problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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Example 18. Solve the equation
$$
x=(\sqrt{1+x}+1)(\sqrt{10+x}-4)
$$ | Solution. The domain of definition (DOD) of equation (6): $x \geqslant-1$. Multiplying both sides of equation (6) by the expression $\sqrt{1+x}-1$, we obtain the equation
$$
x(\sqrt{1+x}-\sqrt{10+x}+3)=0
$$
which is a consequence of equation (6). This equation has two roots: $x_{1}=0$ and $x_{2}=-1$ (note that the DO... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 25. Solve the equation
$$
\sqrt{4-6 x-x^{2}}=x+4
$$ | Solution. The given equation is equivalent to the system
$$
\left\{\begin{array}{l}
x+4 \geqslant 0 \\
4-6 x-x^{2}=(x+4)^{2}
\end{array}\right.
$$
i.e., the system
$$
\left\{\begin{array}{l}
x \geqslant-4 \\
x^{2}+7 x+6=0
\end{array}\right.
$$
## 68
Solving the equation \(x^{2}+7 x+6=0\), we find its two roots: \(... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 26. Solve the equation
$$
\sqrt{x+5}+\sqrt{2 x+8}=7
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{r}
x+5 \geqslant 0 \\
2 x+8 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation (12): $x \geqslant-4$. We have:
$$
\text { (12) } \begin{aligned}
\Leftrightarrow & \left\{\begin{array}{l}
x \geqslant-4 \\
x+5+2 \sqrt{x+5} \sqrt{2 x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 9. Solve the equation
$$
5^{2 x}-2 \cdot 5^{x}-15=0
$$ | Solution. Let $t=5^{x}$. Then $t^{2}-2 t-15=0$. From this, we find $t_{1}=5, t_{2}=-3$. Thus, the given equation is equivalent to the system of equations
$$
5^{x}=5, \quad 5^{x}=-3
$$
## § 3. EXPONENTIAL EQUATIONS
The second equation of this system has no roots, since $-3<0$ for any $x$, while from the first equatio... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 15. Solve the equation
$$
27^{x}+12^{x}=2 \cdot 8^{x}
$$ | Solution. This equation is similar in appearance to equation (6): the exponent of the bases is the same, but the bases 27, 12, and 8 of the three consecutive terms of a geometric progression do not form one.
Consecutive (but four) terms of a geometric progression are the numbers $27, 18, 12$, and 8. Therefore, we can ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 16. Solve the equation
$$
5 \cdot 2^{3 x-3}-3 \cdot 2^{5-3 x}+7=0
$$ | Solution. Using the properties of exponents, rewrite the given equation as
$$
\frac{5}{8} \cdot 2^{3 x}-\frac{96}{2^{3 x}}+7=0
$$
This equation is of the form (12).
Let $t=2^{3 x}$; then we have $\frac{5}{8} t-\frac{96}{t}+7=0$, i.e., $5 t^{2}+56 t-768=0$. From this, we find $t_{1}=-96 / 5, t_{2}=8$.
Thus, equation... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Solve the equation:
a) $\log _{x+1}\left(x^{2}-3 x+1\right)=1$
b) $\log _{x}\left(2 x^{2}-3 x-4\right)=2$. | Solution. a) The given equation is equivalent to the system
$$
\left\{\begin{array} { l }
{ x + 1 > 0 } \\
{ x + 1 \neq 1 , } \\
{ x ^ { 2 } - 3 x + 1 = x + 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>-1 \\
x \neq 0 \\
x^{2}-4 x=0
\end{array} \Leftrightarrow x=4\right.\right.
$$
Therefore, the only roo... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 15. Solve the equation
$$
\log _{1 / 3}\left(x^{2}+3 x-4\right)=\log _{1 / 3}(2 x+2)
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}+3 x-4>0 \\
2 x+2>0
\end{array}\right.
$$
we find the domain of definition (ODZ) of equation (10): $x>1$.
Potentiating equation (10), we obtain the equation
$$
x^{2}+3 x-4=2 x+2
$$
which is a consequence of it.
Equation (11) has two root... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Solve the inequality
$$
\sqrt{6-x}\left(5^{x^{2}-7.2 x+3.9}-25 \sqrt{5}\right) \geqslant 0
$$ | Solution. The domain of definition of inequality (1) is determined by the condition $6-x \geqslant 0$, i.e., $x \leqslant 6$. Inequality (1) is equivalent to a combination consisting of an equation and a system of two inequalities:
$$
\sqrt{6-x}=0, \quad\left\{\begin{array}{l}
5^{x^{2}-7.2 x+3.9}-25 \sqrt{5} \geqslant... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Example 22. Solve the inequality
$$
\left(\sqrt{x^{2}-4 x+3}+1\right) \log _{2} \frac{x}{5}+\frac{1}{x}\left(\sqrt{8 x-2 x^{2}-6}+1\right) \leqslant 0
$$ | Solution. The domain of admissible values of the original inequality consists of all $x$ satisfying the system
$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
8 x-2 x^{2}-6 \geqslant 0
\end{array}\right.
$$
i.e., the system
$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
x^{2}-4 x+3 \leqsla... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
12. Find an integer whose cube is equal to the sum of the cubes of the three preceding consecutive numbers. | 12. Answer. 6. We obtain an interesting identity:
$$
3^{3}+4^{3}+5^{3}=6^{3}
$$ | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
52. Find the remainder when the number $50^{13}$ is divided by 7. | 52. $50^{13}=(49+1)^{13}=49 n+1$ (by the binomial theorem); therefore, the remainder of the division of $50^{13}$ by 7 (and also by 49) is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
98. For what integer values of $x$ is the number $a\left(x^{3}+a^{2} x^{2}+\right.$ $\left.+a^{2}-1\right)$ divisible by 6 for any integer $a$?
As is known, the difference $a^{n}-b^{n}$ is divisible by $a-b$ for any natural number $n$; $a^{n}-b^{n}$ is divisible by $a+b$ for even $n$; $a^{n}+b^{n}$ is divisible by $a+... | 98. $x=3 t$ and $x=3 t-a^{2}$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
162. Find a perfect number of the form $p q$, where $p$ and $q$ are prime numbers. | 162. $\frac{\left(p^{2}-1\right)\left(q^{2}-1\right)}{(p-1)(q-1)}-p q=p q, \quad q=1+\frac{2}{p-1}, \quad$ from here $p_{1}=2, p_{2}=3, q_{1}=3, q_{2}=2$.
Answer. 6 .
42 | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
262. Find a triangular number whose square is also a triangular number. | 262. From the equation $\left[\frac{x(x+1)}{2}\right]^{2}=\frac{y(y+1)}{2}$, we find:
$$
y=\frac{-1+\sqrt{2 x^{2}(x+1)^{2}+1}}{2}
$$
Let $x(x+1)=z$, then $2 z^{2}+1=t^{2}, t^{2}-2 z^{2}=1$.
Solving this Pell's equation, we get an infinite set of values $z=2,12,70,408 \ldots$. From the equation $x(x+1)=2$, we have $x... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
296. The difference between a number and the product of its digits is equal to the sum of the digits of this number. Find this number.
## Chapter $I X$.
## VARIOUS PROBLEMS | 296. Let the desired number contain $n$ digits $a, b, c$, $\ldots, k, l$. According to the condition $a \cdot 10^{n-1}+b \cdot 10^{n-2}+\ldots+$ $+k \cdot 10+l-a b c d \ldots k l=a+b+c+\ldots+k+l, \quad$ from which $9 k=a\left(b c \ldots k l+1-10^{n-1}\right)+b\left(1-10^{n-2}\right)+c\left(1-10^{n-3}\right)+$ $+\ldots... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Charity. A generous person distributed the same amount of money equally among those who asked him for assistance every week. One day he noticed:
- If the number of petitioners is five less next week, each will receive two dollars more.
But, alas, after a week, the number of petitioners did not decrease but increas... | 4. At first, there were 20 people and each received 6 dollars. Fifteen people (five fewer) would have received 8 dollars each. But their number increased to 24 (by four people), and each received only 5 dollars. Thus, the weekly donation amount is 120 dollars. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. The "Reverse" Game. Seven friends decided to play cards by somewhat unusual rules. The one who won had to pay each of the other players as much money as they had in their pocket. The players played seven rounds and, strangely enough, they won in alphabetical order of their names, starting with $A, B, C, D, E, F$ and... | 8. Players \(A, B, C, D, E, F\) and \(G\) had, before the game began, 4 dollars 49 cents, 2 dollars 25 cents, 1 dollar 13 cents, 57 cents, 29 cents, 15 cents, and 8 cents, respectively. The answer can be obtained by working from the end of the problem to the beginning, but a simpler method is as follows: \(7+1=8 ; 2 \t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
46. The Clocks from the Land of Dreams. In a dream, I traveled to a country where wondrous things happened. One incident was so memorable that I didn't forget it even after I woke up. In my dream, I saw a clock and spoke aloud the time it showed, but my guide corrected me. He said:
- Clearly, you don't know that our m... | 46. If these were ordinary clocks, they would show 4 hours \(23 \frac{1}{13}\) minutes. But since the minute hand moved in the direction opposite to the hour hand, the true time was 4 hours \(36 \frac{12}{13}\) minutes. To get the true time, you need to subtract from 60 the number of minutes that the clock shows. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
48. Indistinguishable Clock Hands. A person had a clock on which it was impossible to distinguish the hour hand from the minute hand. If this clock was started at noon, when would it first be impossible to tell the exact time?
The reader should remember that in such clock puzzles, there is a convention that we are abl... | 48. This will first occur at 12 hours \(5 \frac{5}{143}\) minutes, which could be incorrectly interpreted (due to the identical positions of the hands) as 1 hour \(\frac{60}{143}\) minutes. | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 51. Mixed-Up Hands.
- Yesterday, between two and three o'clock, - said Colonel Crackham, - I looked at my watch and, mistaking the hour hand for the minute hand, I was wrong in my estimate of the time. The erroneous time was 55 minutes less than the actual time. What was the actual time? | 51. The true time was 2 hours \(5 \frac{5}{11}\) minutes. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
52. Equal Distances. A few days ago, Professor Rackbrain stunned his students with the following puzzle:
- When between three and four o'clock is the minute hand at the same distance from VIII as the hour hand is from XII? | 52. At 3 hours \(23 \frac{1}{13}\) minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
53. To the right and to the left. At what time between three and four o'clock is the minute hand the same distance to the left of XII as the hour hand is to the right of XII? | 53. At 3 hours \(41 \frac{7}{13}\) minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
54. At a Right Angle. Once during breakfast, Professor Rackbrain asked his young friends a simple question:
- When between five and six o'clock will the hour and minute hands be exactly at a right angle? | 54. For the angle between the hands to be a right angle, the minute hand must be exactly 15 minutes ahead of or behind the hour hand. Each of these positions will occur 11 times in 12 hours, that is, every 1 hour \(5 \frac{5}{11}\) minutes. If eight such intervals pass after 9 o'clock, the clock will show 5 hours \(43 ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
59. A Walk. One man set out for a walk from Appleminster to Boniham at noon, and his friend left Boniham for Appleminster at two o'clock of the same day. On the way, they met. The meeting took place at five minutes to five, after which the friends arrived simultaneously at their final destinations. When did they finish... | 59. It should be noted (and this is the key to the solution) that the person from B. covers 7 km in the same time that the person from E. covers 5 km. Let's assume, for example, that the distance between the cities is 24 km, then they met 14 km from E. The person from E. was moving at a speed of \(3 \frac{3}{7}\) km/h,... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
65. Motorcycle with a sidecar. Atkins, Baldwin, and Clark decided to go on a trip. Their journey will be 52 km. Atkins has a motorcycle with a one-person sidecar. He needs to take one of his companions for some distance, drop him off to walk the rest of the way, return, pick up the other companion who started walking a... | 65. Atkins drives Clark 40 km and drops him off to walk the remaining 12 km. Then he returns and picks up Baldwin 16 km from the start and drives him to the end. The three of them spend 5 hours on the road. Another solution is for Atkins to first drive Baldwin 36 km and return for Clark, who by this time has walked 12 ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
74. Baxter's Dog. Here is an interesting puzzle, complementing the previous one. Anderson left the hotel in San Remo at 9 o'clock and was on the road for a whole hour when Baxter set out after him along the same route. Baxter's dog ran out at the same time as its owner and kept running back and forth between him and An... | 74. It is quite obvious that Baxter will catch up with Anderson in one hour, by which time they will have each traveled 4 km in the same direction. Furthermore, the dog’s speed is 10 km/h; therefore, in this hour, it will have run 10 km! When this puzzle was presented to a French mathematics professor, he exclaimed, “M... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
79. Oncoming Trains. At Wurzltown station, an old lady, looking out the window, shouted:
- Conductor! How long is it from here to Madville?
- All trains take 5 hours in either direction, ma'am, - he replied.
- And how many trains will I meet on the way?
This absurd question puzzled the conductor, but he readily answe... | 79. Since the train travels for 5 hours, we will divide the journey into 5 equal intervals. When the lady departs from Wurzeltown, 4 oncoming trains are already en route, and the fifth one is just leaving the station. She will meet each of these 5 trains. When the lady has traveled \(\frac{1}{5}\) of the distance, a ne... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
85. The Fly and the Cars. The length of the road is 300 km. Car $A$ starts at one end of the road at noon and moves at a constant speed of 50 km/h. At the same time, at the other end of the road, car $B$ starts at a constant speed of 100 km/h and a fly, flying at 150 km/h. Upon meeting car $A$, the fly turns and flies ... | 85. 1) The fly will meet \(B\) in 1 hour 48 minutes.
2) There is no need to determine the distance the fly will fly. This is too difficult a task. Instead, we can simply find the time when the cars could have collided, which is 2 hours. In reality, the fly flies (in kilometers):
\[
\frac{270}{1}+\frac{270}{10}+\frac{... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
88. Transportation Puzzle. Twelve soldiers need to get to a point 20 km away from their location as quickly as possible. To do this, they stopped a small car.
- I drive at a speed of 20 km/h, - the driver said, - but I can only take four of you at a time. How fast do you walk?
- Each of us walks 4 km/h, - one of the s... | 88. The driver must transport four soldiers 12 km and drop them off 8 km from the destination. Then he must return 8 km and pick up another four soldiers (out of eight) who will be there by then, transport them 12 km and drop them off 4 km from the destination. Returning 8 km to pick up the remaining soldiers, who by t... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
99. Fruits for Jam. For making jam, it was necessary to weigh fresh fruits. It turned out that apples, pears, and plums balance each other as shown in the figure.
Could you tell how many p... | 99. On the first scales, we see that an apple and 6 plums weigh the same as a pear, so on the second scales, we can replace the pear with an apple and 6 plums without disrupting the balance. Then we can remove 6 plums from each side and find that 4 apples weigh the same as 4 plums. Therefore, one apple weighs the same ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
122. Digits and Cubes. Professor Rackbrain recently asked his young friends to find all five-digit squares for which the sum of the numbers formed by the first two and the last two digits is a perfect cube. For example, if we take the square of 141, which is 19881, and add 81 to 19, we get 100 - a number that, unfortun... | 122. There are three solutions: \(56169\left(237^{2}\right)\), where \(56+69=125\) \(\left(5^{3}\right) ; 63001\left(251^{2}\right)\), where \(63+01=64\left(4^{3}\right)\) and \(23104\left(152^{2}\right)\), where \(23+04=\) \(27\left(3^{3}\right)\). | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
140. An example of multiplication. One morning at breakfast, the Crackhams were discussing high matters, when suddenly George asked his sister Dora to quickly multiply
$$
1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 0
$$
How long will it take the reader to find this product? | 140. George's question did not catch Dora off guard. She immediately gave the correct answer: 0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 163. The Cat and the Mouse.
- One of these barrels has a mouse in it,- said the dog.
- Which one? - asked the cat.
- Well, the five hundredth.
- What do you mean by that? There are only five barrels here.
- The barrel I mean will be the five hundredth if you start counting forward and backward like this.
And the d... | 163. You just need to divide the given number by 8. If it divides evenly, with no remainder, the mouse is in the second barrel. If the remainder is 1, 2, 3, 4, or 5, the barrel number will match this remainder. If the remainder is greater than 5, subtract it from 10. The resulting difference is the barrel number. The n... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
182. Quarrelsome Children. A man married a widow, and each of them had children from their previous marriages. After 10 years, a battle broke out in which all the children (by then there were 12) participated. The mother ran to the father, shouting:
- Come quickly. Your children and my children are beating our childre... | 182. Each of the parents had 3 children from their first marriage, and 6 children were born from their second marriage. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
216. Blindness in Bats. One naturalist, trying to mystify Colonel Crackham, told him that he had studied the question of blindness in bats.
- I found,- he said, - that the ingrained habit of bats to sleep during the day in dark corners and fly only at night has led to the spread of blindness among them, although some ... | 216. The smallest number of mice is 7, with three possible cases:
1) 2 see well, 1 is blind only in the right eye, and 4 are completely blind
2) 1 sees well, 1 is blind only in the left eye, 2 are blind only in the right eye, and 3 are completely blind;
3) 2 are blind only in the left eye, 3 only in the right eye, and ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
239. Fence Order. One person ordered a fence with a total length of 297 m. The fence was to consist of 16 sections, each containing a whole number of meters. Moreover, 8 sections were to have the maximum length, while the others were to be 1, 2, or 3 meters shorter.
How should this order be carried out? Suppose that t... | 239. 8 sections of 20 m each, 1 section 18 m long, and 7 sections of 17 m each were manufactured. Thus, a total of 16 sections with a total length of \(297 \mathrm{m}\) were obtained, as required by the customer. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
243. Обезьяна и груз. Вот одна забавная задачка, которая представляет собой симбиоз нескольких головоломок, в том числе головоломок Льюиса Кэрролла «Обезьяна и груз» и Сэма Лойда «Сколько лет Мэри?» Хорошенько подумав, вы ее безусловно решите.
Через блок перекинута веревка, на одном конце которой висит обезьяна, а на ... | 243. Сначала мы находим возраст обезьяны ( \(1 \frac{1}{2}\) года) и возраст ее матери ( \(2 \frac{1}{2}\) года). Следовательно, обезьяна весит \(2 \frac{1}{2}\) фунта и столько же весит груз. Затем мы находим, что вес веревки составляет \(1 \frac{1}{4}\) фунта, или 20 унций, а поскольку каждый фут весит 4 унции, то дл... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
270. The Runner's Puzzle. $A B C D$ is a square field with an area of 19.36 hectares. $B E$ is a straight path, and $E$ is 110 meters from $D$. During the competition, Adams ran straight from $A$ to $D$, while Brown started running from $B$, reached $E$, and then continued towards $D$.
 is a right-angled triangle. Therefore, \(A E=330\) m, \(B E=550\) m. If Brown runs 550 m in the same time it takes Adams to run 360 m \((330+30)\), then Brown can run the remaining 100 m in the time it takes Adams to run only \(72 \mathrm{m}\). But \(30+72=102 \mathrm{m}\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
282. Cross of chips. Arrange 20 chips in the shape of a cross, as shown in the figure. How many different cases can you count where four chips form a perfect square?
For example, squares are formed by the chips at the ends of the cross, the chips located in the center, as well as the chips marked with the letters $A$ ... | 282. There are 19 such squares in total. Of these, 9 are the same size as the square marked with the letters \(a\), 4 are the same size as the square marked with the letters \(b\), 4 are the size of \(c\), and 2 are the size of \(d\). If 6 chips marked with the letter \(e\) are removed, it will be impossible to form an... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
368. Black and White. One day over a cup of tea, Professor Rackbrain showed his friends the following old puzzle.
Arrange 4 white and 4 black chips in a row, alternating as shown in the pi... | 368. In the first case, move the pairs in the following order: place 6 and 7 before 1, then 3 and 4, 7 and 1, and 4 and 8 in the free spaces. This will result in the following arrangement of chips: 6,4, 8,2,7,1,5,3.
In the second case, move the chips 3,4 and place them in reverse order (4,3) before chip 1. Then move, ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
377. Difference Squares. Can you arrange 9 digits in a square so that in any row, any column, and on each of the main diagonals, the differences between the sum of two digits and the third digit are the same? In the square provided in our diagram, all rows and columns meet the required condition - the difference in the... | 377. Apparently, there are only three solutions provided here. In each case, the difference is 5.
OTBETM
277
 | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
436. The Crossing. Six relatives need to cross a river in a small boat that can only hold two people at a time. Mr. Webster, who was in charge of the crossing, had a falling out with his father-in-law and son. Unfortunately, I must also note that Mrs. Webster is not speaking to her mother and her daughter-in-law. The t... | 436. The puzzle can be solved in 9 crossings as follows:
1) Mr. and Mrs. Webster cross together;
2) Mrs. Webster returns;
3) the mother and daughter-in-law cross
4) Mr. Webster returns;
5) the father-in-law and son cross
6) the daughter-in-law returns;
7) Mr. Webster and the daughter-in-law cross;
8) Mr. Webste... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
484. Zero from fifty-seven. After the previous puzzle, this one will seem quite simple.
In our drawing, you see 6 cigarettes (matches would work just as well) arranged to form the number 57. The puzzle is to move two of the cigarettes without touching the others to get 0.
, and place them as shown in the figure. We have the square root of 1 minus 1 (i.e., \(1-1\)), which is obviously equal to 0. In the second case, we can move the same two cigarettes, placing one next to V and the other next to the second I, so that the word NIL (noth... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The Absent-Minded Secretary. The typist typed ten letters and addresses on ten envelopes, but the absent-minded secretary placed these letters into the envelopes without any regard for the correspondence between the letter and the addressee. However, she did place only one letter in each envelope. What is the probab... | 1. If nine letters have gone into their intended envelopes, then the tenth letter will certainly do the same. Therefore, the probability that exactly nine letters have gone into their envelopes is zero.
$$
[M . M ., \mathbf{3 3}, 210(\text { March 1950).] }
$$ | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Zero-sum test. A certain test consists of 26 questions. For each incorrect answer, five points are deducted from the test-taker, and for each correct answer, eight points are awarded.
The test-taker answered all the questions. How many questions did he answer correctly if the total number of points he received was ... | 4. The ratio of the number of answers of each type is equal to the inverse ratio of the corresponding points. Therefore, the number of correct answers is $\frac{5}{5+8} \cdot 26=10$.
[M. M., 31, 237 (March 1958).] | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Fourth-order equation. How many negative roots does the equation
$$
x^{4}-5 x^{3}-4 x^{2}-7 x+4=0 ?
$$
have? | 8. The equation $x^{4}-5 x^{3}-4 x^{2}-7 x+4=0$ can be rewritten as $\left(x^{2}-2\right)^{2}=5 x^{3}+7 x$. Since for any negative $x$ the left side of the equation is positive, while the right side is negative, the original equation cannot have negative roots.
[P. E. Horton, M. M., 24, 114 (November 1950).] | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. Flower Seller. A girl bought $x$ roses in the store, paying $y$ dollars for all ( $x$ and $y$ are integers). When she was about to leave, the seller told her: "If you bought 10 more roses, I would give you all the roses for 2 dollars, and you would save 80 cents on each dozen." Find $x$ and $y$. | 20. Since $y$ is an integer $<2, y=1^{\star}$. Then, expressing the cost of one rose in cents, we get
$$
\frac{100}{x}-\frac{200}{x+10}=\frac{80}{12}, \text{ or } x^{2}+25 x-150=0.
$$
The only positive root of this equation is $x=5$. This is the number of roses the girl initially bought. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. Unknown remainder. Find the remainder of the division of $f\left(x^{5}\right)$ by $f(x)$, if
$$
f(x)=x^{4}+x^{3}+x^{2}+x+1
$$ | 30. Since $f(x)=x^{4}+x^{3}+x^{2}+x+1,(x-1) f(x)=x^{5}-1$. Further, $f\left(x^{5}\right)=\left(x^{20}-1\right)+\left(x^{15}-1\right)+\left(x^{10}-1\right)+\left(x^{5}-1\right)+4+1$. But $x^{5}-1$, and therefore $f(x)$, are divisors of each of the brackets. Hence,
$$
f\left(x^{5}\right)=[\text { multiple of } f(x)]+5
$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
35. Simplifying Radicals. Simplify the expression
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}
$$ | 35. Let $\sqrt[3]{2+\sqrt{5}}=a, \quad \sqrt[3]{2-\sqrt{5}}=b, \quad a+b=x$. Then
$$
x^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=a^{3}+b^{3}+3 a b(a+b)=4+3(\sqrt[3]{-1}) x
$$
Thus, $x^{3}+3 x-4=0$, and the only real root of this equation is 1.
$[$ [K. Adler, A. M. M., 59, 328 (May 1952).] | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
41. The Dozing Schoolboy. A schoolboy, waking up at the end of an algebra lesson, heard only a fragment of the teacher's phrase: "… I will only say that all the roots are real and positive." Glancing at the board, he saw there a 20th-degree equation assigned as homework, and tried to quickly write it down. He managed t... | 41. The roots are positive; their arithmetic mean is $-\frac{(-20)}{20}$, and their geometric mean is $(+1)^{\frac{1}{20}}$. Since both these values coincide, it follows that all roots are equal to 1.
[D. S. Greenstein, A. M. M., 63, 493 (September 1956).] | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
51. An equation containing sums. Find $n$, if
$$
\frac{1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3}}{2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3}}=\frac{199}{242}
$$ | 51. First, note that if \(a: b = c: d\), then obviously \((a+b): b = (c+d): d\). Applying such a transformation to the fractions in our equation, we get
\[
\frac{1^{3}+2^{3}+3^{3}+\cdots+(2 n)^{3}}{2^{3}\left(1^{3}+2^{3}+3^{3}+\cdots+n^{3}\right)}=\frac{441}{242}
\]
Then, applying the known formula for the sum of cub... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
56. Probability of divisibility. Find the probability that if the digits $0,1,2, \ldots, 9$ are placed in random order in the empty spaces in the sequence of digits
$$
5-383-8-2-936-5-8-203-9-3-76
$$
then the resulting number will be divisible by 396. | 56. The number 76, formed by the last two digits, is divisible by 4. The difference between 73 (the sum of all digits in even positions) and $17+45$ (the sum of all digits in odd positions) is divisible by 11 regardless of the order in which the empty places are filled*. The sum of all digits, $90+45$, is divisible by ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
58. The son of a mathematics professor. The mathematics professor wrote a polynomial $f(x)$ with integer coefficients on the board and said:
- Today is my son's birthday. If his age $A$ is substituted into this polynomial instead of $x$, then the equation $f(A)=A$ holds. Note also that $f(0)=P$, where $P$ is a prime n... | 58. Since $f(0)=P$,
$$
f(x)=x \cdot q(x)+P \quad \text { and } \quad f(A)=A \cdot q(A)+P=A .
$$
Therefore, $P$ is divisible by $A$. Since $P>A$ and $P$ is prime, $A=1$. Thus, the professor's son is 1 year old. The professor could have written down any polynomial from an infinite class of such polynomials, for example... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
60. Meteorological observations. At a weather station, it was noticed that during a certain period of time, if it rained in the morning, it was clear in the evening, and if it rained in the evening, it was clear in the morning. In total, there were 9 rainy days, with clear evenings 6 times and clear mornings 7 times. H... | 60. There were $\frac{1}{2}(6+7-9)=2$ completely clear days, so the period under consideration covered $9+2=11$ days.
$[$ [M. M., 34, 244 (March 1961).] | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
81. Sheep Buyers. A farmer died, leaving a herd of cows to his two sons. The sons sold the herd, receiving as many dollars for each head as there were heads in the herd. With the money, the brothers bought sheep at 10 dollars each and one lamb, which cost less than 10 dollars. Then they divided the sheep and the lamb b... | 81. Let $x$ be the number of cows in the herd, $y$ the number of sheep, and $z$ the cost of a lamb. Then $x^{2}=10 y+z$, where $y$ is an odd number, and $z<10$. But the second-to-last digit of a square is odd if and only if the last digit is $6^{*}$. Thus, $z=6$, and the luckier son should pay his brother 2 dollars.
[... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
88. Choosing a pair. Among the pairs of numbers listed below, one and only one does not satisfy the equation $187 x-104 y=41$. Which one exactly
1) $x=3, \quad y=5$
2) $x=107, \quad y=192$
3) $x=211, \quad y=379$
4) $x=314, \quad y=565$
5) $x=419, \quad y=753$. | 88. Since the difference between the two terms on the left side of the equation is the odd number 41, one of these terms must be odd and the other even. Since $104 y$ is even, $187 x$ is odd, and therefore $x$ is odd. Thus, the pair $x=314$, $y=565$ does not satisfy our equation.
[D. Woods, S. S. M., 64, 242 (March 19... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
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,
$$
\begin{gathered}
a=2^{\star} ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
105. The Farmer's Task. A certain farmer must buy 100 heads of cattle for 100 dollars. If each calf costs 10 dollars, each lamb 3, and each piglet 0.5 dollars, then how many calves, lambs, and piglets will the farmer buy? | 105. The average cost of one head of cattle is 1 dollar. The cost of each calf differs from the average by +9 dollars, each lamb by +2 dollars, and each piglet by $-\frac{1}{2}$ dollar. Therefore, for each calf, the farmer must buy 18 piglets, and for each lamb, 4 piglets. Consequently, since $5(1+18)+(1+4)=100$, he mu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
108. Unique Square. What square is equal to the product of four consecutive odd numbers? | 108. If $n(n+2)(n+4)(n+6)=m^{2}$, then $\left(n^{2}+6 n+4\right)^{2}=m^{2}+16$. However, among the squares, only 0 and 9 have the form $a^{2}-16$; and since $m^{2}$ is odd, the sought square is $9=(-3)(-1)(1)(3)$.
[D. L. Silverman, M. M., 38, 60 (January 1965).] | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
110. Sum of Cosines. Calculate the sum
$$
\cos 5^{\circ}+\cos 77^{\circ}+\cos 149^{\circ}+\cos 221^{\circ}+\cos 293^{\circ}
$$ | 110. Project the sides of an arbitrary polygon onto a line lying in the plane of this polygon. Then the sum of such projections, taken with the appropriate sign, is zero. Now let's take a regular pentagon with a unit side and note that its exterior angle is $72^{\circ}$. The terms
=5
$$
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 |
157. Ballot Papers. A physical society needed to hold elections for three leadership positions. For each of the three positions, there were 3, 4, and 5 candidates, respectively. To ensure that the number under which each candidate is listed on the ballot does not influence the voting results, it was decided to apply a ... | 157. Perhaps someone will immediately say that it is necessary to take $3 \cdot 4 \cdot 5=60$ different ballots. However, if we add two fictitious names to the group of three candidates and one fictitious name to the group of four candidates, then only 5 different ballots will be required. This technique not only reduc... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
163. Reduced Share. A father gave his children 6 dollars for entertainment, which was to be divided equally. But two young cousins joined the company. The money was divided equally among all the children, so that each child received 25 cents less than originally intended. How many children were there in total? | 163. The share of each child decreased by $\frac{2}{24}$ of the entire sum. Further, $24=2 \cdot 12=3 \cdot 8=4 \cdot 6$. From these representations of the number 24, we choose the pair $q_{1} c_{1}=q_{2} c_{2}$ such that $q_{1}+1=q_{2}$, and $c_{1}=c_{2}+2$. Thus, there were initially 6 children, and then there were 8... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
209. "Fibonacci Tetrahedron". Find the volume of the tetrahedron whose vertices are located at the points with coordinates $\left(F_{n}, F_{n+1}, F_{n+2}\right), \quad\left(F_{n+3}, F_{n+4}, F_{n+5}\right), \quad\left(F_{n+6}, F_{n+7}, F_{n+8}\right)$ and $\left(F_{n+9}, F_{n+10}, F_{n+11}\right)$, where $F_{i}$ is the... | 209. The Fibonacci sequence satisfies the recurrence relation $F_{n}+F_{n+1}=F_{n+2}$, so any three consecutive Fibonacci numbers satisfy the equation $x+y=z$. Consequently, all four vertices of our tetrahedron lie in the same plane, and its volume is 0.
We can observe that all the reasoning remains valid even if the ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
231. Henry's Walk. Henry went for a walk in the countryside sometime between 8 and 9 in the morning, when the hands of his clock were aligned. He arrived at his destination between 2 and 3 in the afternoon; at this time, the hands of his clock were pointing in exactly opposite directions. How long did Henry's walk last... | 231. The walk lasted 6 hours. Suppose we extended the hour hand in the opposite direction. If at the beginning of the walk the hour and minute hands were coincident, then after six hours the hour hand and its extension will simply have swapped places, while the minute hand, having completed exactly six full circles, wi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
251. Packing Cylinders. Forty cylinders with a diameter of 1 cm and the same height were tightly packed in a box in 5 rows of 8 cylinders each so that they would not "rattle" during transportation. How many cylinders need to be removed from the box so that, by moving the remaining cylinders and adding the removed cylin... | 251. Only two cylinders need to be removed from the box. Let's renumber the cylinders as shown in the figure. Remove cylinders 6 and 16.
Move cylinders $7-10$ to the right and up (see the figure). Move cylinders 11-15 to the left, cylinders 17-20 up and to the left, and cylinders 21-25 to the left. The remaining cylin... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
266. When division is exact. For which positive integer $n$ will the quantity $n^{4}+n^{2}$ be divisible by $2 n+1$? | 266.
$$
\begin{aligned}
f(n) & =\frac{n^{4}+n^{2}}{2 n+1}=\frac{n^{2}\left(n^{2}+1\right)}{2 n+1}=\frac{n^{2}}{4}\left[\frac{4 n^{2}+4}{2 n+1}\right]= \\
& =\left(\frac{n}{2}\right)^{2} \cdot\left[2 n-1+\frac{5}{2 n+1}\right]
\end{aligned}
$$
It is obvious that the greatest common divisor of the numbers $n$ and $2n+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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|
$$[^16]
Find the maximum possible $n$ and for this $n$ all sets satisfying the condition of the problem. | 275. Let $\left|z_{m}\right|=\max _{i}\left|z_{i}\right|$. Then on the complex plane, all points $z_{i}$ will be located inside the circle $R$ of radius $\left|z_{m}\right|$ centered at the point $z=0$. Clearly, 6 points $z_{i}$, located on the circumference of $R$ and forming a regular hexagon, together with the point... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
322. Limit of a Sum. Calculate the value of the expression
$$
\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 |
Ex. 104. A regular triangle $ABC$ is inscribed in a circle. Another, smaller circle, is inscribed in the sector bounded by the chord $BC$, touching the larger circle at point $M$, and the chord $BC$ at point $K$. The ray $MK$ intersects the larger circle a second time at point $N$. Find the length of $MN$, if the sum o... | Ex. 104. 6. Hint. Show that point $N$ coincides with $A$ and use the result of Ex. 28. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ex. 118. A circle with center on side $AB$ of triangle $ABC$ touches sides $AC$ and $BC$. Find the radius of the circle, given that it is expressed as an integer, and sides $AC$ and $BC$ are equal to 5 and 3. | Ex. 118. Answer: $r=1$. Solution. By equating the expressions for the area of triangle $ABC$, we get that $\frac{a+b}{2} r=\frac{a b}{2} \sin C$. From this, it follows that $r=\frac{a b}{a+b} \sin C$. In our case, $r=\frac{15}{8} \sin C$. If $r \geq 2$, then $\sin C>1$. This solves the problem. | 1 | 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 |
## 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.
 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 |
## 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
$$
1979-\left(1000 m^{\prime... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 27. Large Families
The Martens have more children than the Duponts. Suppose the difference of the squares of these two numbers is 24 and that both families have more than one child. How many children do the Martens have?
70 | 27. Let $m$ and $d$ be the number of children in the Marten and Dupont families, respectively. According to the problem,
$$
m^{2}-d^{2}=24 \text {, i.e., }(m+d)(m-d)=24\left(=2^{3} \cdot 3\right) \text {. }
$$
Therefore, the following values for $m+d, m-d$, and thus $2 m[=(m+d)+(m-d)]$, $m$, and $d$ are possible:
| ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 29. Leonie and Cats
When old lady Leonie is asked how many cats she has, she melancholically replies: “Four fifths of my cats plus four fifths of a cat.” How many cats does she have?
 | 29. Let $n$ be the number of cats Leonie has. From her last words, we can write the relationship
$$
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
$$
Si... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 36. San Salvador Embankment
Would you come to have dinner with me tonight? I live in one of the eleven houses on San Salvador Embankment; however, to find out which one, you will have to think.
When, from my home, I look at the sea and multiply the number of houses to my left by the number of houses to my right, I... | 36. Let $d$ be the number of houses to the right of my house when I look at the sea, and $g$ be the number of houses to the left of it. Then
$$
d+g=10,
$$
and
$$
d g-(d+1)(g-1)=5
$$
or
$$
d+g=10, \quad d-g=4 .
$$
From this, $2 d=14, d=7$ and, therefore, $g=3$.
Thus, my house is the fourth from the left when look... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 43. Rue Saint-Nicaise
On December 24, 1800, First Consul Bonaparte was heading to the Opera along Rue Saint-Nicaise. A bomb exploded on his route with a delay of a few seconds. Many were killed and wounded. Bonaparte accused the Republicans of the plot; 98 of them were exiled to the Seychelles and Guiana. Several p... | 43. Let $x$ be the number of those executed; then, according to the last condition of the problem, the number of those killed in the explosion is $2 x+4$, and the number of wounded is
$$
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 |
## 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 |
## 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)
$$
where $h=0,1,2, \ldots$ or 12 is the number of hours ${ }^{1}$. If the angles $p_{0}$
${ }^{1}$ The angles $p$ and $g$ a... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 6. Coffee is served
Coffee is always served to me between 1 and 2 o'clock, at a moment when the bisector of the angle formed by the two hands of my clock points exactly at the "12" (hour) mark. At what exact time is my coffee served? | 6. Let's denote this unknown moment of time as 1 hour $x$ minutes (since it is between 1 hour and 2 hours!). The angle formed by the hour hand with the bisector of the angle between the two hands is (in degrees)
$$
\frac{360}{12}+\frac{x}{60} \cdot \frac{360}{12}=30+\frac{x}{2}
$$
(here we use the fact that the bisec... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## 10. Confession
The confessor has a habit of assigning penances "by the tariff" - in exact proportion to the gravity of the sins committed. Thus, before absolving the sin of pride, he requires the penitent to recite the "Te Deum" once and the "Pater Noster" twice. Slander is "priced" at two "Pater Noster" and seven ... | 10. Since I will have to repeat the "Te Deum" only nine times, it is clear that I am not guilty of adultery.
I confess to one instance of slander, for ten "Credo" would be too few for two instances of slander, and if I had not slandered at all, the "Credo" would correspond to such a number of sins of pride and slander... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 18. Seals
Once I entered a maritime museum and saw some seals there. There were not many of them - just seven eighths of all the seals plus another seven eighths of a seal that I counted in the pool. How many seals were there in the pool in total?
dili in two rounds). According to the French constitution, a second... | 18. Let $n$ be the total number of seals. In this case,
$$
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 |
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