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## 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 | 42,071 |
## 16. Navigation
A ship is sailing along a river. After 6 hours, it returns to the starting point, having traveled a distance of 36 km on the map (naturally, the ship had to move in different directions at different times).
What is the speed of the ship, assuming it did not spend any time turning around, and the spe... | 16. Let $x$ (km/h) be the speed of the ship in still water. Then the speed of its movement downstream is $x+3$, and the speed of movement upstream is $x-3$. Since the total time spent is 6 hours, and the distances traveled downstream and upstream are clearly the same, we get
$$
18 /(x+3)+18 /(x-3)=6
$$
or
$$
x^{2}-6... | 7.24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,072 |
## 17. Paris - Deauville
Monsieur and Madame Dubois are traveling from Paris to Deauville, where their children live. Each is driving their own car. They leave together and arrive in Deauville simultaneously. However, Monsieur Dubois spent one-third of the time his wife continued driving on stops, while Madame Dubois ... | 17. Let $a$ and $a^{\prime}$ be the stopping times of Madame and Monsieur Dubois, respectively, and $r$ and $r^{\prime}$ be the times they spent driving. Since they both left and arrived in Deauville at the same time,
$$
a+r=a^{\prime}+r^{\prime}
$$
On the other hand,
$$
a^{\prime}=r / 3 \quad \text { and } \quad a=... | \frac{9}{8}r^{\} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,073 |
## 19. Picnic
At nine o'clock in the morning, Paul set off on his bicycle from point $A$ to point $B$ at a speed of $15 \mathrm{km} /$ h. At a quarter to ten, Pierre in turn set off from $B$ to $A$ at a speed of 20 km/h. They agreed to have a picnic halfway and strictly adhered to this condition. At what time did they... | 19. Let $x$ be half the distance from $A$ to $B$. Then the time Paul spent traveling from $A$ to the meeting point is $x / 15$. The time it took Pierre to cover the same distance is $x / 20$. But since Pierre left three-quarters of an hour later, then
$$
x / 15 = x / 20 + 0.75
$$
and, therefore, $x = 45$.
Thus, Paul... | 12:00 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,075 |
## 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 | 42,077 |
## 22. On the way from college
Francine leaves college at noon every day, and at the same time her father sets out to meet her in his car. One day, Francine left earlier than usual and started walking home. After a quarter of an hour, she met her father, got into the car, and they drove back; that day they arrived hom... | 22. The father drove the car 10 minutes less than usual: 5 minutes less in the direction from home to college and 5 minutes less in the return direction. Therefore, he met Francine 5 minutes earlier than usual. By that time, she had been walking for a quarter of an hour. Thus, Francine left 20 minutes earlier, i.e., at... | 11:40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,078 |
## 23. Skiing
A skier can't decide between two routes from $A$ to $B$ of equal length. The first route goes entirely on flat ground, while the second, on the contrary, is half uphill and half downhill. The skier knows that his speed on an uphill is three times slower than on flat ground, but on a downhill, it is three... | 23. No, not correct. Let $t$ be the time it takes for the skier to complete the flat route. The part of the second route consisting of ascents is half as long as the entire first route; it will take the skier $3t / 2$ time, which is already more than $t$. Without calculating the time required for the part of the second... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,079 |
## 24. Two Whales
Two whales are swimming in Antarctic waters in a straight line at a speed of 6 km/h. Suddenly, one of them, without changing direction, started swimming faster, at a speed of 10 km/h. Then he suddenly turned around and swam back towards the second whale, which did not change its speed or direction. S... | 24. Let $t_{1}$ be the time that has passed since the whales parted until the moment the faster whale turns around, and $t_{2}$ be the time that has passed from the moment of the turn until the whales meet again (if we take one hour as the unit of time, then $\left.t_{1}+t_{2}=3 / 4\right)$.
The distance between the p... | 9:51 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,080 |
## 25. Subsonic and Supersonic
Two pilots, Dupont and Durand, simultaneously leave the French airport Roissy and head for Kennedy Airport in New York. One of the pilots is flying a subsonic aircraft, while the other is flying a supersonic aircraft. After some time, it turned out that if the aircraft piloted by Dupont ... | 25. Let $x$ be the distance Dupont has flown, and $x^{\prime}$ be the distance he has left to fly; similarly, $y$ be the distance Durand has flown, and $y^{\prime}$ be the distance he has left to fly.
The total distance between the airports can be expressed in four different ways:
$$
x+x^{\prime} ; \quad 2 x+x^{\prim... | Dur | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,081 |
## 26. Moving Walkway
When I walk on a moving walkway ${ }^{1}$ in one direction, I cover it in 6 s. In the opposite direction, I cover it in 6 min, which is the time it usually takes me to walk a distance of $500 \mathrm{~m}$.
What is the length of the moving walkway?[^5] | 26. $v$ (m/min) - the speed of the track; $V$ (m/min) - my own speed, and finally, $x$ (m) - the length of the track. Based on the condition of the problem, we have
$$
x=(V+v) / 10, \quad x=(V-v) \cdot 6
$$
Or
$$
61 x=12 V \quad \text { and } \quad V=500 / 6 .
$$
From this,
$$
x=12 / 6 \cdot 500 / 61 \approx 16.39... | 16.39 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,082 |
## 1. Angle Bisector
In any triangle, the bisector of any angle coincides with the bisector of the angle formed by the altitude and the diameter of the circumscribed circle emanating from the same vertex. Why does this happen? | 1. Let $A B C$ be an arbitrary triangle; $S$ - the circumcircle and $A^{\prime}$ - the point on the circle $S$, diametrically opposite to $A$; $H$ - the foot of the altitude dropped from $A$. In this case, $\angle H A C = 90^{\circ} - \angle B C A$ and $\angle B A A^{\prime} = 90^{\circ} - \angle B A^{\prime} A$. But $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,083 |
## 4. Construction
Given two fixed points $A$ and $B$. How to draw two parallel lines through these points, such that they are at a given distance $l$ from each other (of course, less than the distance from $A$ to $B$)? | 4. Construct a circle with diameter $A B$ and an arc with center $A$ and radius $l$. Let $C$ be the point of intersection of this arc with the circle. Draw the line $B C$, and through $A$ draw a line parallel to $B C$. The length $A C$ (i.e., $l$)
; thereby we determine the midpoint of $A B$. Then construct a semicircle with d... | AH+HB=AB= | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,087 |
## 6. In the Desert
Two people got lost in the desert in a jeep. After many hours, using a rather complex observation system, they discovered that the sum of the squares of the distances from their location to the oases of SidiBen and Ben-Sidi is equal to twice the square of their distance to the oasis of Sidi-Sidi.
... | 6. The driver of the jeep was right. Indeed, let $S$ be the oasis of Sidi-Ben, $B$ - the oasis of Ben-Sidi; $I$ - the oasis of SidiSidi, $M$ - the midpoint of the path connecting Sidi-Ben with Ben-Sidi, $J$ - the location of the jeep, and $H$ - the foot of the perpendicular dropped from $J$ to $I M$. Then
$$
(J B)^{2}... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,088 |
## 7. Cutting
Draw two concentric semicircles with radii of 2 and 4 cm on a sheet of paper. Cut out the area of paper enclosed between these semicircles and their common diameter, and join the straight-line edges of the resulting paper "half-ring" together. What is the volume of the resulting truncated cone? | 7. A truncated cone is bounded by two circles with radii of 1 and 2 cm, respectively (half of 2 cm and 4 cm). The distance between corresponding points of these circles is 2 cm $(4-2=2)$. Using the Pythagorean theorem, we find that the distance between the planes of these circles is $\sqrt{3}$ cm (since $2^{2}-1^{2}=3$... | 12.697\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,089 |
## 8. Dublin, 1856
In 1856, the Irishman Berwick showed that the area of any right-angled triangle is equal to the area of a rectangle whose sides are the segments cut off on the hypotenuse by the point of tangency of the inscribed circle. Why is this so? | 8. Let $ABC$ be a right triangle with a right angle at vertex $A$, and let $D, E$, and $F$ be the points of tangency of the inscribed circle (with center at $O$) with the sides $BC, AC$, and $AB$ respectively; here the radius of the circle is $r = AE = AF$. Let further $m = CD = CE$, and $n = BD = BF$. Clearly, the are... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,090 |
## 9. Water, Oil, and Mercury
In a conical measuring cylinder, mercury (density 13.59), water (density 1), and oil (density 0.915) are poured sequentially. The three liquids fill the glass without mixing and form three layers of equal thickness. Which of the liquids poured into the measuring cylinder has the greatest ... | 9. The volume of a cone is equal to one third of the product of its height and the area of its base. But given the angle at the vertex of the cone, the radius of the circle lying at the base is proportional to the height of the cone. Therefore, the volume of the cone is proportional to the cube of its height.
![](http... | 17.385V_{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,091 |
## 11. "Wipers"
On the windshield of a car, there are two "wipers" of length $L$ each, rotating around two points that are also $L$ apart. Each "wiper" "sweeps" one semicircle. What area do both wipers sweep? | 11. The area swept by the "windshield wipers" is equal to the sum of the areas of two semicircles with radius $L$ minus the area of the "curvilinear triangle" bounded by the straight base $O O^{\prime}$ of length $L$ and two arcs $O A$
L^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,093 |
## 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 | 42,094 |
## 13. The Magi's Pancake
On Christmas Eve, the mother served a round "Magi's pancake" on the table. She began to cut it along the diameter but stopped when she felt the knife hit a bean. Then she made a new straight cut at a $45^{\circ}$ angle to the previous one - but, unfortunately, she hit the bean again. The chil... | 13. No, not by chance. Indeed, let $O$ be the center of the pancake, and $F$ be the bean; $AB$ is the chord representing the second
cut, $I$ is its midpoint. Consider triangles $OAF$ and $... | FA^{2}+FB^{2}=2OA^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,095 |
## 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
The area of the land that went to Pierre is
$$
(1... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,098 |
## 17. Hexagon
Consider a hexagon. How many diagonals, i.e., straight lines connecting two non-adjacent vertices, do you think it has?
If this question seems too easy to you, then extend the six sides of the original hexagon and determine how many points, different from the vertices of the hexagon, all these lines (e... | 17. 18) Through six different points, one can draw as many lines as there are ways to choose two out of six elements, i.e.,
$$
C_{6}^{2}=\frac{6!}{2!4!}=15
$$
Among these 15 lines, 6 are the sides of the hexagon; the remaining 9 lines are diagonals.
2) Generally speaking, the number of intersection points of 15 line... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,099 |
## 18. Intersections
Consider an arbitrary quadrilateral. Show that the two segments connecting the midpoints of opposite sides, as well as the segment connecting the midpoints of the diagonals, all intersect at one point, and that this point bisects each of the considered segments. | 18. Let $ABCD$ be the given quadrilateral; $M, N, P, Q$ be the midpoints of sides $AB, BC, CD, DA$. In triangle $ACB$, the segment $MN$, connecting the midpoints of sides $AB$
and $BC$, is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,100 |
## 19. Italy, 1678
Perhaps you have heard of Giovanni Ceva's theorem, an Italian mathematician of the 17th century. Here is what Ceva proved in 1678: if $P$ is any interior point of triangle $A B C$ and the lines $A P$, $B P$, and $C P$ intersect the sides of the triangle at points $X, Y$, and $Z$ respectively, then
... | 19. Let $S(*, *, *)$ be the area of a triangle with vertices at three points, provisionally denoted by asterisks. Then
$$
\frac{S(B P X)}{S(C P X)}=\frac{1 / 2 \cdot B X \cdot \text { distance from } P \text { to } B C}{1 / 2 \cdot C X \cdot \text { distance from } P \text { to } B C}=\frac{B X}{C X} .
$$
th, the 2nd with the ($p+2$)th, and so on, until you return to the starting point. How many sides will the resulting polygon have?
Remark. It is useful to check the general formula obtained for... | 23. If $p$ is a divisor of $n$, then the number of sides of the polygon is $n / p$. If $p$ is not a divisor of $n$, then the fraction $n / p$ should be multiplied by $x$ - the number of corresponding circumferences: $x=\inf (y$ such that $(y n / p)$ is an integer). In other words,
$$
\text { number of sides }=\frac{n ... | \frac{\operatorname{LCM}(n,p)}{p} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,105 |
## 24. Walking the Dog
Monsieur and Madame Dubois went for a walk with their dog. Each of them wanted to hold the leash themselves. Therefore, in the end, they attached two leashes, each 1 m long, to the poor animal's collar. Suppose Monsieur and Madame Dubois always walk 1 m apart from each other.
What is the area o... | 24. Let $O$ and $O^{\prime}$ be the points where Monsieur and Madame Dubois are located at a certain moment in time. If the dog were held on a leash by only Monsieur Dubois, the area of the region would be $\pi$ (the area of a circle with a radius of 1 m). The same would be true if the dog were held by only Madame Dubo... | 1.228 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,106 |
## 25. Quadrilateral
A quadrilateral inscribed in a circle is not quite ordinary: one of its diagonals coincides with the diameter of the circle.
What can you say about the projections of the four sides of the quadrilateral onto the other diagonal? | 25. Let $ABCD$ be a given quadrilateral, $AC$ the diameter of the circumscribed circle, $H$ and $K$ the corresponding projections of vertices $A$ and $C$ onto $BD$, and finally, $E$ the point of intersection of the circle with the line $CK$.
$\angle ABD = \angle ECD$ as angles with mutually perpendicular sides. Theref... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,107 |
## 26. Fortress Wall
A small town, which has the shape of a semicircle, is surrounded by a fortress wall. There are three gates through which one can enter the town; gates $P_{1}$ and $P_{2}$ are located at the ends of the curved section of the wall, while gate $P_{3}$ is located somewhere else on the same curved sect... | 26. Let $O$ be the center of the semicircle, and let $A$ and $B$ be the points of intersection of the tangents at points $P_{1}$ and $P_{2}$ with the tangent at point $P_{3}$. Note that $\angle P_{1} A P_{3}$ and $\angle P_{2} B P_{3}$ are obviously supplementary:
\cdot(P_{3}B+BP_{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,108 |
## 27. Equal Areas
Consider an isosceles right triangle \(ABC\) with hypotenuse \(BC\). Describe a semicircle around it with endpoints \(B\) and \(C\); then draw an arc inside triangle \(ABC\) that is tangent to side \(AB\) at point \(B\) and to side \(AC\) at point \(C\). The area of the region enclosed between this ... | 27. Let $a$ be the length of the legs $AB$ and $AC$ of triangle $ABC$; then its area is $a^2 / 2$, and the length of the hypotenuse $BC$ is $a \sqrt{2}$. The area of the semicircle with diameter $BC$ is
$$
\frac{1}{2} \pi \cdot \frac{BC^2}{4} = \pi \frac{a^2}{4}
$$
Let $O$ be the center of the circle that touches $AB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,109 |
## 28. Tangency
Two circles touch each other externally. Through the point of tangency, two lines are drawn, intersecting these two circles at four additional points. If we connect these four points, what kind of quadrilateral will be formed? | 28. Let $O$ and $O^{\prime}$ be the centers of the original circles, and $P$ be their point of tangency. The line $O O^{\prime}$ passes through the point $P$. Furthermore, let $A A^{\prime}$ and $B B^{\prime}$ be two lines passing through the point $P$, such that points $A$ and $B$ belong to the circle with center $O$,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,110 |
## 29. Triangle
How to construct a triangle, knowing its perimeter, one angle, and the height dropped from this angle? | 29. Construct an angle with vertex $A$, equal to the given one. Let $D$ and $E$ be points on the sides of the angle, at a distance from $A$ equal to half the given perimeter,
and let $O$ be... | AB+BC+CA=AD+AE=\text{perimeter.} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,111 |
## 30. Right Triangle
In any right triangle, the sum of the legs is equal to the sum of the diameters of the inscribed and circumscribed circles. Why? | 30. Let $A B C$ be a right triangle with a right angle at vertex $A$, and let $D, E$, and $F$ be the points of tangency of the sides $B C, C A$, and $A B$ of the triangle with its inscribed circle. Then, obviously,
$A F=A E=r$ (the radius of the inscribed circle);
$$
C E=C D ; \quad B F=B D
$$
## Therefore,
$$
\beg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,112 |
## 31. If the Earth Were an Orange
Wrap a perfectly round orange with a red string. Then lengthen the string so that it encircles the orange, passing 1 meter away from its surface. Now wrap the Earth (we assume it is spherical) with a blue string and lengthen this string so that it encircles the Earth, passing 1 meter... | 31. Let $r$ be the radius of an orange, and $R$ be the radius of the Earth, both expressed in meters. The red rope was previously equal to $2 \pi r$, and then we extended it to the length of $2 \pi(r+1)$, i.e., approximately 6.283 m (more precisely, $2 \pi$ m). The blue rope
![](https://cdn.mathpix.com/cropped/2024_05_... | 2\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,113 |
## 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 | 42,114 |
## 33. Statues
In a very large park, a straight alley passes sequentially through four small bridges, which are at unequal distances from each other. In the middle of the segments separating adjacent bridges, statues of Vercingetorix, Charlemagne, and Henry IV stand in sequence, and in the middle of the section betwee... | 33. Let $P_{1}, P_{2}, P_{3}$ and $P_{4}$ be four consecutive bridges. Then the distance from $P_{1}$ to the statue of Vercingetorix
7
is $P_{1} P_{2} / 2$, and the distance from $P_{1}$... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,115 |
## 34. Alsatian Village
The main street of the village (it is called Main) is perpendicular to General de Gaulle Street; they intersect at the Town Hall Square. The Catholic cathedral faces the main street, while the Protestant one faces General de Gaulle Street. The school is located on the street connecting these tw... | 34. Let $E$ be the school, $M$ the town hall, $P$ the Protestant church, and $C$ the Catholic church. Since the area of triangle $M C P$ is equal to the sum of the areas of triangles $M C E$ and $M P E$,
. For example, the fourth street corresponds to
| J | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | | | | | | |
| | 0 | | 1 | | | |
| | | 0 | | | | |
| | 1 | | 0 | | | |
| | | |... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,118 |
## 37. Two Circles
Two circles touch each other internally at point $A$. Let $B$ be the point on the larger circle that is diametrically opposite to $A$, and $B D$ be a chord of this circle that touches the smaller circle at point $C$. Then $A C$ must be the bisector of angle $B A D$. Why?
60 | 37. Let $O$ be the center of the smaller circle. The lines $CO$ and $DA$ are parallel, as both are perpendicular to $DB$. Therefore, angles $DAB$ and $COB$ are equal.
But in the smaller cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,119 |
## 38. Two Lunes
Consider a right triangle $A B C$ with a right angle at vertex $A$. Describe a semicircle around it with diameter $B C$; then construct two semicircles with diameters $A B$ and $A C$, external to the triangle $A B C$. The sum of the areas of the two resulting lunes is equal to the area of triangle $A ... | 38. The area of triangle $ABC=$ (area of a semicircle with diameter $BC$) - (area of a segment with chord $AB$) - (area of a segment with chord $AC) = \pi \cdot BC^{2} / 8$ - (area of the two specified segments).
The area of the lune external to $AB=$ (area of a semicircle with diameter $AB$) - (area of a segment with... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,120 |
## 39. Two Bridges
A straight channel runs between two villages located on opposite banks, with one village being closer to the channel than the other. It was decided to build two bridges perpendicular to the banks of the channel. The first bridge was to be built such that the distance from each village to the nearest... | 39. Let $A$ and $B$ be two villages, $B^{\prime}$ be a point such that the segment $B B^{\prime}$ is equal and parallel to each of the future bridges (i.e., this segment is perpendicular to the banks of the canal), and the point $B^{\prime}$ is closer to the canal than the point $B$; finally, $d$ is the bank of the can... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,121 |
## 40. Three Perpendiculars
Try to show that the sum of the lengths of three perpendiculars dropped from an internal point of an equilateral triangle to its sides does not depend on the position of this point.[^10] | 40. Let $A B C$ be an equilateral triangle, $P$ its interior point, and $H, K$ and $L$ the feet of the perpendiculars dropped from $P$ to the sides $A B, B C$ and $C A$. The lengths of the segments $P H, P K$ and $P L$ we denote by $h, k$ and $l$.
; such iron was once used to make knight's armor. The museum keeper decided that this exhibit would look more interesting (perhaps even reminiscent of a shield of an ancient knight) if the four corners of this sheet were... | 41. Each niche in the cross-section represents an isosceles right triangle with a leg length of $x$ (meters). The hypotenuse, therefore, is $\sqrt{2} x$. This hypotenuse is equal to one of the sides of the regular octagon.
![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-176.jpg?height=748&width=73... | x\approx1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,123 |
## 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 | 42,124 |
## 3. Year of Birth
Subtract the sum of the four digits that make up your birth year from the year itself. You will get a number that is divisible by 9. Why? | Let $m, c, d$ and $u$ be four digits forming the year of your birth. Then the year of your birth can be written as
$$
1000 m+100 c+10 d+u
$$
If you subtract $m+c+d+u$ from this, you get
$$
999 m+99 c+9 d=9(111 m+11 c+d)
$$
which is clearly divisible by 9. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,126 |
## 4. "English Arithmetic"
$20+20+20+10+10=80$. In English, this can be written as:
| TWENTY |
| ---: |
| + TWENTY |
| + TWENTY |
| $+\quad T E N$ |
| $+\quad T E N$ |
What digits should the eight letters $E, G, H, I, N, T, W, Y$ represent to make the last record truly correct? | 4. Renumber the columns
| 654321 |
| ---: |
| $+T W E N T Y$ |
| $+T W E N T Y$ |
| $+T W E N T Y$ |
| $+\quad T E N$ |
Let's denote the sum of the digits in the $i$-th column, including the carry from the $(i-1)$-th column, by $s_{i}$.
We will denote by $r(x)$ the number of tens in the integer $x$, so $r\left(s_{i}... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,127 |
## 5. Historical Arithmetic
One historian once noted that the dates of the establishment of the 5th Republic ${ }^{1}$ and the date of Madagascar's inclusion in the French colonial empire are composed of the same digits, written in a different order, and that, in addition, the remainders from dividing each of these tw... | 5. Let $N$ be an arbitrary date, $m, n, d$ and $u$ be the digits representing the thousands, hundreds, tens, and units in the number $N$. Then
$$
N=1000 m+100 c+10 d+u \text {, }
$$
i.e., $N=(900+90+9+1) m+(90+9+1) c+(9+1) d+u=$ $=(999 m+99 c+9 d)+(m+c+d+u)$.
The first term is clearly divisible by 9 for any permutat... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,128 |
## 7. At the Lyceum
The director of one of the lyceums, to his surprise, discovered one day that the total number of his students is equal to the product of the difference of the squares of the number of physical education teachers and the number of Russian language teachers, and these two latter numbers themselves.
... | 7. Let $a$ be the number of physical education teachers, $b$ the number of Russian language teachers, and $N$ the total number of students in the lyceum. Then
$$
N=a b\left(a^{2}-b^{2}\right)=a b(a+b)(a-b) .
$$
If $a$ or $b$ is divisible by 3, the director can obviously arrange all students in a column with 3 student... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,130 |
## 8. At the Restaurant
Monsieur Dupont remembered that today was their wedding anniversary and invited his wife to have lunch at a good restaurant. Leaving the restaurant, he found that he had only one-fifth of the money he had brought with him left, and the number of centimes left was the same as the number of franc... | 8. Let $F$ be the initial number of francs, $C$ be the initial number of centimes, $F^{\prime}$ be the remaining number of francs, and $C^{\prime}$ be the remaining number of centimes.
Since after dining at the restaurant, Monsieur Dupont had only one-fifth of his money left, we have
$$
100 F + C = 5 \left(100 F^{\pr... | 79 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,131 |
## 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 | 42,133 |
## 11. Ones and Twos
Write down an even number of ones in a row. Subtract from the resulting number a similar number composed of a series of twos, which is half as long as the first series of ones. In doing so, you will get a perfect square. (For example, $1111-22=1089=33^{2}$.) Could you explain why? | $$
\begin{aligned}
\underbrace{11 \ldots 1}_{2 n \text { digits }}-\underbrace{22 \ldots 2}_{n \text { digits }} & =\underbrace{11 \ldots 1}_{n} \underbrace{11 \ldots 1}_{n}-2(\underbrace{11 \ldots 1}_{n})= \\
& =\underbrace{11 \ldots 1}_{n} \underbrace{00 \ldots 0}_{n}-\underbrace{11 \ldots 1}_{n}= \\
& =\underbrace{1... | (\underbrace{33\ldots3}_{n})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,134 |
## 12. Glorious Great-grandfather
“I am 91 years old, I have four children, eleven grandchildren, and many great-grandchildren. If you ask me how many exactly, I will only answer you that the product of their number multiplied by[^13]
66
the number of my grandchildren and by my age equals a number that is written as ... | 12. Let $n$ be the number of great-grandchildren we are looking for. Based on the words of the great-grandfather, we can write
$$
11 \cdot n \cdot 91 = 1001 n = \ll n 0 n \gg.
$$
From this, it follows that $n$ is a two-digit number, i.e., it is between 10 and 99. The information provided by the great-grandfather, to ... | 1001n=n0n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,135 |
## 13. Caramels
Martina distributed 26 caramels to her four little brothers. Each of the brothers ate a few caramels, and an hour later, Martina found that each of them had the same number of caramels left. Suppose the oldest brother ate as many as the third, the second ate half of his caramels, and the fourth ate as ... | 13. Let $x$ be the number of candies left with each of the brothers, and $y$ be the number of candies eaten by the elder (or third) brother. Then the second brother ate $x$ candies, and the fourth brother ate $-(y+x+y)$ candies.
In total, the number of candies eaten is
$$
y+x+y+(x+2 y)=2 x+4 y
$$
candies. On the oth... | 3,2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,136 |
## 14. Square
Can the square of an integer end with three identical non-zero digits? | 14. First, note that the square of any integer must end in one of the following digits: $0,1,4,5,6$ or 9. Moreover, the square of any integer is either divisible by 4 (if the number is even) or equal to a number divisible by 4 increased by 1 (if the original number is odd). However, numbers ending in 11, 55, 66, or 99 ... | 1444=(38)^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,137 |
## 15. Puzzle
Martin once told Martina: “I am three times as old as you were when I was as old as you are now.”
And in response, Martina noted: “When I am as old as you are now, together we will be 77 years old.”
How old are Martin and Martina? | 15. Let $x$ be Martin's age and $y$ be Martina's age. What Martin said to Martina can be expressed by the equation
$$
x=3[y-(x-y)]
$$
from which
$$
4 x=6 y \text{, or } 2 x=3 y \text{. }
$$
Martina's response can be written as the equation
$$
x+[x+(x-y)]=77
$$
i.e.
$$
3 x=y+77
$$
Substitute $y$ with $3 x-77$ in... | Martinis33old,Martinais22old | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,138 |
## 16. How many of you were there, children?
If you had asked me such a question, I would have answered you only that my mother dreamed of having no fewer than 19 children, but she did not manage to fulfill her dream; however, I had three times as many sisters as cousins, and brothers - half as many as sisters. How ma... | 16. From the conditions of the problem, it follows that the number of my sisters is divisible by both 3 and 2. Therefore, it is divisible by 6. Hence, the total number of children is
$$
[(\text { number divisible by } 6) \cdot(1+1 / 2)]+1
$$
(the last 1 corresponds to myself). This number must be strictly less than 1... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,139 |
## 17. Airline
One American airline serves several major cities, establishing direct flights between each pair of these cities. Next year, it plans to increase the number of flights by 76, which will allow it to serve some additional cities under the same conditions.
How many cities does this airline currently serve,... | 17. Let $n$ be the number of cities currently served. The corresponding number of flights is $n(n-1)$, since each city is connected to $n-1$ cities served by the airline. For the next year, the airline plans
$$
(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 | 42,140 |
## 18. Essay in Latin
An essay in Latin is scored on a scale from 0 to $20^{1}$. Michel's score is above the average, while Claude's score is below the average. What score did each of them receive, if it is known that when one third of the smaller of these two scores is subtracted from each of them, one of the resulti... | 18. Let $g$ be the larger of the two estimates, and $p$ be the smaller of them. Then
$$
g-p / 3=3(p-p / 3)
$$
from which it follows that
$$
g=7 p / 3
$$
If $p=3$ and $g=7$, Michelle would have received a score below average, which contradicts the problem's condition. Therefore, the only possible solution is:
$$
g=... | =14p=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,141 |
## 19. Count on Your Fingers
If you have trouble remembering the multiplication table and struggle with multiplying by 9, you can use the following system. To find the product of $9 \cdot n$ (where $n$ is any single-digit number), place both hands on the table. Then raise the $n$-th finger, counting from the left. The... | 19. The number of fingers lying to the left of the raised finger is $n-1$, and the number of fingers to the right of it is $10-n$. However,
$$
9 n=10(n-1)+(10-n)
$$
Thus, indeed, $n-1$ equals the number of tens, and ($10-n$) - the number of units of the result. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,142 |
## 21. At the Races
One fine spring evening I went to the Longchamp racecourse. I bet on the first horse and doubled the amount of money I had with me. Inspired by this example, I bet 60 francs on the second horse and lost it all. However, thanks to the third horse, I was able to double my remaining cash again. Howeve... | 21. After the race in which the sixth horse participated, I had no money left at all. After the race with the participation of the fifth horse, I had 60 francs more, i.e., I had a total of 60 francs. After the race with the participation of the fourth horse, I had half as much money, i.e., 30 francs. After the race wit... | 52.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,144 |
## 22. Age Difference
The sums of the digits that make up the birth years of Jean and Jacques are equal to each other, and the age of each of them starts with the same digit. Could you determine the difference in their ages? | 22. Let $(m, c, d, u)$ be the number of thousands, hundreds, tens, and units in Jean's birth year, and let $\left(m^{\prime}, c^{\prime}, d^{\prime}, u^{\prime}\right)$ be the corresponding digits of Jacques' birth year. Then Jean's age is
$$
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 | 42,145 |
## 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 | 42,146 |
## 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
$$
\begin{gathered}
9 d-9 u+45=... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,147 |
## 25. School
In all the classes of one school, the same number of students studied. After a fire, six classes of the school became unsuitable for lessons, and therefore, five more students had to be added to each class where lessons were held. But then, due to water damage from the fire hoses, another ten classes wer... | 25. Let $c$ be the number of students in each class before the fire, and $n$ be the number of classes at that time. Then the following relationships hold:
$$
\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 | 42,148 |
## 26. Brother and Sister
- Sister, you have as many brothers as sisters.
- Brother, you have twice as many sisters as brothers.
Could you determine the number of children in this family from this conversation? | 26. Let $x$ be the number of boys, and $y$ be the number of girls. The brother's words mean that
$$
x=y-1 \text {. }
$$
The sister's response can be written as the following equation:
$$
y=2(x-1), \text { or } y=2 x-2
$$
Substituting $x$ with $y-1$, we find that
$$
y=4
$$
and, consequently,
$$
x=3 \text {. } \qu... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,149 |
## 27. Large Families
The Martens have more children than the Duponts. Suppose the difference of the squares of these two numbers is 24 and that both families have more than one child. How many children do the Martens have?
70 | 27. Let $m$ and $d$ be the number of children in the Marten and Dupont families, respectively. According to the problem,
$$
m^{2}-d^{2}=24 \text {, i.e., }(m+d)(m-d)=24\left(=2^{3} \cdot 3\right) \text {. }
$$
Therefore, the following values for $m+d, m-d$, and thus $2 m[=(m+d)+(m-d)]$, $m$, and $d$ are possible:
| ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,150 |
## 28. At the Ball
When I saw Eleanor, I found her very pretty. After a brief banal conversation, I told her how old I was and asked about her age. She replied:
- When you were as old as I am now, you were three times as old as I was. When I am three times as old as I am now, together we will be exactly a century old... | 28. So, let's turn to the conversation of two individuals. Let $e-$ be Eleonora's age, and $m$ - my age. When I was as old as she is now, Eleonora's age was
$$
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{. }
$$
When Eleonora become... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,151 |
## 29. Leonie and Cats
When old lady Leonie is asked how many cats she has, she melancholically replies: “Four fifths of my cats plus four fifths of a cat.” How many cats does she have?
 | 29. Let $n$ be the number of cats Leonie has. From her last words, we can write the relationship
$$
n=(4 / 5) n+4 / 5, \text { i.e. } \quad n=4 .
$$
Thus, Leonie has 4 cats. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,152 |
## 30. New Mathematics
My son has learned to count in a numeral system different from the decimal system, and instead of 136, he writes 253 in this system. In what numeral system is my son counting? | 30. Let $a$ be the base of an unknown numeral system. When my son writes "253" in this system, it represents $2 a^{2}+5 a+3$; according to the problem, this corresponds to the number 136 in the decimal system. Therefore, we can write
$$
2 a^{2}+5 a+3=136
$$
i.e.
$$
2 a^{2}+5 a-133=0
$$
or
$$
(a-7)(2 a+19)=0
$$
Si... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,153 |
## 32. Chinese Numbers
Think of any integer from 1 to 26. Then consider the following six square tables one by one. If the thought-of number is in this table, write down the number in the top left corner of the table. Then add up all the numbers you have written down.
| 1 | 4 | 7 | 2 | 5 | 8 | 3 | 4 | 5 |
| ---: | --... | 32. Any number $n$, less than or equal to 26, is of course strictly less than $3^{3}=27$. Therefore, in the ternary numeral system, it can be represented as follows:
$$
\begin{gathered}
n=(0, \text { or } 1, \text { or } 2) \cdot 3^{0}+(0, \text { or } 1, \text { or } 2) \cdot 3^{1}+ \\
+(0, \text { or } 1, \text { or... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,154 |
## 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 | 42,155 |
## 34. For Those Under Sixteen
Tell me in which of these columns your age appears, and I will name it by adding the digits in the top row of these columns.
| 2 | 8 | 4 | 1 |
| ---: | ---: | ---: | ---: |
| 3 | 9 | 5 | 3 |
| 6 | 10 | 6 | 5 |
| 7 | 11 | 7 | 7 |
| 10 | 12 | 12 | 9 |
| 11 | 13 | 13 | 11 |
| 14 | 14 | 14 ... | 34. Of course, black magic has nothing to do with it, but writing numbers in the binary system of numeration is directly related to this problem. Any integer less than 16 can be written as:
$$
x_{0} \cdot 2^{0}+x_{1} \cdot 2^{1}+x_{2} \cdot 2^{2}+x_{3} \cdot 2^{3}
$$
where each "digit" $x_{i}$ (here $i=0,1,2$ or 3) h... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,156 |
## 35. Product
The product of four consecutive numbers is 3024. What are these numbers? | 35. The number 3024 does not end in 5 or 0. Therefore, none of these four numbers are divisible by 5 or 10. But if the four numbers in question were greater than 10, their product would definitely exceed 10000. In our case, this is not so, which means the four numbers must either be $1,2,3,4$ or $6,7,8,9$. However, in ... | 6,7,8,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,157 |
## 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 | 42,158 |
## 37. How old is Florence?
Try to determine this yourself, and we will only indicate that the difference between the fifth power of the desired number of years and the number itself is divisible by 10. | 37. Let $a$ be the age of Florence, which we want to determine. Then
$$
a^{5}-a=10 k
$$
where $k$ is some integer. Factoring the left side of the equation, we get
$$
a\left(a^{2}-1\right)\left(a^{2}+1\right)=10 k
$$
i.e.
$$
a(a+1)(a-1)[(a-2)(a+2)+5]=10 k .
$$
No matter what $a$ is, one of the two values, $a$ or $... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,159 |
## 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
$$
2 \l... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,160 |
## 39. How old am I?
Try to answer this question if it is known that Alfred's age is obtained by rearranging the digits of my age, that the quotient of the number equal to my age divided by the sum of these two digits differs from the analogous quotient for Alfred by a number equal to the difference between these two ... | 39. Let $x$ be my age, $d$ be the number of tens in it, and $u$ be the number of units. Then $x=10 d+u$, and Alfred's age is $10 u+d$. The first condition of the problem can be written as:
$$
\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 | 42,161 |
## 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 | 42,162 |
## 41. Lift
Three sisters are standing in line at a ski lift on a mountain slope together with their instructor. Out of boredom, he asked the oldest sister how old they are.
- I'll only say, - she replied, - that if you subtract nine times the age of my youngest sister from the product of my age and the age of my mid... | 41. So, the elder sister is over 10 years old, the middle one is exactly 10 years old, and the younger one is less than 10 years old. Let $a$ be the age of the elder sister, and $b$ be the age of the younger sister.
According to the problem,
$$
10 a-9 b=89
$$
or
$$
10(a-b)+b=10 \cdot 8+9
$$
Therefore,
$$
a-b=8 \t... | 17,10,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,163 |
## 42. Meeting on the Tower
Two delegations were to meet on the top floor of a tower that had several elevators, each capable of holding nine people. The first delegation ascended to the meeting place using a certain number of elevators that were fully occupied, plus one last elevator where there were still five empty... | 42. Let $d_{1}$ be the number of members of the first delegation, and $d_{2}$ be the number of members of the second delegation. We know that
$$
\left.d_{1}=(\text { a number divisible by } 9)+4 ; d_{2}=\text { (a number divisible by } 9\right)+6 .
$$
The number of frames shot is
$$
\begin{aligned}
d_{1} \cdot d_{2}... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,164 |
## 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
$$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,165 |
## 44. Bag of Balls
Children are dividing a bag of balls among themselves. The first child took one ball and a tenth of the remaining balls, then the second took 2 balls and a tenth of the remaining, then the third took 3 balls and a tenth of the remaining, and so on, until the last child took all that was left.
How ... | 44. Let $n$ be the number of children, $x$ be the share of balloons taken by each of them, and $N$ be the total number of balloons. Obviously, $N=n x$. The share of balloons taken by the first child is
$$
1+(N-1) / 10=x
$$
The share of the last child is $x$. The share of the second-to-last child can be written as:
$... | 9,N=81,n=9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,166 |
## 45. St. Margaret
At St. Margaret's educational institution, boys are only admitted in the final years. Their number is exactly equal to the sum of the digits of the total number of students.
In the chapel of St. Margaret, numerous ceremonies are held. Can all the girls be seated on the benches, with nine per bench... | 45. Let us first assume that the total number of students at St. Margaret's institution is less than 10,000, which is not too restrictive a condition ${ }^{1}$. In this case, this number is written using four digits: $m$, $c, d, u$, where $m$ is the number of thousands, $c$ is the number of hundreds, $d$ is the number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,167 |
## 46. The Emperor's Soldiers
General Lassalle knew no fear of death. "If a hussar is over 30, it means he often walked with wet pants," the general used to say. Before one brutal battle, he counted his soldiers: the sum of the digits in the number of soldiers was 17. After the battle, he counted the dead and wounded:... | 46. Suppose first that the number $N$ of General Lassalle's soldiers is less than 10000 (not too restrictive ${ }^{2}$[^29] condition). Let «mсdu» be the number
$$
N=1000 m+100 c+10 d+u
$$
Then by the condition
$$
m+c+d+u=17
$$
The number of killed and wounded
$$
N^{\prime}=\ll m^{\prime} c^{\prime} d^{\prime} u^{... | V | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,168 |
## 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 | 42,169 |
## 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 | 42,170 |
## 49. Trick
Maurice always disappointed his teachers, especially his math teachers. However, when he needed to calculate the square of 35, or 75, or finally 85, he always gave the correct answer in one second. The thing is, he had one trick up his sleeve that he was very proud of: to determine the square of any two-d... | 49. Let $d$ be the number of tens in the number that needs to be squared. This means the last number has the form
$$
10 d+5 \text{. }
$$
Squaring it, we get
$$
100 d^{2}+100 d+25
$$
or
$$
100 d(d+1)+25
$$
For example,
$$
(85)^{2}="(8 \cdot 9) 25 "=7225
$$
Thus, Maurice's trick always gives the correct answer. | 100(+1)+25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,171 |
## 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 | 42,173 |
## 52. Five Numbers
Try to find five consecutive integers such that the sum of the squares of the two largest of them equals the sum of the squares of the other three? | 52. Let $n$ be the average of our five numbers; then the other numbers are $(n+1)$ and $(n+2); (n-1)$ and $(n-2)$. According to the problem, the following relationship holds:
$$
(n+1)^{2}+(n+2)^{2}=n^{2}+(n-1)^{2}+(n-2)^{2}
$$
Simplifying this equation, we get
$$
n(n-12)=0
$$
The only acceptable solution is
$$
n=1... | 10,11,12,13,14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,174 |
## 53. Two Sisters
The age difference between two sisters is 4 years. If the cube of the age of the first one minus the cube of the age of the second one equals 988, how old is each sister? | 53. Let $x$ be the age of the older sister, and $y$ be the age of the younger sister. Then
$$
x=y+4
$$
and
$$
x^{3}-y^{3}=988
$$
or
$$
(x-y)\left(x^{2}+x y+y^{2}\right)=988
$$
188
From this, we easily obtain
$$
12 y^{2}+48 y+64=988
$$
Therefore,
$$
y=7, \quad \text { and } x=11
$$ | 7,\quad11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,175 |
## 54. Bastille Day ${ }^{1}$
A certain general once noticed that the number of soldiers in one of the barracks could be obtained by adding the number of regiments stationed there to three times the square of this number and twice its cube. On this basis, he decided that during the upcoming Bastille Day celebration, a... | 54. Let $s$ be the number of soldiers in a given barracks, and $b$ be the number of regiments stationed there. Then
$$
s=b+3 b^{2}+2 b^{3}=b(b+1)(2 b+1) .
$$
It is clear that $b(b+1)$ is divisible by 2 for any $b$. On the other hand, if $b$ is divisible by 3, then $s$ is divisible by 3; if $b$ gives a remainder of 1 ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,176 |
## 55. 100 years among the three
Patrick was walking with his father and grandfather. He wanted to imagine the time when all of them together would be 100 years old.
His father said on this matter: “I will be 28 years older than you, and you will be six-fifths of your current age.” And his grandfather added: “I will ... | 55. Let $e$ be Patrick's age, $p$ be the father's age, and $g$ be the grandfather's age; let $x$ be the time interval after which all three will be 100 years old. Then the following equations hold:
1) $(g+x)+(p+x)+(e+x)=100$
2) $e+x=(6 / 5) e$,
3) $(p+x)-(e+x)=28$,
4) $g+x=2(p-e+1.5)$.
From the last two equations, we ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,177 |
## 1. British Hospital
In a British hospital, the following announcement can be read: "Surgeon $X$ and Nurse $Y$ have the pleasure of announcing their forthcoming marriage." Suppose that at the time the hospital was opened, the groom was as old as the bride is now, and that the product of the ages of the groom and the... | 1. Let $x$ be the age (number of full years) of the surgeon (groom or bride), $y$ be the number of years of the paramedic (groom or bride), and $h$ be the number of years the hospital has been in existence.
The conditions of the problem can be written as the following two equations:
$$
\begin{gathered}
|x-y|=h \\
x(y... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,180 |
## 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 | 42,181 |
## 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 | 42,182 |
## 5. In the Zoo
Let's assume that in all zoos where there are hippos and rhinos, there are no giraffes; in all zoos where there are rhinos and no giraffes, there are hippos; finally, in all zoos where there are hippos and giraffes, there are also rhinos. How do you think, can there be a zoo where there are hippos, bu... | 5. Yes, such a zoo can exist. Indeed, let $Z$ be the set of zoos, $H$ be the subset of those that have hippos, $R$ be the subset
of those that have rhinos, and $G$ be the subset of those th... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,184 |
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