammunition to Europe. At the head was an escort ship, followed by three cargo ships, then an aircraft carrier, and last... | 12. The number of ways to choose the first escort ship is 4. The selection of three cargo ships can be carried out
$$
C_{3}^{3}=\frac{7!}{3!4!}=35
$$
ways.
The number of ways to choose an aircraft carrier is 3. The last escort ship can be chosen in 3 ways (since only three unused ships remain).
Thus, there are $4 \... | 1260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 18. Expected Lifespan
An ethnographer found that in a certain primitive tribe he studied, the distribution of the lifespan of the tribe members could be described as follows: $25 \%$ of them only lived up to 40 years, $50 \%$ died at 50 years, and $25 \%$ - at 60 years. He then randomly selected two individuals to ... | 18. If the one of the two members of the tribe who lives longer reaches 40 years, then obviously both "subjects" will live only to 40 years; the corresponding probability is
$$
25 \cdot 25 \% = 1 / 4 \cdot 1 / 4 = 1 / 16
$$
If the "long-liver" reaches 50 years, this means that either the first of the two will live on... | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 33. In Search of a Job
To find a job after demobilization, soldier Maurice began sending letters to various companies where people of his specialty can be employed. He believes that each of his applications has a one in five chance of being accepted, and he stops sending letters as soon as he finds that he has at l... | 33. The probability that Maurice will remain unemployed after sending $n$ letters is $(1-1 / 5)^{n}=(4 / 5)^{n}$. Therefore, the probability of finding a job is $1-(4 / 5)^{n}$.
Maurice will stop writing when $n$ becomes such that this probability is not less than $3 / 4$, i.e., when the inequality
$$
(4 / 5)^{n} \le... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 36. Daily Newspapers
In a small town where many vacationers spend their holidays, $28 \%$ of the adult vacationers read "Mond", $25 \%$ read "Figaro", and $20 \%$ read "Orore". Additionally, $11 \%$ of the vacationers read both "Mond" and "Figaro", $3 \%$ read both "Mond" and "Orore", and $2 \%$ read both "Figaro" ... | 36. Let $x$ be the desired percentage. Denote by $V$ the set of all vacationers, by $M$ the set of "Mond" readers, by $F$ the set of "Figaro" readers, and finally by $A$ the set of "Aurore" readers. To determine the desired percentage, we will use the Venn diagram provided below.
We can observe that for each of these configurations, the number of mailboxes is 26. In fact, this number doe... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 4. From Bordeaux to Saint-Jean-de-Luz
Two cyclists set out simultaneously from Bordeaux to Saint-Jean-de-Luz (the distance between these cities is approximately 195 km). One of the cyclists, whose average speed is 4 km/h faster than the second cyclist, arrives at the destination 1 hour earlier. What is the speed of... | 4. Let $v$ (km/h) be the desired speed, and $t$ (h) be the time it takes for the cyclist moving at the higher speed $v$ to cover the entire distance.
The speed of the second cyclist, who is moving slower, is $v-4$. The time spent by him on the entire journey is $t+1$. Thus, we have
$$
195=v t=(v-4)(t+1),
$$
from whi... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 5. The Story of the Vagabonds
At the very moment when Pierrot left the "Commercial" bar, heading for the "Theatrical" bar, Jeanno was leaving the "Theatrical" bar, making his way to the "Commercial" bar. They were walking at a constant (but different) speed. When the vagabonds met, Pierrot proudly noted that he had... | 5. Let $x$ (m) - be the distance between the bars we are looking for; $d$ - the distance traveled by Pierrot by the time of the meeting; $V$ (m/s) - Pierrot's speed, and $v$ - Jeanno's speed before the fight.
The sum of the distances traveled by the vagrants is
$$
d+(d-200)=x, \quad \text { from which } \quad x=2 d-2... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 6. Invited Evening
To celebrate her husband being awarded the Legion of Honor, his wife, a well-known lady from the Paris region, decided to order cookies from the best pastry shop in Paris.
- Our evening, - she clarifies on the phone, - will start exactly at 6 PM. I want the fresh cookies to be delivered precisel... | 6. Let $v$ be the speed of interest (but currently unknown), $t$ the corresponding delivery time, and $d$ (km) the distance the car must travel. We obtain the following system of equations:
$$
\text { 1) } t=d / v, \text { 2) } t+1 / 4=d / 20, \text { 3) } t-1 / 4=d / 60 \text {. }
$$
From equations (2) and (3), it f... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 8. The Emir and His Chauffeur
Emir Ben Sidi Mohammed always drives from his palace to the airport at the same speed along a magnificent highway that runs through the desert. If his chauffeur increased the average speed of the car by $20 \mathrm{km} /$ h, the Emir would save 2 minutes, and if the chauffeur drove at ... | 8. Let $I$ (km) be the unknown distance; $v$ (km/h) the speed of the emir's car, and finally, $t$ (h) the time it takes for the emir to travel from the palace to the airport. We obtain the following system of equations:
$$
v=I / t ; \quad v+20=I /(t-2 / 60) ; \quad v-20=I /(t+3 / 60) .
$$
Excluding the first equation... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 10. Escalator in the Metro
Usually, I go up the escalator in the metro. I calculated that when I walk up the stairs of the moving upward escalator, I climb 20 steps myself, and the entire ascent takes me exactly $60 \mathrm{c}$. My wife walks up the stairs more slowly and only climbs 16 steps; therefore, the total ... | 10. Let $x$ be the unknown number of steps on the stationary escalator. When the escalator is working, it rises at a speed of $x-20$ steps in 60 seconds (this follows from calculations related to my ascent) or at a speed of $x-16$ steps in 72 seconds (which follows from how my wife ascends). Thus, the speed of the esca... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 11. Polar Expedition
A polar explorer set out from one point to another on a sled pulled by a team of five dogs. But after 24 hours, two of the dogs died, causing the explorer's speed to decrease to $3 / 5$ of the original speed, and he was delayed by two days.
- Oh! - exclaimed the polar explorer. - If the two do... | 11. If two dogs had run 120 km more, the polar explorer's delay would have been reduced by 24 hours (48-24). Therefore, if these dogs had run 240 km further, the polar explorer's delay would have been 48-24-24=0 hours, i.e., he would have arrived on time; hence, the distance from the place where the dogs fell to the de... | 320 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 12. Punctual Wife
Every Saturday between 15 and 16, I play tennis with my friend Philippe. My wife arrives to pick me up in the car precisely at 16:10. One day, Philippe fell ill. Unaware of this, I set out for tennis as usual. However, at 15:05, realizing that Philippe wasn't coming, I gathered my things and start... | 12. On the day when Philip fell ill, my wife drove the car for 10 minutes less than usual; that means she drove 5 minutes less in each direction. Therefore,
I met her at 4:05 PM, instead of 4:10 PM as usual. But then I walked for 60 minutes to cover the distance my wife drives in 5 minutes. Thus, I walk 12 times slowe... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 13. The Scout and the Drummer
On the occasion of the village festival, a procession was organized that stretched for 250 m; the scouts led the procession, and the musicians brought up the rear. Soon after the march began, the youngest scout remembered that he had not tied his neckerchief, which was left with his fr... | 13. Let $x$ be the unknown speed of the procession. When the scout ran to the end of the procession, he moved relative to the procession (considered stationary!) at a speed of $10+x$; when he returned, he moved relative to the procession at a speed of $10-x$. Since the length of the procession is 250 m, the total time ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 15. Metro in Mexico City
A worker is walking along the tracks in one of the tunnels of the capital's subway, on a section between two stations that are far apart. Every 5 minutes a train passes him coming from the opposite direction, and every 6 minutes another train overtakes him. The worker walks at a constant sp... | 15. Let $x$ be the time between the passage of two consecutive trains moving in the same direction; $I$ be the distance between two consecutive trains; $V$ be the speed of each train, and finally, $v$ be the speed of the worker. Then, relative to the worker, each following train moves at a speed of $V-v$, and each onco... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 21. Rue Sainte-Catherine
I was walking with my mother-in-law. We were slowly, at a speed of 3 km/h, walking down Rue Sainte-Catherine in Bordeaux, which, as everyone knows (or maybe doesn't know), is straight. Suddenly, I remembered that I needed to drop a letter in a mailbox located a bit further down the street. ... | 21. Let's choose meters and minutes as units of measurement. Then my speed after I left my mother-in-law is $5000 / 60 = 250 / 3 \, \text{m} /$ min, and my mother-in-law's speed (and our common speed during the walk) is 3000/60 = $50$ m $/$ min. Let $S$ be the point where we parted, $R$ the point where we met again, an... | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 12. Figure
On a plane, there are five points such that no three of them lie on the same line. The points are connected by straight lines. What is the number of intersection points of the lines (excluding the original five points), if none of the drawn lines are parallel to each other? | 12. Let $A, B, C, D$ and $E$ be five given points. The line $A B$ is intersected by the lines forming three sides of the triangle $C D E$. This gives three points of intersection. The same applies to each other line. The total number of lines is equal to the number of ways to choose two points from five, i.e., 10. Thus... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 14. "Revolutionary" ${ }^{1}$ Geometry
Let in triangle $A B C$ angle $B$ be a right angle, and $M$ be a point on the hypotenuse, equidistant from the two sides of the triangle. Could you find the value of the following expression:
$$
\begin{aligned}
E & =\sqrt{1830}\left(A C-\sqrt{A B^{2}+B C^{2}}\right) \\
& +178... | 14. 15) By the Pythagorean theorem $A C^{2}=A B^{2}+B C^{2}$; therefore,
$$
A C-\sqrt{\overline{A B^{2}+B C^{2}}}=0
$$
2) The area of $A B C=$ area of $A B M+$ area of $B M C$. Therefore, if point $M$ is at a distance $d$ from the legs of the triangle, then
![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad... | 1789 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 22. Lighthouse
From the tower of the lighthouse, located 125.7 m above sea level, the horizon line is visible. At what approximate distance from the tower is this line, if we assume that the Earth's surface is spherical and its great circle is 40,000 km?
![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad8... | 22. Let $S$ be the top of the lighthouse, $A$ be the point on the horizon, and $O$ be the center of the Earth. Triangle $O A S$ is a right triangle because line $S A$ is tangent to the spherical surface of the Earth. Therefore,
![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-163.jpg?height=452&wid... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 32. Siesta
Albert lay lazily on the sun-drenched terrace, drifting into a sweet doze. This did not prevent him, however, from observing his right hand. "My index and middle fingers," he thought, "form an angle of about $20^{\circ}$. Can I, while keeping their relative position, turn my hand so that the shadows of m... | 32. Albert placed the tips of his index and middle fingers on the terrace (a horizontal plane), maintaining a constant angle between them ($20^{\circ}$). When his hand rotates around these two points of support, the angle formed by the shadows of his fingers changes: it is $20^{\circ}$ when the hand lies on the terrace... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 1. Bark
One smart dog knew how to count in the quaternary number system. She conveyed zero in her dog language with the sound "o", one with the sound "u", two with the sound "v", and finally, three with the sound "a". What number, in this case, did her bark "ouavoouav" represent? | 1. Ouoouo $=\left(2 \cdot 4^{0}\right)+\left(3 \cdot 4^{1}\right)+\left(1 \cdot 4^{2}\right)+\left(0 \cdot 4^{3}\right)+\left(2 \cdot 4^{4}\right)+$ $+\left(3 \cdot 4^{5}\right)+\left(1 \cdot 4^{6}\right)+\left(0 \cdot 4^{7}\right)=2+12+16+0+512+3072+0+$ $+4096+0=7710$. | 7710 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 8. At the Restaurant
Monsieur Dupont remembered that today was their wedding anniversary and invited his wife to have lunch at a good restaurant. Leaving the restaurant, he found that he had only one-fifth of the money he had brought with him left, and the number of centimes left was the same as the number of franc... | 8. Let $F$ be the initial number of francs, $C$ be the initial number of centimes, $F^{\prime}$ be the remaining number of francs, and $C^{\prime}$ be the remaining number of centimes.
Since after dining at the restaurant, Monsieur Dupont had only one-fifth of his money left, we have
$$
100 F + C = 5 \left(100 F^{\pr... | 79 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 10. Anniversary
"Do you remember Leonitina's fortieth birthday? It was December 28, $19 \ldots$ Since then, Leonitina has aged. I noticed that half of her age equals twice the sum of its digits."
Please fill in the missing digits in the date of Leonitina's fortieth birthday. | 10. So, Leontina's age is equal to four times the sum of its digits. Therefore, she is less than 100 years old (and not less than 10 years old). Let $d$ be the number of tens, and $u$ be the number of units in the notation of her age. Then
$$
10 d+u=4(d+u)
$$
from which
$$
2 d=u \text {. } \quad \text {. }
$$
Thus,... | 1970 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 16. How many of you were there, children?
If you had asked me such a question, I would have answered you only that my mother dreamed of having no fewer than 19 children, but she did not manage to fulfill her dream; however, I had three times as many sisters as cousins, and brothers - half as many as sisters. How ma... | 16. From the conditions of the problem, it follows that the number of my sisters is divisible by both 3 and 2. Therefore, it is divisible by 6. Hence, the total number of children is
$$
[(\text { number divisible by } 6) \cdot(1+1 / 2)]+1
$$
(the last 1 corresponds to myself). This number must be strictly less than 1... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 17. Airline
One American airline serves several major cities, establishing direct flights between each pair of these cities. Next year, it plans to increase the number of flights by 76, which will allow it to serve some additional cities under the same conditions.
How many cities does this airline currently serve,... | 17. Let $n$ be the number of cities currently served. The corresponding number of flights is $n(n-1)$, since each city is connected to $n-1$ cities served by the airline. For the next year, the airline plans
$$
(n+k)(n+k-1)
$$
flights, where $k$ is the additional number of cities. Therefore,
$$
(n+k)(n+k-1)-n(n-1)=7... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 22. Age Difference
The sums of the digits that make up the birth years of Jean and Jacques are equal to each other, and the age of each of them starts with the same digit. Could you determine the difference in their ages? | 22. Let $(m, c, d, u)$ be the number of thousands, hundreds, tens, and units in Jean's birth year, and let $\left(m^{\prime}, c^{\prime}, d^{\prime}, u^{\prime}\right)$ be the corresponding digits of Jacques' birth year. Then Jean's age is
$$
1979-(1000 m+100 c+10 d+u)
$$
Jacques' age is
$$
1979-\left(1000 m^{\prime... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 23. Division
Division of integers is performed. If the dividend is increased by 65 and the divisor is increased by 5, both the quotient and the remainder remain unchanged. What is this quotient? | 23. Let $D$ - the dividend, $d$ - the divisor, $q$ - the quotient, and $r$ - the remainder. The conditions of the problem can be written as two equations:
$$
D=q d+r, \quad D+65=q(d+5)+r
$$
Subtracting the first equation from the second, we get
$$
65=5 q
$$
from which
$$
q=13
$$ | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 24. Permutations
A certain three-digit number will increase by 45 if the two rightmost digits are swapped, and it will decrease by 270 if the two leftmost digits are swapped. What will happen to this number if the two outermost digits are swapped? | 24. Let $c, d$ and $u$ be the number of hundreds, tens, and units of a given number, respectively. According to the problem, we have the following equations:
$$
\begin{aligned}
& (100 c+10 d+u)=(100 c+10 u+d)-45 \\
& (100 c+10 d+u)=(100 d+10 c+u)+270
\end{aligned}
$$
From this, we get
$$
\begin{gathered}
9 d-9 u+45=... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 25. School
In all the classes of one school, the same number of students studied. After a fire, six classes of the school became unsuitable for lessons, and therefore, five more students had to be added to each class where lessons were held. But then, due to water damage from the fire hoses, another ten classes wer... | 25. Let $c$ be the number of students in each class before the fire, and $n$ be the number of classes at that time. Then the following relationships hold:
$$
\begin{gathered}
n c=(n-6)(c+5) \\
n c=(n-16)(c+20)
\end{gathered}
$$
From these two equations, it follows that
$$
\begin{gathered}
-6 c+5 n-30=0 \\
-16 c+20 n... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 27. Large Families
The Martens have more children than the Duponts. Suppose the difference of the squares of these two numbers is 24 and that both families have more than one child. How many children do the Martens have?
70 | 27. Let $m$ and $d$ be the number of children in the Marten and Dupont families, respectively. According to the problem,
$$
m^{2}-d^{2}=24 \text {, i.e., }(m+d)(m-d)=24\left(=2^{3} \cdot 3\right) \text {. }
$$
Therefore, the following values for $m+d, m-d$, and thus $2 m[=(m+d)+(m-d)]$, $m$, and $d$ are possible:
| ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 28. At the Ball
When I saw Eleanor, I found her very pretty. After a brief banal conversation, I told her how old I was and asked about her age. She replied:
- When you were as old as I am now, you were three times as old as I was. When I am three times as old as I am now, together we will be exactly a century old... | 28. So, let's turn to the conversation of two individuals. Let $e-$ be Eleonora's age, and $m$ - my age. When I was as old as she is now, Eleonora's age was
$$
e-(m-e)=2 e-m .
$$
Therefore, from her first statement, the following equality follows:
$$
e=3(2 e-m) \text{, or } 3 m=5 e \text{. }
$$
When Eleonora become... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 29. Leonie and Cats
When old lady Leonie is asked how many cats she has, she melancholically replies: “Four fifths of my cats plus four fifths of a cat.” How many cats does she have?
 | 29. Let $n$ be the number of cats Leonie has. From her last words, we can write the relationship
$$
n=(4 / 5) n+4 / 5, \text { i.e. } \quad n=4 .
$$
Thus, Leonie has 4 cats. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 30. New Mathematics
My son has learned to count in a numeral system different from the decimal system, and instead of 136, he writes 253 in this system. In what numeral system is my son counting? | 30. Let $a$ be the base of an unknown numeral system. When my son writes "253" in this system, it represents $2 a^{2}+5 a+3$; according to the problem, this corresponds to the number 136 in the decimal system. Therefore, we can write
$$
2 a^{2}+5 a+3=136
$$
i.e.
$$
2 a^{2}+5 a-133=0
$$
or
$$
(a-7)(2 a+19)=0
$$
Si... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 33. Least Number
What is the least number that, when divided by $2, 3, 4, 5$, and 6, gives remainders of $1, 2, 3, 4$, and 5 respectively? | 33. Let $n$ be an unknown number. Since $n$ when divided by 2 leaves a remainder of 1, the number $n+1$ is divisible by 2. Since $n$ when divided by 3 leaves a remainder of 2, the number $n+1$ is divisible by 3, and so on. Similarly, $n+1$ is divisible by 4, 5, and 6. But the least common multiple of 2, 3, 4, 5, and 6 ... | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 36. San Salvador Embankment
Would you come to have dinner with me tonight? I live in one of the eleven houses on San Salvador Embankment; however, to find out which one, you will have to think.
When, from my home, I look at the sea and multiply the number of houses to my left by the number of houses to my right, I... | 36. Let $d$ be the number of houses to the right of my house when I look at the sea, and $g$ be the number of houses to the left of it. Then
$$
d+g=10,
$$
and
$$
d g-(d+1)(g-1)=5
$$
or
$$
d+g=10, \quad d-g=4 .
$$
From this, $2 d=14, d=7$ and, therefore, $g=3$.
Thus, my house is the fourth from the left when look... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 38. How old is the eldest brother?
Determine this yourself, if it is known that the age of the middle brother is equal to the product of the ages of his two brothers, that the sum of the ages of all three brothers is 35, while the sum of the decimal logarithms of their ages is 3. | 38. Let $a$ be the age of the oldest, $c$ the age of the middle, and $b$ the age of the youngest brother. Then, based on the conditions of the problem, the following relationships hold:
$$
\begin{gathered}
c^{2}=a b \\
a+b+c=35 \\
\lg a+\lg b+\lg c=3
\end{gathered}
$$
From the first equation, it follows that
$$
2 \l... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 39. How old am I?
Try to answer this question if it is known that Alfred's age is obtained by rearranging the digits of my age, that the quotient of the number equal to my age divided by the sum of these two digits differs from the analogous quotient for Alfred by a number equal to the difference between these two ... | 39. Let $x$ be my age, $d$ be the number of tens in it, and $u$ be the number of units. Then $x=10 d+u$, and Alfred's age is $10 u+d$. The first condition of the problem can be written as:
$$
\left|\frac{10 d+u}{d+u}-\frac{10 u+d}{d+u}\right|=|d-u| \cdot
$$
Simplifying this equality, we get
$$
9|d-u|=(d+u) \cdot|d-u... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 40. Question about age
One fine spring Sunday morning, the head of the family went for a walk with his sons.
- Have you noticed,- he said to them,- that the age of the oldest of you is equal to the sum of the ages of the other two brothers?
- Yes. And we also noticed,一 they replied in unison,一 that the product of ... | 40. The head of the family has three sons. The product of the ages of the father and all his sons is
$$
3^{3} \cdot 1000 + 3^{2} \cdot 10 = 27090 = 43 \cdot 7 \cdot 5 \cdot 3^{2} \cdot 2
$$
Since the age of the oldest son is equal to the sum of the ages of his brothers, the only possible factorization of the obtained... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 43. Rue Saint-Nicaise
On December 24, 1800, First Consul Bonaparte was heading to the Opera along Rue Saint-Nicaise. A bomb exploded on his route with a delay of a few seconds. Many were killed and wounded. Bonaparte accused the Republicans of the plot; 98 of them were exiled to the Seychelles and Guiana. Several p... | 43. Let $x$ be the number of those executed; then, according to the last condition of the problem, the number of those killed in the explosion is $2 x+4$, and the number of wounded is
$$
2(2 x+4)+(4 / 3) x=5 x+x / 3+8
$$
(hence, $x$ is a multiple of 3).
It is also known that
$$
(2 x+4)+(5 x+x / 3+8)+x<98
$$
or
$$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 47. Suez and Panama
A meeting took place between Egyptians and Panamanians, where issues regarding the operation of the Suez and Panama Canals were discussed. In total, there were twelve participants from both sides, with more Egyptians than Panamanians. Upon arriving at the meeting place, the Egyptians greeted eac... | 47. If $n$ people greet each other in pairs, then a total of $n(n-1) / 2$ greetings take place. Therefore, we can create the following table:
In order for there to be exactly 31 greetings, there must have been seven Egyptians and five Panamanians at the meeting. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 48. Sequence of Numbers
To write the sequence of numbers from 1 to $n$, it took 2893 digits. What is the number $n$? | 48. To write the first nine single-digit numbers, it is necessary
$$
9 \cdot 1=9 \text{ digits. }
$$
To write the next 90 two-digit numbers, 180 digits are required, and to write the next 900 three-digit numbers, 2700 digits are required.
In total, we get 2889 digits. This is four digits less than the number of digi... | 1000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 51. Congratulations from Methuselah
Every New Year, starting from the first year of our era, Methuselah, who is still alive to this day, sends a greeting to his best friend, who, naturally, has changed many times over the centuries and decades. However, the formula for the greeting, on the contrary, has remained un... | 51. From the 1st to the 999th year, all digits were used an equal number of times, except for 0 (all digits were used exactly the same number of times if the first years were recorded by Methuselah as: year 0001, year 0002, ..., year 0999; however, since he did not do this, the digit 0 was used 111 times less frequentl... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 55. 100 years among the three
Patrick was walking with his father and grandfather. He wanted to imagine the time when all of them together would be 100 years old.
His father said on this matter: “I will be 28 years older than you, and you will be six-fifths of your current age.” And his grandfather added: “I will ... | 55. Let $e$ be Patrick's age, $p$ be the father's age, and $g$ be the grandfather's age; let $x$ be the time interval after which all three will be 100 years old. Then the following equations hold:
1) $(g+x)+(p+x)+(e+x)=100$
2) $e+x=(6 / 5) e$,
3) $(p+x)-(e+x)=28$,
4) $g+x=2(p-e+1.5)$.
From the last two equations, we ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 1. British Hospital
In a British hospital, the following announcement can be read: "Surgeon $X$ and Nurse $Y$ have the pleasure of announcing their forthcoming marriage." Suppose that at the time the hospital was opened, the groom was as old as the bride is now, and that the product of the ages of the groom and the... | 1. Let $x$ be the age (number of full years) of the surgeon (groom or bride), $y$ be the number of years of the paramedic (groom or bride), and $h$ be the number of years the hospital has been in existence.
The conditions of the problem can be written as the following two equations:
$$
\begin{gathered}
|x-y|=h \\
x(y... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 2. Clock
At the moment, according to my watch, it is more than 3:20 but less than 3:25. I observe the exact position of the hands on my watch, and then I move the hands so that the minute hand is in the position previously occupied by the hour hand, and the hour hand is in the position previously occupied by the mi... | 2. If $p$ and $g$ are the angles (in radians) that the hour and minute hands form with the direction from the center of the clock face to the 12 o'clock mark, then
$$
g=12(p-h \cdot 2 \pi / 12)
$$
where $h=0,1,2, \ldots$ or 12 is the number of hours ${ }^{1}$. If the angles $p_{0}$
${ }^{1}$ The angles $p$ and $g$ a... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 3. In the Cafeteria
Every day while having breakfast in the cafeteria, I noticed that snacks are always served between 7 and 8 o'clock, when both the hour and minute hands are equally distant from the number 6, and coffee is placed on the table at the moment when the minute hand catches up with the hour hand.
How ... | 3. Let 7 hours $x$ minutes be the time when the appetizer is served. The angle formed by the hour hand and the segment connecting the center of the clock face with the number 6 is (in degrees
$$
\frac{360}{12}+\left(\frac{x}{60} \cdot \frac{360}{12}\right)=30\left(1+\frac{x}{60}\right)=30+\frac{x}{2}
$$
The angle for... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 6. Coffee is served
Coffee is always served to me between 1 and 2 o'clock, at a moment when the bisector of the angle formed by the two hands of my clock points exactly at the "12" (hour) mark. At what exact time is my coffee served? | 6. Let's denote this unknown moment of time as 1 hour $x$ minutes (since it is between 1 hour and 2 hours!). The angle formed by the hour hand with the bisector of the angle between the two hands is (in degrees)
$$
\frac{360}{12}+\frac{x}{60} \cdot \frac{360}{12}=30+\frac{x}{2}
$$
(here we use the fact that the bisec... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 10. Confession
The confessor has a habit of assigning penances "by the tariff" - in exact proportion to the gravity of the sins committed. Thus, before absolving the sin of pride, he requires the penitent to recite the "Te Deum" once and the "Pater Noster" twice. Slander is "priced" at two "Pater Noster" and seven ... | 10. Since I will have to repeat the "Te Deum" only nine times, it is clear that I am not guilty of adultery.
I confess to one instance of slander, for ten "Credo" would be too few for two instances of slander, and if I had not slandered at all, the "Credo" would correspond to such a number of sins of pride and slander... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 13. Friendly Lunch
Ten married couples, who were friends with each other, met to have lunch together. First, they had an aperitif in the living room, and then all twenty people followed each other into the dining room. What is the minimum number of people that must go into the dining room so that among them there a... | 13. It is clear that as soon as thirteen people enter the dining room, there will inevitably be at least one married couple among them.
As soon as three people enter the dining room, there will certainly be at least two people of the same gender among them. | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 14. Dromedaries ${ }^{1}$
Some of the dromedaries are sleeping, while others are awake; the number of sleeping dromedaries is seven eighths of the number of awake dromedaries plus another seven eighths of a dromedary. If half of the sleeping dromedaries woke up, the number of awake dromedaries would be between 25 a... | 14. Let $d$ be the number of sleeping, and $r$ be the number of awake dromedaries. According to the problem, we have
$$
d=(7 / 8) r+7 / 8
$$
or
$$
8 d=7 r+7
$$
Thus, $d$ must be a multiple of 7.
On the other hand, this number must be such that it can be divided into two equal parts, i.e., it must be even; therefor... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 16. Grandmother, the Mark, and the Kittens
If you write down my mark, which I received in school for Latin ${ }^{1}$, twice in a row, you will get my grandmother's age. And what would you get if you divide this age by the number of my kittens? Imagine this - you would get my morning mark, increased by fourteen thir... | 16. Let $a$ be the age of my grandmother, $c$ be the number of my kittens, and $l$ be my grade in Latin. From the first condition of the problem, it follows that $l$ is an integer less than 10 (since it is clear that the age of my grandmother is less than 1010 years), and that $a=11 l$.
The second condition of the pro... | 77 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 17. Mathematical Puzzle
This week's mathematical puzzle in the "Modern Values" magazine was particularly difficult. When I started thinking about it (it was between 9 and 10 o'clock), I noticed that the two hands of my clock were in a straight line, one continuing the other. And I found the solution only at the mom... | 17. Let 9 hours $x$ minutes be the time when I started solving the problem. The angle formed by the hour hand with the common direction of the clock hands at noon is (in degrees)
$$
2 \cdot 360 / 12 + [(60 - x) / 60] \cdot 30
$$
The angle formed by the minute hand with the same direction is
$$
(360 / 60) \cdot x
$$
... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 18. Seals
Once I entered a maritime museum and saw some seals there. There were not many of them - just seven eighths of all the seals plus another seven eighths of a seal that I counted in the pool. How many seals were there in the pool in total?
dili in two rounds). According to the French constitution, a second... | 18. Let $n$ be the total number of seals. In this case,
$$
n=(7 / 8) n+7 / 8
$$
from which it follows that $n=7$.
There were seven seals in the marine museum. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 19. Tram
On one tram line, trams depart regularly every 10 minutes throughout the day. A tram takes one hour to travel from one end of the line to the other. A passenger boards a tram at one terminal stop and rides to the last stop of the tram; out of boredom, he looks out the window and counts the oncoming trams o... | 19. The passenger will encounter all the tram cars that departed from the other end of the tram line less than an hour before his departure; generally speaking, there are six such trains. He will also encounter all the trains that will depart from the opposite end of the line within the next hour; generally speaking, t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 20. Nicolas plays
## with tin soldiers
Nicolas has the same number of tin Indians, Arabs, cowboys, and Eskimos. After his cousin Sebastian's visit, he indignantly discovered the disappearance of a third of his soldiers. Suppose the number of Eskimos remaining is the same as the number of cowboys that disappeared, ... | 20. Let $x$ be the number of soldiers of each type that Nicolas had initially, and $y$ be the number of cowboys taken by Sebastian (or the number of remaining Eskimos); in this case, the number of Eskimos taken is $x-y$.
On the other hand, it is known that the number of Indians taken is $x / 3$. Let $z$ be the number ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 22. General Cleaning
A solid joint-stock company occupies three floors in a tower, which has the shape of a parallelepiped and is located in Paris in the La Défense district: the 13th and 14th floors, where the company's offices are located, and the 25th floor, where its board of directors is situated.
To clean th... | 22. Let $n$ be the unknown number of cleaners. The number of man-hours required to clean two floors of the bureau is
$$
4 \cdot(n+n / 2)
$$
The number of man-hours required to clean the 25th floor is
$$
4 \cdot(n / 2)+8
$$
But this last number is half of the previous one, since the area of one floor is half the are... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 29. Apples
Amelie picked apples, ate a quarter of them herself, and gave the rest to her little sisters Bertha, Charlotte, and Dorothy. Bertha ate a quarter of her apples and gave the rest to her three sisters. Charlotte ate one apple and gave the rest to her three sisters. As for little Dorothy, she also ate a qua... | 29. Let $a$ be the number of apples Amelie ends up with, $c$ be the number of apples Charlotte has at this stage, and $2c - y$ be the number of apples Berta has (0 being the number of apples Dorothy has). In the following table, we will represent all events in reverse order of their actual occurrence:
| | Amelie | Be... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 35. Parent Meeting
At the end of the school year, the third-grade teachers met with some of their students' parents; exactly 31 people were present at this meeting. The Latin teacher was asked questions by 16 parents, the French teacher by 17, the English teacher by 18, and so on up to the math teacher, who was ask... | 35. Let $n$ be the number of parents present at the meeting, and $m$ be the number of teachers. The first teacher talked to $15+1$ parents, the second to $15+2$, and so on, ..., the $m$-th teacher talked to $15+m$ parents. But the last teacher was the math teacher, with whom all the parents talked. Therefore,
$$
15+m=... | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 36. Sacks of Flour
Two trucks are transporting identical sacks of flour from France to Spain. The first one carries 118 sacks, while the second one carries only 40. Since the drivers of these trucks do not have enough pesetas to pay the customs duty, the first one leaves 10 sacks for the customs officials, as a res... | 36. Let $x$ be the cost of one sack of flour, and $y$ be the duty charged per sack.
If the first driver left 10 sacks at the customs, he should pay duty only for 108 sacks. Similarly, the second driver should pay duty only for 36 sacks. As a result, we get the following system of equations:
$$
\begin{aligned}
10 x+80... | 1600 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 38. Simone and Her Complexes
Simone suffered from numerous complexes. Therefore, she decided to go to one psychoanalyst and after a course of treatment, she got rid of half of her complexes and half of one of the remaining complexes. Then she went to another psychoanalyst, thanks to which she got rid of half of the... | 38. Consider the following table:
| Stage of Treatment | Number of Complexes |
| :--- | :---: |
| | 1 |
| At the end | $(1+1 / 2) \cdot 2=3$ |
| Before the 3rd psychoanalyst | $(3+1 / 2) \cdot 2=7$ |
| Before the 2nd psychoanalyst | $(7+1 / 2) \cdot 2=15$ |
| Before the 1st psychoanalyst | $(7+1)$ |
Thus, at the beg... | 2758 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 39. Square Plot
Mathieu has a square plot of land. Surrounding this plot is an alley of constant width with an area of $464 \mathrm{~m}^{2}$. Walking around his plot, Mathieu noticed that the difference in length between the outer and inner edges of the alley is 32 m.
What is the total area of Mathieu's plot, incl... | 39. Let $x$ be the length of the outer edge of the alley, and $y$ be the length of the inner edge of the alley. We have the following system of equations:
$$
\begin{gathered}
x^{2}-y^{2}=464 \\
4 x-4 y=32
\end{gathered}
$$
From the second equation, it follows that $x=y+8$. Then the first equation gives $x=33$ and $y=... | 1089 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 40. Tennis Tournament
199 people have registered to participate in a tennis tournament. In the first round, pairs of opponents are selected by lottery. The same process is used to select pairs in the second, third, and all subsequent rounds. After each match, one of the two opponents is eliminated, and whenever the... | 40. After each match, one of the opponents is eliminated, and since 199 people are participating in the tournament, by the end, 198 athletes must be eliminated. Therefore, for the entire tournament, 198 boxes of balls will be needed. | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 42. Barrels
In how many ways can a 10-liter barrel be emptied using two containers with capacities of 1 liter and 2 liters? | 42. Let $u_{n}$ be the number of ways to empty a barrel of capacity $n$ liters using two vessels of capacity 1 and 2 liters.
A barrel of $n$ liters will be emptied if we first use either the 1-liter vessel [and then we need to choose $(n-1)$ liters more from the barrel], or the 2-liter vessel [and then we need to choo... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## 43. Calves, Cows, Pigs, Chickens
The dowry of a young peasant girl consists of a cow, a calf, a pig, and a chicken. Her fiancé knew that five cows, seven calves, nine pigs, and one chicken together cost 108210 francs, and also that a cow is 4000 francs more expensive than a calf, that three calves cost as much as t... | 43. Let $c$ be the price of a pig. Then the price of a calf is $(10 / 3) c$, and therefore, the price of a cow is $(10/3)c+4000$ and the price of a chicken is $(5 / 3000) \cdot(10 / 3) \cdot c$.
Considering the first piece of information the suitor had, we get
$$
(50 / 3) c+20000+(70 / 3) c+9 c+c / 180=108210
$$
fro... | 17810 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 2. Marel
Take a piece of cardboard and draw the following diagram on it:
Also, take three silver and three gold coins (if you are on the beach, draw the diagram directly in the sand and... | 2. Let's renumber our nine circles (nine positions of the scheme):
Let $C$ be the player who starts, and $F$ be the player who finishes placing their coins.
I. The player who starts the ga... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
150*. Prove that if $a+b+c=0$, then
$$
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9
$$
where $a \neq 0, \quad b \neq 0, c \neq 0, a \neq b, \quad a \neq c, b \neq c$. | Proof.
$$
\begin{gathered}
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)= \\
=1+\frac{(b-c) c}{a(a-b)}+\frac{(c-a) c}{b(a-b)}+\frac{a(a-b)}{c(b-c)}+ \\
+1+\frac{(c-a) a}{b(b-c)}+\frac{b(a-b)}{c(c-a)}+\frac{(b-c) b}{a(c-a)}+1= \\
=3+\frac{c}{a-b}\left... | 9 | Algebra | proof | Yes | Yes | olympiads | false |
233. If $x_{1}$ and $x_{2}$ are the roots of the equation
$$
\frac{3 a-b}{c} x^{2}+\frac{c(3 a+b)}{3 a-b}=0
$$
then, without solving it, find $x_{1}^{117}+x_{2}^{117}$. | Solution.
$x_{1}^{117}+x_{2}^{117}=\left(x_{1}+x_{2}\right)\left(x_{1}^{116}-x_{1}^{115} x_{2}+\ldots+x_{2}^{116}\right), \quad$ but $\quad x_{1}+x_{2}=0$, so, $x_{1}^{117}+x_{2}^{117}=0$.
Note. It is useful to solve problem 233 when reviewing, after discussing the equality:
$$
a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
352. Find a natural number that is equal to the sum of all its preceding natural numbers. Does more than one such number exist, or is there only one? | Let $n$ be the desired natural number; the sum of the numbers preceding this natural number is expressed by the formula
$$
S_{n-1}=\frac{n(n-1)}{2}
$$
According to the problem, $\frac{n(n-1)}{2}=n$, from which $n^{2}-3 n=0$ or $n_{1}=3, n_{2}=0$. The desired number is 3. It is the only one, as the second root of the ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. What will be the result if we add:
a) the smallest three-digit and the largest two-digit number;
b) the smallest odd one-digit and the largest even two-digit number. | a) The smallest three-digit number is 100, and the largest two-digit number is 99.
$$
100+99=199
$$
b) The smallest odd one-digit number is 1, and the largest even two-digit number is 98.
$$
1+98=99
$$
Answer: a) 199 ; b) 99 . | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The perimeter of a cardboard sheet in the shape of a square is 28 dm. How many square centimeters does its area contain? | 4. 5) $28: 4=7$ (dm) - the length of the side of the square;
2) $7 \times 7=49$ (sq. dm) - the area of the square;
3) $100 \times 49=4900$ (sq. cm).
Answer: the area of the square is 4900 sq. cm. | 4900 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A part is machined from a metal blank. The chips obtained from machining 8 parts can be remelted into one blank. How many parts can be made from 64 blanks? | 5. 6) $64: 8=8$ (blanks) - can be manufactured from the waste when machining 64 parts;
2) $8: 8=1$ (blank) - can be manufactured from the waste when machining 8 parts;
3) $64+8+1=73$ (parts).
Answer: from 64 blanks, 73 parts can be manufactured. | 73 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. How many strikes do the clocks make in a day if they strike once every half hour, and at each hour $1,2,3 \ldots 12$ times? | 1. $(1+2+3+4+5+6+7+8+9+10+11+12) \times 2+24=180$.
Answer: 180 hits. | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. What is the least number greater than 1992 that gives a remainder of $7$ when divided by $9$? | 5. Divide 1992 by 9 with a remainder
| $-1992\llcorner 9$ |
| :--- |
| $\frac{18}{-19}$ |
| $\frac{-18}{221}$ |
| $\frac{-12}{9}$ |
| 3 |
We obtained a remainder of 3. To make the remainder equal to 7, the divisor needs to be increased by four:
$$
1992+4=1996 \text {. }
$$
Answer: 1996. | 1996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the sum of all even numbers from 10 to 31. Calculate in different ways. | 6. Let's list all even numbers from 10 to 31:
$$
10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
$$
a) Let's add the obtained numbers:
$$
10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 = 220.
$$
b) We can notice that the sum of numbers equally distant from the ends is 40:
$$
10 + 30 = 12 + 28 = 14 + 26 = \ldots =... | 220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. What will be the result if we add the largest odd two-digit number and the smallest even three-digit number? | 1. The largest odd two-digit number is 99, and the smallest even three-digit number is 100.
$$
99+100=199
$$
Answer: The sum of the largest odd two-digit number and the smallest even three-digit number is 199. | 199 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Write in digits the number equal to the sum of 22 million, 22 thousand, 22 hundred, and 22 units. | 2. Many believe that this number will be 22022222.
In fact, this number is 22024222, because if you add 22 thousand, which is 22000, and 22 hundred, which is 2200, and 22 units to 22 million, which is 22000000, you get 22024222.
Answer: 22024222. | 22024222 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a two-digit number, the number of tens is two times less than the number of units. If you subtract the sum of its digits from this two-digit number, you get 18. Find this number. | 5. According to the condition, in a two-digit number, the number of tens is twice less than the number of units, i.e., the number of units is twice the number of tens.
Two-digit numbers that satisfy this condition:
$$
12 ; \quad 24 ; \quad 36 ; \quad 48 .
$$
The sum of the digits of each of the obtained numbers:
$$... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the greatest number of Saturdays that can be in a year? | 2. Since a year can have 365 or 366 days, we will choose the larger of these numbers - 366. Saturday occurs once every seven days. Therefore, to find the number of Saturdays in a year, we need to divide 366 by 7 with a remainder. We get
, 20 numbers are divisible by 5 \((100: 5=20)\).
Of these, 4 nu... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Find the sum of all possible different two-digit numbers, all digits of which are odd. | 7. To find out how many such numbers there are, let's try to determine which digits can stand in the tens place. There are five such digits: $1,3,5,7,9$.
The second digit must also be odd, so it can also be chosen in five ways.
As can be seen from the diagram, the problem is about three ducks.
Answer: 3 ducks. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There were plates on the shelf. First, from all the plates except two, $1 / 3$ part was taken, and then $1 / 2$ of the remaining plates. After this, 9 plates were left on the shelf. How many plates were on the shelf? | 4. Let's represent the condition of the problem in a drawing
1) $9 \times 2=18$ (plates) — remain after the first time plates were taken
2) $18-2=16$ (plates) - corresponds to $2 / 3$;
3) ... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.