problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
233. Multiplication of dates. In 1928, there were four dates with a remarkable property: when written in the usual way, the product of the day and the month gives the year of the XX century. Here are these dates: $28 / 1-28, 14 / 2-28, 7 / 4-28$ and $4 / 7-28$.
How many times in the XX century (from 1901 to 2000 inclu... | 233. In the XX century, there are 215 dates with the specified property, if cases like \(\frac{25}{4}-00\) are included. The most "productive" in this regard turned out to be 1924, in which there were 7 such dates: \(24 / 1-24\), \(12 / 2-24, 2 / 12-24, 8 / 3-24, 3 / 8-24, 6 / 4-24, 4 / 6-24\). To solve the problem, on... | 215 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
235. Another curious example of multiplication. Here is another puzzle from Professor Rackbrain.
What number, when multiplied by $18, 27, 36, 45, 54, 63, 72$, 81 or 99, gives a product whose first and last digits match the corresponding digits of the multiplier, and when multiplied by 90, gives a product whose last tw... | 235. The desired number is 987654321, which when multiplied by 18 gives 17777777778 with 1 and 8 at the beginning and end, respectively. The same is true for other multipliers, except for 90, when we get 88888888890 with 90 at the end.
[The author did not notice numbers like 1001, 10101, and 100101, composed of 0 and ... | 987654321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
237. Counting Losses. An English officer recounted that he was part of a detachment that initially numbered 1000 people. During one of the operations, the detachment suffered heavy losses, and those who survived were captured and sent to a prisoner-of-war camp.
On the first day of the journey, \(\frac{1}{6}\) of all t... | 237. During the fighting, 472 people were killed. By making calculations, the reader will find that each of the four camp groups had 72 people.
The general solution can be obtained from the indeterminate equation
\[
\frac{35 x-48}{768}=\text { an Integer, }
\]
where \(x\) is the number of survivors. Solving it in th... | 472 | Algebra | 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 |
241. Paving. Two square sections of pavement need to be laid with square tiles measuring $1 \mathrm{~m}^{2}$. A total of 2120 tiles are required for this, and the side of one section is 12 m longer than the side of the other.
What are the dimensions of each section? | 241. One side of one plot measures 38 m (1444 tiles), one side of the other - 26 m (676 tiles). | 38 | Algebra | 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 |
244. Annoying Breakdowns. On the occasion of the holiday, many residents of the town gathered to spend the day in the great outdoors. For this purpose, they hired all the available vans, with the same number of people in each van. Halfway through, 10 vans broke down, so each remaining van had to take one extra person. ... | 244. There were 900 people in total. Initially, 100 vans, each carrying 9 people, set off. After 10 vans broke down, the remaining vans had 10 people each (one extra person). When 15 more vans broke down during the return trip, each of the 75 remaining vans had 12 people (three more people than in the morning when they... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
247. Сколько весит рыба? Крэкхэмы задумали остановиться во время своего путешествия в каком-нибудь месте, где есть хорошая рыбная ловля, поскольку дядя Джейбз был заядлым рыболовом и они хотели доставить ему удовольствие. Они выбрали очаровательное местечко и, воспользовавшись случаем, устроили там пикник. Когда дядя п... | 247. Рыба весит 72 унции, или \(4 \frac{1}{2}\) фунта. Хвост весит 9 унций, туловище 36 и голова 27 унций. | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
248. Cats and Mice. One morning at Professor Rackbrain's table, the lively discussion was about the extermination of rodents when suddenly the professor said:
- If a certain number of cats ate a total of 999,919 mice, and all the cats ate the same number of mice, how many cats were there in total?
Someone suggested t... | 248. It is clear that 999919 cannot be a prime number and that, since we need to find a unique solution, it must factorize into a product of two prime factors. These factors are 991 and 1009. We know that each cat caught more mice than there were cats. Therefore, there were 991 cats, and each of them caught 1009 mice. | 991 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
249. Egg Cabinet. A person has a cabinet where he stores a collection of bird eggs. This cabinet has 12 drawers, and all of them (except the top one, where the catalog is stored) are divided into cells by wooden partitions, each of which extends the entire length or width of the corresponding drawer. In each subsequent... | 249. Let the box number be \(n\). Then it will have \(2 n-1\) partitions in one direction and \(2 n-3\) in the other, which will give \(4 n^{2}-4 n\) cells and \(4 n-4\) partitions. Thus, in the twelfth box, there are 23 and 21 partitions (a total of 44) and 528 cells. This rule applies to all boxes except the second, ... | 262 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
256. Lines and Squares. Here is one simple question. What is the smallest number of straight lines needed to construct exactly 100 squares? In the figure on the left, nine lines are used to construct 20 squares (12 with side length $A B$, 6 with side length $A C$, and 2 with side length $A D$). In the figure on the rig... | 256. If 15 lines are drawn as shown in the figure, exactly 100 squares will be formed. Forty of them have a side equal to \(A B\), twenty have a side equal to \(-A C\), eighteen have a side equal to \(-A D\), ten have a side equal to \(-A E\), and four have a side equal to \(-A F\). With 15 lines, it is even possible t... | 15 | Combinatorics | 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 |
271. Three Tablecloths. One morning at breakfast, Mrs. Crackham announced to all present that a friend had given her three wonderful tablecloths, all square with a side of 144 cm. Mrs. Crackham asked those present to name the maximum dimensions of a square table that could be covered with all three tablecloths at the s... | 271. Three tablecloths measuring \(144 \times 144\) cm will cover a table measuring \(183 \times 183\) cm if they are placed as shown in the figure. Square \(A B C D\) is the tabletop, and squares 1, 2, and 3 are the tablecloths. Parts of the second and third tablecloths, of course, will hang off the table.
, then by adding two poles at each end, as in case \(B\), you will get double the area. It should be noted that, firstly, the problem does not specify the shape of the enclosure. Secondly, even if it were required that the original enclosure had... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
279. Motorcyclists again. Here is another case involving the motorcyclists mentioned in the previous puzzle. On the map segment (see figure), three roads form a right-angled triangle. When asked about the distance between $A$ and $B$, one of the motorcyclists replied that after he traveled from $A$ to $B$, then to $C$,... | 279. The distances are shown in the figure. The person asking the question only needed to square the 60 km traveled by the first motorcyclist (3600), and divide the result by twice the sum of these 60 and 12 km, which is the distance from road \(A B\) to \(C\), i.e., by 144. Doing the calculations in his head, he, of c... | 25 | 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 |
322. Cigarette Packaging. Cigarettes are shipped from the factory in boxes of 160. They are arranged in 8 rows of 20 each, completely filling the box.
Can more than 160 cigarettes be placed in the same box with a different packaging method? If so, how many more cigarettes can be added?
At first glance, it seems absur... | 322. Let the diameter of a cigarette be 2 units, and let 8 rows of 20 cigarettes each (see case \(A\)) completely fill a box. In this case, the internal length of the box is 40, and the depth is 16 units. Now, if we place 20 cigarettes in the bottom row and if instead of 20 in the next row we place 19, as shown in case... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
339. Three Greek Crosses from One. In the figure shown here, you see an elegant solution to the problem of cutting two smaller Greek crosses of the same shape from a larger symmetrical Greek cross. Part $A$ is cut out as a whole, and it is not difficult to assemble a similar cross from the remaining 4 parts.
However, ... | 339. Cut off the top and bottom parts of the cross and place them in positions \(A\) and \(B\) (case \(I\)), and cut the remaining larger part into 3 pieces so that the 5 pieces obtained can be used to form the rectangle shown in case II. It can be said that this rectangle is composed of 15 squares - 5 squares for each... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
348. The Cabinetmaker's Problem. A cabinetmaker had a $7 \times 7$ chessboard piece made of excellent plywood, which he wanted to cut into 6 pieces so that they could be used to form 3 new squares (all of different sizes).
How should he proceed, without losing any material and making cuts strictly along the lines? | 348. The figure shows how a piece of plywood can be cut. Squares \(A\) and \(B\) are cut out entirely (1), and from the four pieces \(C, D\), \(E\) and \(F\), a third square can be formed (2).
, with the numbers going from top to bottom. The puzzle is to move the entire stack to square $F$ in the fewest possible moves. You can move one chip at a time to any square, bu... | 361. In 9 moves, form a stack of five chips (from 1 to 5) in square \(B\). In 7 moves, build a stack of four chips (from 6 to 9) in square \(C\). Form a stack of three chips (from 10 to 12) in \(D\) in 5 moves. Place a stack of two chips (13 and 14) in \(E\) in 3 moves. Move one chip (15) to \(F\) in 1 move. Move 13 an... | 49 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
364. Even and Odd Chips. Place a stack of eight chips in the central circle, as shown in the figure,
such that the numbers from top to bottom are in order from 1 to 8. The goal is to move t... | 364. The least number of moves is 24. You should act as follows. (It is only necessary to indicate with letters from which circle to which the chip is moved. Only one chip can be moved at a time.) So, \(E\) to \(A\), \(E\) to \(B\), \(E\) to \(C\), \(E\) to \(D\), \(B\) to \(D\), \(E\) to \(B\), \(C\) to \(B\), \(A\) t... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
365. Railway Switch. How can two trains pass each other using the switch depicted here and continue moving forward with their locomotives? The small side spur is only sufficient to accommodate either a locomotive or one car at a time. No tricks with ropes or flying are allowed. Each change of direction made by one loco... | 365. Draw the path diagram as shown in the figure, take 5 chips marked \(X, L, R, A\) and \(B\). The locomotives are \(L\) and \(R\), the two cars on the right are \(-A\) and \(B\). The three cars on the left should not be separated, so we will denote them as \(X\). The dead-end is marked as \(S\). Next,
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 |
378. Is it that simple? Before you is a simple magic square, in which the sums of the numbers in any row, any column, and on the main diagonals are equal to 72. The puzzle con-
| 27 | 20 | 25 |
| :--- | :--- | :--- |
| 22 | 24 | 26 |
| 23 | 28 | 21 |
sists in transforming it into a multiplicative magic square, in whi... | 378. To solve the puzzle, you just need to move the right digit up in each cell to get powers of 2. Revealing these powers, you will find that the resulting square meets the required condition with a product of 4096. Of course, anyone familiar with arithmetic knows that 20 equals 1. | 4096 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
393. By the Stream. There is a general opinion that puzzles requiring the measurement of a certain amount of liquid can only be solved by a series of trials, but in such cases, general formulas for solutions can be found. Taking advantage of an unexpected leisure, I examined this question more closely. As a result, som... | 393.
\(A\)
\(B\)
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 15 l & 16 l & \(15 \pi\) & 16 l & 15 l & 16 l & 15 l & 16 l \\
\hline 0 & \(16^{*}\) & 15 & \(5^{*}\) & \(15^{*}\) & 0 & 0 & 11 \\
\hline 15 & \(1^{*}\) & 0 & 5 & 0 & 15 & 15 & 11 \\
\hline 0 & 1 & 5 & 0 & 15 & 15 & \(10^{*}\) & 16 \\
\hline 1 & 0 & 5 & 16 &... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
397. Measuring Water. A maid was sent to a spring with two vessels of 7 and 11 pints capacity. She needs to bring back exactly 2 pints of water.
What is the minimum number of operations in this case? By "operation" we mean either filling a vessel, emptying it, or pouring water from one vessel to another. | 397. Two pints of water can be measured in 14 operations, if the vessels above the line are empty, and each line corresponds to one operation.
\begin{tabular}{cr}
7 l & 11 l \\
\hline 7 & 0 \\
0 & 7 \\
7 & 7 \\
3 & 11 \\
3 & 0 \\
0 & 3 \\
7 & 3 \\
0 & 10 \\
7 & 10 \\
6 & 11 \\
6 & 0 \\
0 & 6 \\
7 & 6 \\
2 & 11
\end{ta... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
401. The Road to Tipperary. A popular bard assures us that "the road is long to Tipperary." Look at the attached map and tell us if you can find the best route there. The straight segments represent transitions from city to city. You need to get from London to Tipperary in an even number of transitions. It is not diffi... | 401. The thick line on the diagram shows the route from London to Tipperary, made in 18 moves. To reach the destination in an even number of moves, it is absolutely necessary to include in the route the move marked by the words Irish Sea.
 to \(E\), without passing through the same segment more than once on any route.
 to \(D\), specifically:
\begin{tabular}{cccr}
\begin{tabular}{c}
Number of \\
segments
\end{tabular} & \begin{tabular}{c}
Number of \\
variations
\end{tabular} & \begin{tabular}{c}
Number of \\
routes
\end{tabular} & \\
1 & 1 & 2 & 2 \\
2 & 1 & 9 & 9 \\
3 & 2 & 12 & 24 \\
4 & 5... | 2501 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
433. Gifted Paintings. A wealthy collector had 10 valuable paintings. He wanted to make a gift to a museum, but the collector couldn't figure out how many gift options he had: after all, he could give any one painting, any two, any three paintings, and so on, he could even give all ten paintings.
The reader might thin... | 433. Multiply 2 by itself as many times as there are pictures, and subtract 1. Thus, 2 to the tenth power is 1024. Subtracting 1, we get 1023, which is the correct answer. Suppose we have only three pictures. Then one of them can be chosen in three ways,
\footnotetext{
* It can be said that Dudeney proved a local, not ... | 1023 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
434. Parliamentary Elections. How many different ways are there to elect 615 members of parliament if there are only four parties: Conservatives, Liberals, Socialists, and Independents? The mandates can be distributed, for example, as follows: Conservatives - 310, Liberals - 152, Socialists - 150, Independents - 3. Oth... | 434. There are 39147416 different ways in total. Add 3 to the number of members (which gives 618) and subtract 1 from the number of parties (which gives 3). Then the answer is the number of ways to choose 3 items from 618, that is,
\[
\frac{618 \times 617 \times 616}{1 \times 2 \times 3}=39147416 \text { ways }
\]
Th... | 39147416 | Combinatorics | 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 |
438. Escape across the river. During the flight of the Turkish troops at Treise, a small detachment found itself on the bank of a wide and deep river. Here they found a boat in which two boys were boating. The boat was so small that it could only hold two children or one adult.
How did the officer manage to cross the ... | 438. Two children are rowing to the other shore. One of them gets out, and the other returns. A soldier ferries across, gets out, and the boy returns. Thus, to ferry one adult across, the boat has to make 4 trips from shore to shore. Therefore, it had to make \(4 \times 358=1432\) trips to ferry the officer and 357 sol... | 1432 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
443. Four chips in a straight line. Before you is a board of 36 squares, on which 4 chips are arranged in a straight line such that any square on the board is on the same row, column, or diagonal as at least one of the chips. In other words, if we consider our chips as chess queens, then every square on the board is un... | 443. There are 9 main solutions presented in the figure. Solution \(A\) is the one given in the problem statement. Of these 9 solutions, \(D\), \(E\), and \(J\) each generate 8 solutions through rotations and reflections, as explained earlier, while the others yield only 4 solutions each. Therefore, there are a total o... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
445. At Lunch. The clerks of the firm "Pilkings and Popinjay" decided that they would sit three at a time at the same table every day until any 3 people would be forced to sit at this table again. The same number of clerks of the firm "Redson, Robson and Ross" decided to do the same, but with 4 people at a time. When t... | 445. If Pilkins had 11 clerks and Redson 12, they could have sat at a table in 165 and 495 ways respectively, which would have been the solution to the problem. However, we know that both firms had an equal number of clerks. Therefore, the answer is 15 clerks, sitting three at a time for 455 days, and 15 clerks, sittin... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
449. Ten Barrels. A merchant had 10 barrels of sugar, which he arranged into a pyramid as shown in the figure. Each barrel, except one, was marked with its own number. It turned out that the merchant had accidentally placed the barrels in such a way that the sum of the numbers along each row was 16.
By changing the position of the numbers (but not... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
450. Signal Lights. Two spies on opposite banks of a river came up with a way to signal at night using a frame (similar to the one shown in the picture) and three lamps. Each lamp could emit white, red, or green light. The spies developed a code in which each signal meant something.
. With one red and two white lamps, we can also obtain 15 combinations, and for each of them, there are 3 more combinations of the order of colors; in total, 45 combinations. The same result will be obtained with one red and two gre... | 471 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
469. Groups of Dominoes. Is it known to any of my readers that if all 28 dominoes are laid out in a single line according to the usual rule (6 to 6, 2 to 2, blank to blank, etc.), the numbers at the ends will always match, so that the dominoes can actually be arranged in a circle? A very old trick involves hiding one o... | 469. The figure shows one of the solutions. The chain of dominoes is broken into 4 parts, each containing 7 pieces, and the sum of the points in each part is 22.
 | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
478. Setting up dominoes. One day someone reminded Professor Rackbrain about his promise to say how many ways there are to arrange 28 domino tiles in a single line according to the usual rules of the game, if arrangements from left to right and from right to left are considered different. After some time, he reported t... | 478. There are 126760 different ways to arrange 15 dominoes in a line, if the two directions are distinguished. | 126760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
479. Matchstick Puzzle. Taking a box of matches, I found that I could form any pair of regular polygons shown in our diagram using all the matches each time. So, if I had 11 matches, I could use them to form, as shown, either a triangle and a pentagon, or a pentagon and a hexagon, or a square and a triangle (using only... | 479. The smallest possible number is 36 matches. We can form a triangle and a square from 12 and 24 matches, a triangle and a pentagon from 6 and 30 matches, a triangle and a hexagon from 6 and 30 matches, a square and a pentagon from 16 and 20 matches, a square and a hexagon from 12 and 24 matches, and a pentagon and ... | 36 | 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 |
495. Count the Matches. A friend writes that he bought a small box of short matches, each an inch long. He found that he could arrange them in the form of a triangle, the area of which contained as many square inches as there were matches. Then he used 6 matches, and it turned out that from the remaining ones, he could... | 495. There were 36 matches in the box, from which my friend could form a triangle \((17,10,9)\) with an area of 36 square inches. After using 6 matches, the remaining 30 formed a triangle \((13,12,5)\) with an area of 30 square inches, and using another 6 matches, he was able to form a triangle \((10,8,6)\) with an are... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
497. Shuffling Cards. An elementary method of shuffling cards consists of taking a deck face down in the left hand and transferring the cards one by one to the right hand; each successive card is placed on top of the previous one: the second on top of the first, the fourth on top of the third, and so on until all the c... | 497. To shuffle 14 cards in the manner described above and return them to their original order, it takes 14 shuffles, although in the case of 16 cards, only 5 are required. We cannot delve into the nature of this phenomenon here, but the reader may find it interesting to conduct an independent investigation of this que... | 14 | Number Theory | 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 |
7. Mental arithmetic. Square 85 in your mind. | 7. Since $(10 a+5)^{2}=100 a^{2}+100 a+25=a(a+1) 100+25$, we get that $(85)^{2}=8 \cdot 9 \cdot 100+25=7225 \star$.
$$
[\text { M. M., 24, } 273 \text { (May 1951).] }
$$ | 7225 | Other | 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 |
23. The End of the World. On April 1, 1946, the newspaper "Jericho Daily Lapse" reported: "A famous astrologer and numerologist from Guyasuela, Professor Euclid Paracelsus Bombast Umbugio, predicted that the end of the world will occur in 2141. His prediction is based on deep mathematical and historical research. Profe... | 23. All that is needed to solve the problem is that $x^{n}-y^{n}$ for $n=0,1,2, \ldots$ is divisible by $x-y$. Let the quantity the professor was calculating be $F(n)$. Then, since $2141-1863=$ $1770-1492=278, F(n)$ is divisible by 278 for any $n$. Similarly, $2141-1770=1863-1492=371$, a number coprime with 278. Thus, ... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
25. Length of a spiral. A piece of wire is wound in a spiral around a cylindrical tube, forming 10 turns. The length of the tube is 9 cm, and the length of its outer circumference is 4 cm. The ends of the spiral lie on the same generatrix of the cylinder. Find the length of the wire.
, the tenfold repeated circumference ($10 \cdot 4$ cm), and the wire ($L$) now form a right triangle. Therefore, $L=(81+1600)^{\frac{1}{2}}=41$ cm.
\left(q^{2}+q+1\right)\left(r^{2}+r+1\right)\left(s^{2}+s+1\right)}{p q r s}
$$
is not less than 81. | 26. The given fraction can be rewritten as
$$
\left(p+1+\frac{1}{p}\right)\left(q+1+\frac{1}{q}\right)\left(r+1+\frac{1}{r}\right)\left(s+1+\frac{1}{s}\right)
$$
Further, the sum of two positive reciprocal numbers $\geqslant 2$. Therefore, each bracket $\geqslant 3$, and the entire product $\geqslant 81$.
[R. L. Moe... | 81 | Inequalities | proof | Yes | Yes | olympiads | false |
30. Unknown remainder. Find the remainder of the division of $f\left(x^{5}\right)$ by $f(x)$, if
$$
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 |
33. Simple multiplication. Multiply 5746320819 by 125. | 33. Since $125=1000: 8, \quad 5746320819 \cdot 125=$ $5746320819000: 8=718290102375$
$[M$. M., 25, 289 (May 1952).] | 718290102375 | Logic and Puzzles | 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 |
48. Book Series. A certain series of books was published at seven-year intervals. When the seventh book was released, the sum of all the years in which the books of this series were published was 13524. When was the first book of the series published? | 48. The arithmetic mean of the years of publication is $\frac{13524}{7}=$ 1932, or the middle term of our arithmetic progression. The first term of this progression differs from the middle term by three differences. Therefore, the first book was published in 1932 $-3 \cdot 7=1911$.
$[M$. M., 34, 372 (September 1961).] | 1911 | 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 |
53. Dice. A die, on the faces of which the numbers $0,1,2,3,4,5$ are depicted, is rolled until the total sum of the points rolled exceeds 12. What is the most likely value of this sum? | 53. Consider the penultimate throw. After this throw, the sum of points can be $12, 11, 10, 9$ or 8. If it is 12, then after the last throw, the final sum of points can, with equal probability, take the values $13, 14, 15, 16$ or 17. Similarly, if this sum is 11, then the final sum can, with equal probability, take the... | 13 | Combinatorics | 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 |
71. Cutting a Sphere. Into what maximum number of congruent pieces can a sphere be cut so that each side of each piece represents an arc of a great circle that is less than a quarter of such a circle? | 71. Inscribed in this sphere is a regular dodecahedron or icosahedron, and perpendiculars are dropped from the center of the sphere to each face. Construct 60 isosceles triangles whose vertices are at the bases of these perpendiculars and whose bases are the sides of the corresponding faces. Now project these triangles... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
74. A curious number. Find such a positive number that $\frac{1}{5}$ of it, multiplied by its $\frac{1}{7}$, equals this number. | 74. If $(N: 5)(N: 7)=N$, then $N(N-35)=0$, so $N=35$. In other words, if the given number is multiplied by itself, the result will be 35 times greater than if we multiply $\frac{1}{5}$ of this number by $\frac{1}{7}$ of it. Therefore, the number we are looking for is 35.
In the general case, if a number equals $\prod_... | 35 | Algebra | 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 |
85. Contest. Euclid Paracelsus Bombast Umbugio tries to supplement his meager professor's salary by participating in company-sponsored soap production contests. In one such competition, it was required to determine the number of paths along which the word MATHEMATICIAN* could be read on a given diagram. Umbugio counted... | 85. A path can be traced by moving "backwards" from $N$. If we consider the left half of the diagram, including the central column, then at each step backward we have a choice between two possible directions, giving us $2^{12}$ paths. Doubling this number and subtracting 1 (to avoid counting the central column twice), ... | 8191 | Combinatorics | 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 |
97. The Fibonacci sequence. Consider the Fibonacci sequence $1,1,2,3,5,13, \ldots$, whose terms $F_{n}$ (Fibonacci numbers) satisfy the relation $F_{n+2}=F_{n}+F_{n+1}$, $F_{1}=F_{2}=1$. Now consider the sequence of digits in the units place of the Fibonacci numbers. Will this sequence be cyclic, that is, can it be obt... | 97. Yes, it is possible. Each member of the desired sequence $11235831 \ldots$ can be obtained by adding the previous two members and taking the digit in the units place of this sum.
In this sequence, two odd and one even members alternate. There are $5 \cdot 5=25$ ordered pairs composed of odd digits. Therefore, afte... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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}
$$ | 110. Project the sides of an arbitrary polygon onto a line lying in the plane of this polygon. Then the sum of such projections, taken with the appropriate sign, is zero. Now let's take a regular pentagon with a unit side and note that its exterior angle is $72^{\circ}$. The terms
. Restore all the numbers an... | 123. We need to find a three-digit and a one-digit number, the product of which is a four-digit number, and only the digits $2,3,5$ or 7 are allowed. There are only four possible solutions: $3 \cdot 775=2325, 5 \cdot 555=2775$, $5 \cdot 755=3775$ and $7 \cdot 325=2275$.
Since none of the three-digit numbers in these f... | 25575 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
125. Dear club. Ten people decided to found a club. If there were 5 more people, each would contribute 100 dollars less. How much money did each member contribute? | 125. If the original number of participants increases by $50 \%$, then each person's share is only $\frac{2}{3}$ of the previous individual contribution. Thus, 100 dollars constitute $\frac{1}{3}$ of the original contribution, and each club member paid 300 dollars.
$[$ [ M. M., 32, 229 (March 1959).] | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
136. Joint system. For which values of $k$ is the system given below consistent?
$$
\left\{\begin{aligned}
x+y & =1 \\
k x+y & =2 \\
x+k u & =3
\end{aligned}\right.
$$ | 136. Adding the second and third equations of the given system, we get
$$
x+y+k(x+y)=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 |
139. Moving Digits. A certain number, which is written using fewer than 30 digits, starts (if we move from left to right) with the digits 15, that is, it has the form $15 \ldots$. If we multiply it by 5, the result can be obtained simply by moving these two digits to the right end; in the end, we get a number of the fo... | 139. Let $f$ be a proper fraction, in the decimal periodic expansion of which one period coincides exactly with the initial number $15 \ldots$ By the conditions of this problem, $5 f=$ $0, \ldots 15 \ldots 15 \ldots$, and $100 f=15, \ldots 15 \ldots 15$, from which $95 f=15$ and $f=\frac{3}{19}$. Expanding $\frac{3}{19... | 157894736842105263 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
143. Mental arithmetic. Multiply 96 by 104. | 143. Applying the identity $(a-b)(a+b)=a^{2}-b^{2}$, we get: (96) $(104)=(100-4)(100+4)=10000-16=9984$. | 9984 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
149. Scattered set. In a printing house, a multiplication line of the form $\overline{a b c} \cdot \overline{b c a} \cdot \overline{c a b}^{*}$ was being set; but the set scattered, and the digits of the product got mixed up. As a result, the product was printed as 2342355286. It is known that $a>b>c$ and that the unit... | 149. Let $n=a b c$, and $N$ be the desired product. If $c=1$, then the largest possible $n$, namely 981, would yield a product $N=159080922$, which is too small. Therefore, $982 \leqslant n \leqslant$ 987. The sum of the digits of $N$ is $35 \equiv 2(\bmod 3)$; hence, $n$ and the numbers obtained by permuting the digit... | 328245326 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
153. The product of three prime numbers. A certain number is the product of three prime factors, the sum of the squares of which is 2331. There are 7560 numbers (including 1) less than this number and coprime with it. The sum of all divisors of this number (including 1 and the number itself) is 10560. Find this number. | 153. Let $N=p q r$. Then $p^{2}+q^{2}+r^{2}=2331$; hence, each of these prime numbers is less than $(2331)^{\frac{1}{2}}<49$, and all prime numbers are odd.
The sum of all divisors of the number $N$ is $(1+p) \cdot(1+q) \cdot(1+r)=$ $=10560=11 \cdot 960$. The only multiple of 11 not exceeding 49 and greater than some ... | 8987 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
162. In a nonagon. Let $A B$ and $B C$ be two adjacent sides of a regular nonagon inscribed in a circle with center $O$. Let, further, $M$ be the midpoint of $A B$, and $N$ be the midpoint of the radius perpendicular to $B C$. Show that the angle $O M N=30^{\circ}$.
(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 |
175. A divisor of its palindrome. In what base does 792 divide 297?
Note: The original problem statement is in Russian, but the provided translation is in English as requested. | 175. In any number system with base $B \geqslant 10$ the following inequalities hold:
$$
2(297)<2(300)=600<792<800=4(200)<4(297)
$$
Therefore, $792=3(297)$, so $7 B^{2}+9 B+2=3\left(2 B^{2}+9 B+7\right)$, or $B^{2}-18 B-19=0$. Discarding the negative root of this quadratic equation, we get $B=19$.
[D. L. Silverman, ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
202. Invariant remainder. Find a number that, when dividing the numbers $1108, 1453, 1844$, and 2281, gives the same remainder. | 202. Since the remainder does not change, the divisor must be odd. Further, $1453-1108=345,1844-1453=391,2281-1844=$ 437. Now notice that $437-391=391-345=46=2 \cdot(23)$. Since it is always true that
$$
\left(N_{1} d+r\right)-\left(N_{2} d+r\right)=d\left(N_{1}-N_{2}\right)
$$
the desired number is 23, and the corre... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
203. Nine digits. Find a nine-digit number, all digits of which are different and do not contain zero, and the square root of which has the form $\overline{a b a b c}$, where $\overline{a b}=c^{3}$. | 203. The digit $c$ can only coincide with 3 or 4. However, the number $(64644)^{2}$ is ten-digit. Therefore, the only solution to our problem will be $(27273)^{2}=743816529$.
[N. Farnum, S. S. M., 63, 603 (October 1963).]
This solution will be unique even if we drop the restriction $a b=c^{3}$. Moreover, $27273=3(909... | 743816529 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
206. Characteristics of men. In a certain group of men, $70 \%$ have brown eyes; $70 \%$ have dark hair, $85 \%$ are taller than 5 feet 8 inches, and $90 \%$ weigh more than 140 pounds. What percentage of men definitely have all four of these characteristics? | 206. Let's take 100 men and record each of the four characteristics for each man who possesses the characteristic. As a result, we will have $70+75+85+90=320$ records. Since there are 320 records for 100 men, we can observe that with the most even distribution of characteristics among the men, each man would have at le... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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 |
227. Isosceles Triangle. In isosceles triangle \( A B C \), the angle at vertex \( C = 20^{\circ} \). Points \( M \) and \( N \) are chosen on the lateral sides \( A C \) and \( B C \) respectively, such that angle \( A B M = 60^{\circ} \) and angle \( B A N = 50^{\circ} \). Prove, without using trigonometry, that angl... | 227. Since the sum of the interior angles of a triangle is two right angles, or \(180^{\circ}\), \(\angle CBA = \angle CAB = 80^{\circ}\), \(\angle CBM = 20^{\circ}\), and \(\angle BAN = 50^{\circ} = \angle BNA\) (so \(BN = AB\)). Draw \(MR\)
, the polynomia... | 236. Willie's friend, relying too much on the "trial and error" method, overlooked one fact that could have been very useful. The fact is that if we take a polynomial \( P(x) \) with integer coefficients and two different integers \( a \) and \( b \), then \( P(a) - P(b) \) will be divisible by \( a - b \). Let's denot... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
249. Bonus Fund. In a certain institution, there was a bonus fund. It was planned to distribute it so that each employee of the institution would receive 50 dollars. But it turned out that the last person on the list would receive only 45 dollars. To maintain fairness, it was then decided to give each employee 45 dolla... | 249. In fact, $5 was seized from $95: 5=19$ people; therefore, the original amount of the fund was $20 \cdot 50-5=$ $=995$ dollars. | 995 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
250. The Court Mathematician's Salary. Once, the court mathematician received his entire annual salary in silver talers, from which he formed nine piles, together constituting a magic square. The king, upon seeing this, was impressed but regretfully noted that in none of the piles was the number of coins a prime number... | 250. As established by L. S. Fryer, the elements of an arbitrary magic square of the third order can actually be represented using three parameters:
$$
\begin{array}{ccc}
(e+x), & (e-x-y) & (e+y) \\
(e-x+y), & e, & (e+x-y) \\
(e-y), & (e+x+y) & (e-x)
\end{array}
$$
It is clear that these numbers can be rearranged so ... | 1350 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
256. Palindromic Squares. A certain square is written in the base-6 numeral system using 5 different non-zero digits. If the digit in the units place is moved from the end of this number to the beginning, the square root of the resulting number matches the square root of the original number, written in reverse order. F... | 256. $12345 \leqslant N^{2} \leqslant 54321$, so $113 \leqslant N \leqslant 221$. If we write the digits of the number $N$ in reverse order, the resulting number will also satisfy these inequalities. The number $N^{2}$ is divisible by 5, so $N$ is also divisible by 5. Let's list all three-digit numbers divisible by 5 a... | 53241 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.