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204. How much wood. Once at breakfast, Colonel Crackham said that two workers he knew could saw 5 cubic meters of wood in a day. They could chop the sawn wood into firewood at a rate of 8 cubic meters per day. The colonel wanted to know how many cubic meters of wood the workers needed to saw so that they could chop it ... | 204. Two workers need to saw \(3 \frac{1}{13} \mathrm{~m}^{3}\) of firewood. | 3\frac{1}{13}\mathrm{~}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,120 |
205. Nut Packages. George Crackham placed five paper packages on the table for breakfast. When asked what was in them, he replied:
- I put a hundred nuts in these five packages. In the first and second packages, there are 52 nuts, in the second and third - 43, in the third and fourth - 34; in the fourth and fifth - 30... | 205. In five bags, there are \(27,25,18,16,14\) nuts. The contents of each bag can be found by subtracting from 100 the total number of nuts in the pairs of bags that do not include the given bag. Thus, in the third bag, there are \(100-(52+30)=18\) nuts. | 27,25,18,16,14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,121 |
206. Распределение орехов. Тетушка Марта купила орехов. Томми она дала один орех и четверть оставшихся, и Бесси получила один орех и четверть оставшихся, Боб тоже получил один орех и четверть оставшихся, и, наконец, Джесси получила один орех и четверть оставшихся. Оказалось, что мальчики получили на 100 орехов больше, ... | 206. Первоначально было 1021 орех. Томми получил 256 , Бесси 192, Боб 144 и Джесси 108 орехов. Всего девочки получили 300, а мальчики 400 орехов. Тетушка Марта оставила себе 321 орех. | 321 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,122 |
207. Young Bandits. Three young "highwaymen," returning from the cinema, met a vendor with apples. Tom grabbed half of all the apples, but threw 10 back into the basket. Ben took a third of the remaining apples, but returned 2 apples he didn't like. Jim took half of the remaining apples, but threw back one wormy one. T... | 207. The vendor had 40 apples. Tom left her 30, Bob 22 and Jim 12 apples. | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,123 |
208. Biscuits. A trader packed his biscuits (all of the same quality) into boxes weighing $16, 17, 23, 39$, and 40 pounds respectively and was unwilling to sell them in any other way than whole boxes. A customer asked him to supply 100 pounds of biscuits.
Could you fulfill this order? If not, how close can you get to ... | 208. You need to give the customer four boxes at 17 pounds each and two at 16 pounds, which will make exactly 100 pounds. | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,124 |
## 209. Three Workers.
- We, Bill, and I, - said Kayzi, - can do this job for you in 10 days, and if Bill is replaced by Alec, we can manage it in 9 days.
- And even better, - said Alec, - give me Bill as a helper, and we will do your job in 8 days.
How long would it take each worker to complete this job alone? | 209. Alek can complete the work in \(14 \frac{34}{49}\) days, Bill in \(17 \frac{23}{41}\) days, and Casey in \(23 \frac{7}{31}\) days. | 14\frac{34}{49},17\frac{23}{41}, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,125 |
211. "Boomerang". I call "boomerang" one of the oldest types of arithmetic puzzles. Someone is asked to think of a number and after a series of calculations, to state the result. Upon hearing the result, the person who asked the question immediately announces the thought-of number. There are hundreds of different versi... | 211. Get the remainder of the division by 3, multiply it by 70, get the remainder of the division by 5 and multiply it by 21, and get the remainder of the division by 7 and multiply it by 15. Add the results, and you will get either the number you thought of or a number that differs from the thought number by an intege... | 79 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,127 |
213. Lilavati. Here is a small puzzle borrowed from "Lilavati" (1150 AD) by Bhaskara
"A beautiful maiden with beaming eyes told me a number. If this number is multiplied by 3, $\frac{3}{4}$ of the product is added, the sum is divided by 7, $\frac{1}{3}$ of the quotient is subtracted, the result is multiplied by itself... | 213. The maiden named the number 28. The trick is to reverse the entire process of calculations: multiply 2 by 10, subtract 8, square the result, and so on. For example, remember that increasing a product by \(\frac{3}{4}\) means taking \(\frac{7}{4}\) of it. The reverse action involves taking \(\frac{4}{7}\). | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,129 |
216. Blindness in Bats. One naturalist, trying to mystify Colonel Crackham, told him that he had studied the question of blindness in bats.
- I found,- he said, - that the ingrained habit of bats to sleep during the day in dark corners and fly only at night has led to the spread of blindness among them, although some ... | 216. The smallest number of mice is 7, with three possible cases:
1) 2 see well, 1 is blind only in the right eye, and 4 are completely blind
2) 1 sees well, 1 is blind only in the left eye, 2 are blind only in the right eye, and 3 are completely blind;
3) 2 are blind only in the left eye, 3 only in the right eye, and ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,132 |
219. Sheep Division. A certain Australian farmer, on his deathbed, left his sheep to his three sons. Alfred is to receive $20 \%$ more than John and $25 \%$ more than Charles. John's share is 3600 sheep.
How many sheep will Charles receive? Perhaps the reader will be able to solve the problem in a few seconds. | 219. Charles's share amounts to 3456 sheep. Some readers probably first found Alfred's share and then subtracted \(25 \%\), but such a solution is, of course, incorrect. | 3456 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,135 |
222. Marching Column. A military unit was moving in a marching column, in which the number of ranks exceeded the number of soldiers in a rank by 5. When the enemy appeared, a reformation into 5 ranks took place, as a result of which the number of soldiers in each rank increased by 845.
How many people were in the unit | 222. There were a total of 4550 people in the unit. Initially, the soldiers marched in a column of 70 ranks with 65 people in each; then they reformed into 5 ranks with 910 soldiers in each. | 4550 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,138 |
224. Boxes with Shells. Shells for six-inch howitzers were packed in boxes of 15, 18, and 20.
- Why do you have different boxes? - I asked the officer at the warehouse.
- You see, - he replied, - this allows us to deliver the required number of shells to the battery without opening the boxes.
Indeed, this system work... | 224. An officer at the warehouse must issue the required number of shells in boxes of 18 shells until the remaining number of shells is a multiple of 5. If the number of shells is not 5, 10, or 25, the remaining shells should be issued in boxes of 15 and 20 shells. The largest number of shells for which the system fail... | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,140 |
225. Fruit Garden. The gardener decided to create a new fruit garden. He planted young trees in rows so that a square was formed. In doing so, he had 146 extra seedlings left. But to increase the square by adding an extra row, the gardener had to buy 31 more trees.
How many trees are there in the garden at the end of ... | 225. Initially, there were 7890 saplings, from which a square of \(88 \times 88\) was formed, and 146 trees were left over. By purchasing an additional 31 trees, the gardener was able to increase the square to \(89 \times 89\), and the number of trees in the garden became 7921. | 7921 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,141 |
226. Cubes and Squares. Here is an interesting, though not easy puzzle, the author of which could not be identified.
Three children had identical boxes of cubes. The first girl made a square frame from all her cubes, marked with the letter $A$ on the diagram. The second girl made a larger square $-B$. The third girl m... | 226. The smallest number of cubes in the box is 1344. By building a frame around an empty square \(34 \times 34\), the first girl formed a square \(50 \times 50\), the second - a square \(62 \times 62\) and the third - a square \(72 \times 72\) with four extra cubes in the corners. | 1344 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,142 |
227. Find the triangle. The sides and height of a certain triangle are expressed by four consecutive integers. What is the area of this triangle? | 227. The sides of a triangle are 13, 14, and 15, with the base being 14, the height 12, and the area 84. There are infinitely many rational triangles whose sides are expressed as consecutive integers, such as 3, 4, and 5 or 13, 14, and 15, but only one of them satisfies our conditions for the height.
Triangles with si... | 84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,143 |
228. Cow, Goat, and Goose. A certain farmer found out that his cow and goat eat the grass on the meadow in 45 days, the cow and goose in 60 days, and the goat and goose in 90 days. If he releases the cow, goat, and goose on the field at the same time, how many days will it take them to eat all the grass on the meadow?
... | 228. Since the cow and the goat eat \(\frac{1}{45}\), the cow and the goose \(\frac{1}{60}\), and the goat and the goose \(\frac{1}{90}\) of all the grass in a day, we easily find that the cow eats \(\frac{5}{360}\), the goat \(\frac{3}{360}\), and the goose \(\frac{1}{360}\) of all the grass in a day. Therefore, all t... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,144 |
230. Mental Arithmetic. In order to test his pupils' ability in mental arithmetic, Rakebrain one morning asked them to do the following:
- Find two whole numbers (each less than 10), the sum of whose squares plus their product would give a perfect square.
The answer was soon found. | 230. There are two solutions not exceeding ten: 3 and 5, 7 and 8.
The general solution is obtained as follows. Denoting the numbers by \(a\) and \(b\), we get
\[
a^{2}+b^{2}+a b=\square=(a-m b)^{2}=a^{2}-2 a m b+b^{2} m^{2}
\]
Therefore,
\[
b+\mathrm{a}=-2 a m+b m^{2}
\]
from which
\[
b=\frac{a(2 m+1)}{m^{2}-1}
\... | (3,5)(7,8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,146 |
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 | 41,149 |
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 | 41,151 |
236. Numerical Crossword. In the image, you see a numerical crossword. It is similar to a regular crossword, but with the difference that instead of letters, digits are entered into the cells. The following conditions must be met.
, etc. Number 26 down is the square of number 18 across. Therefore, number 18 across is either 111 or 222, since the squar... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,152 |
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 | 41,153 |
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 | 41,155 |
240. Geometric progression. One morning, Professor Rackbrain challenged his friends to find no fewer than three integers that form a geometric progression starting with 1, and the sum of which must be a perfect square. (For example, $1+2+4+8+16+32=63$. However, the last number is one short of being a square. I know onl... | 240.
\[
1+3+9+27+81=121=11^{2}
\]
and
\[
1+7+49+343=400=20^{2}
\] | 121=11^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,156 |
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 | 41,157 |
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 | 41,159 |
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 | 41,160 |
245. Pat Murphy. Many years ago, such an incident occurred. The participants of one expedition fell into the clutches of bloodthirsty savages. The chief, after receiving rich gifts, finally softened and allowed the captives to leave, but on the condition that half of them would be flogged. The expedition consisted of 5... | 245. Pete said: "No matter what number you name, it's all the same, but since there are ten people here plus myself, I'll name eleven and start counting from myself." Naturally, he was the first to be sent to be flogged. Therefore, if starting from number 1, the smallest number that would be fatal for the Englishmen wo... | 11和29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,161 |
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 | 41,163 |
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 | 41,164 |
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 | 41,165 |
250. Iron Chain. On the battlefield, two pieces of iron chain were found. How it got there and for what purposes it was used is unknown, but we don't need to know that. The chain was made up of round links (all of the same size) made from an iron rod with a thickness of $\frac{1}{2}$ cm. One piece of the chain was $36 ... | 250. If the inner diameter of a link is multiplied by the number of links and the double thickness of the iron rod is added, then the length of the chain is obtained. Each link, when attached to the chain, loses its length by twice the thickness of the rod. The inner diameter is \(2 \frac{1}{3} \mathrm{~cm}\). If we mu... | 915 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,166 |
251. Guess the Coin. "Do you know this trick?" Dora asked her brother. "Put a ten-cent coin in one pocket and a five-cent coin in the other. Now multiply the number of cents in the right pocket by 3, and the number in the left pocket by 2; add what you get, and tell me whether the result is even or odd."
Her brother r... | 251. If the brother answered Dora with "even," then the ten-cent coin was in the right pocket, and the five-cent coin in the left. If he said "odd," then the five-cent coin was in the right pocket, and the ten-cent coin in the left. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,167 |
255. With one compass. Can you construct the 4 vertices of a square using only one compass? You have only a sheet of paper and a compass. Resorting to various tricks, such as folding the paper, is not allowed. | 255. To mark the vertices of a square using a single compass, first draw a circle. Then, fixing the compass opening and starting from any arbitrarily chosen point \(A\) on the circumference, mark points \(B, C\), and \(D\). From points \(A\) and \(D\) as centers with the radius \(A C\), describe two arcs intersecting a... | A,F,D,G | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,171 |
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 | 41,172 |
265. Garden. The four sides of the garden are 20, 16, 12, and 10 meters, and its area is maximized with these dimensions. What is the area? | 265. This trapezoid can be inscribed in a circle. The semi-perimeter of its sides is 29. By subtracting each side from this number in turn, we get \(9,13,17,19\). The product of these numbers is 37791. The square root of the obtained number is 194.4, which coincides with the size of the desired area. | 194.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,181 |
268. Square Window. Once over a cup of tea, Colonel Crackham told a story about a person who had a square window with an area of 1 m $^{2}$, which let in too much light. The owner of the window blocked off half of it, but in doing so, he was left with a square window that was one meter wide and one meter high.
How cou... | 268. In the figure, the original window with an area of \(1 \mathrm{~m}^{2}\) is shown with dashed lines. After the owner blocked off the four corners, he was left with a square window of half the area, but one meter wide and one meter high.
 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 | 41,186 |
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.
 cm in size, the width of the miniature will be 6 cm, and the height 12 cm. It is easy to verify that the excess will be as required by the condition of the problem. | 10\times20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,188 |
273. In the Garden. Once, over a cup of tea, Colonel Crackham said:
- My friend Tompkins loves to challenge us with unexpected puzzles at every opportunity, but they are not too profound. One day, while we were strolling in the garden, he suddenly pointed to a rectangular flower bed and said:
- If I made it 2 meters w... | 273. The flower bed was \(14 \mathrm{m}\) long and \(10 \mathrm{m}\) wide. | 14\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,189 |
275. The Pen Puzzle. Answers to well-known puzzles given in old books are often completely wrong. Nevertheless, it seems that no one ever notices these errors. Here is one such example.
A farmer had a pen with a fence that was 50 rails long, in which only 100 sheep could fit. Suppose the farmer wanted to expand the pe... | 275. The old answer is that if you arrange the poles as shown in case \(A\), 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 | 41,191 |
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 | 41,195 |
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 | 41,198 |
284. Circles and Disks. Once at a fair, we saw a man sitting at a table covered with an oilcloth featuring a large red circle in the center. The man offered the public to cover the circle with five thin disks lying nearby, promising a valuable prize to anyone who could do it. All the disks were of the same size, of cou... | 284. In the figure, the circle bounding the red circle is shown with a dashed line, and a regular pentagon inscribed in it. The common center of the circle and the pentagon is marked with the letter \(C\). We will find a point \(D\) equidistant from \(A, B\), and \(C\), and draw a circle \(A B C\) with radius \(A D\). ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,200 |
285. Three Fences. One day over a cup of tea, Colonel Crackham said:
- A man had a circular field, and he wanted to divide it into 4 equal parts with three fences of equal length. How can this be done?
- And why did he need fences of the same length? asked Dora.
- No information on this has been preserved, replied the... | 285. To divide a circular field into 4 equal parts with three fences of equal length, the diameter of the circle should initially be divided into 4 equal parts, and then semicircles should be described on both sides of it, as shown in the figure. The curved lines will then represent the required fences.
![](https://cd... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,201 |
287. Car and Circle. A car is driving in a circle. The wheels on the outer side of the circle move twice as fast as the wheels on the inner side.
What is the length of the circumference traveled by the outer wheels if the distance between the wheels on both axles is 1.5 m? | 287. Since the outer wheels move twice as fast as the inner ones, the length of the circumference they describe is twice the length of the inner circumference. Therefore, the diameter of one circle is twice the diameter of the other. Since the distance between the wheels is 1.5 m, the diameter of the larger circle is 6... | 18.85 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,203 |
288. Grinding Wheel. Three people bought a grinding wheel with a diameter of $20 \mathrm{~cm}$. How much should each of the companions grind off so that the wheel is divided equally, if 4 cm of the diameter is excluded for the hole? The practical value of each share is not considered, only the equal division of the tot... | 288. The first companion must use the grinding wheel until its radius is reduced by 1.754 cm. The second must reduce the radius by another 2.246 cm, leaving 4 cm and a hole for the third. This is a very good approximation. | 1.754 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,204 |
289. Automobile Wheels. "You see, sir," said the car salesman, "the front wheel of the car you are buying makes 4 more revolutions than the rear wheel every 360 feet; but if you were to reduce the circumference of each wheel by 3 feet, the front wheel would make 6 more revolutions than the rear wheel over the same dist... | 289. The circumferences of the front and rear wheels are 15 and 18 feet, respectively. Thus, every 360 feet, the front wheel makes 24 revolutions, and the rear wheel makes 20 revolutions, with a difference of 4 revolutions. If the circumference length is reduced by 3 feet, then 12 will fit 30 times in 360 feet, and 15 ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,205 |
292. Another paradox with the wheel. Two cyclists stopped on a railway bridge somewhere in Sussex when a train passed by.
- This train is going from London to Brighton, - said Henderson.
- The greater part of it, - remarked Banks, - and the rest is moving in the direction of London.
- What on earth do you mean by that... | 292. I have already mentioned that if you mark a point on the rim of a bicycle wheel, it will trace a curve in space called a simple cycloid. If, however, you mark a point on the flange of a locomotive or railway car wheel, it will trace a trochoid, a curve ending in loops. In the diagram, I have illustrated the flange... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,208 |
293. Mechanical Paradox. The famous mechanical paradox devised by James Ferguson * around 1751 should be known to everyone. He proposed it to a skeptical watchmaker during an argument.
- Suppose, said Ferguson, that I make one wheel three times as thick as the other three and cut teeth on all of them. Then I will free... | 293. The mechanism depicted in the figure consists of two wooden boards \(B\) and \(C\), connected at the corners to form a frame. The frame is rotated around axis \(a\) using handle \(n\), which passes through the frame and is rigidly attached to the board or table \(A\). Inside the frame, a gear wheel \(D\) is rigidl... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,209 |
301. Draw an ellipse. I think many readers are familiar with the method of constructing an ellipse that I am about to describe. It is very useful if you want to make a frame for a portrait or lay out an oval flower bed. You hammer in two nails or two pins (or two stakes if you are making a flower bed) and put a loop of... | 301. Draw two perpendicular segments \(C D\) and \(E F\) (the length of \(C D\) is 12 cm, the length of \(E F\) is 8 cm), intersecting each other at their midpoints. Find points \(A\) and \(B\) such that \(A F\) and \(F B\) equal half of \(C D\), that is, 6 cm, and place your pins at \(A\) and \(B\), \(\cdot\) taking a... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,217 | |
305. The Ladder. As soon as the conversation turned to a ladder that was needed for some household purposes, Professor Rackbrain suddenly interrupted the discussion, offering the participants a little puzzle:
- A ladder is standing vertically against a high wall of a house. Someone pulls its lower end 4 m away from th... | 305. The distance from the top end to the ground is \(\frac{4}{5}\) of the entire length of the ladder. Multiply the distance from the wall (4 m) by the denominator of this fraction (5), and you get 20. Now subtract the square of the numerator of the fraction \(\frac{4}{5}\) from the square of its denominator. This res... | 6\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,221 |
306. Lightning Rod. A strong gust of wind broke the lightning rod's pole, causing its top to hit the ground 20 m from the base of the pole. The pole was repaired, but it broke again under a gust of wind, 5 m lower than before, and its top hit the ground 30 m from the base.
What is the height of the pole? In both cases... | 306. The height of the pole above the ground was \(50 \mathrm{m}\). In the first case, it broke at 29 m, and in the second case at 34 m from the top. | 50\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,222 |
307. Rope. A rope hangs from the ceiling, touching the floor. If, while keeping the rope taut, it touches the wall, the end of the rope will be 3 cm from the floor. The distance from the freely hanging rope to the wall is 48 cm.
What is the length of the rope? | 307. The length of a freely hanging rope is 3 m \(88 \frac{1}{2}\) cm. | 388.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,223 |
313. Cardboard Box. The reader has probably noticed that there are many problems and questions whose answers, it seems, should have been known to many generations before us, but which, however, have apparently never even been considered. Here is an example of such a problem that came to my mind.
 | 316. A smart fly would choose the path marked on the diagram with a solid line, and it would take 2.236 minutes to overcome it. The path marked with a dashed line is longer, and it would take more time.
. Therefore, climbing a tree that is 8 m tall, it will travel a distance of \(10 \mathrm{m}\). | 10\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,237 |
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 | 41,238 |
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 | 41,255 |
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).
![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-269.jpg?height=442&width=397&top_left_y=961&top_left_x=65... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,264 |
351. The Ruined Carpet. A lady had an expensive Persian carpet measuring $12 \times 9 \mathrm{~m}$, which was severely damaged in a fire. Therefore, she had to cut out a hole measuring $8 \times 1 \mathrm{~m}$ in the middle of the carpet, and then cut the remaining part into two pieces, from which she sewed a square ca... | 351. If the carpet is cut into two parts as shown in case 1, and the pieces are sewn together as depicted in case 2, a square will be formed. The width of the step is 2, and the height is 1 m.
![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-270.jpg?height=406&width=862&top_left_y=1124&top_left_x=60... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,267 | |
352. How to fold a hexagon? Puzzles that require folding something out of paper are both interesting and instructive. I mean not the various boxes, boats, and frogs folded from paper, as these are more toys than puzzles, but the solution of certain geometric problems, so to speak, "with bare hands."
 and \(C O D\). We also make folds \(E H\) and \(F G\), which bisect \(A O\) and \(O B\). Flip \(A K\) so that \(K\) lands on the line \(E H\) at point \(E\), and then make folds along \(A E\) and \(E O G\). Similarly, find poi... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,268 |
354. How to fold an octagon? Can you cut a regular octagon (see figure) from a square sheet of paper using only scissors, without using a compass and ruler? You are allowed to pre-fold the paper to then cut along the creases.
 and \(C D\), you can mark the midpoints \(E\) and \(G\) with folds. Similarly, you can find points \(F\) and \(H\), and then fold the square \(E H G F\). Next, align \(C H\) with \(E H\) and \(E C\) with \(E H\), which will give you the intersection point \(I\). Do the same with the re... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,270 |
360. Another jumping puzzle. Draw the board and place 17 chips on it as shown in the figure. The puzzle is to remove all chips except one, by performing a series of jumps similar to those in simplified solitaire. One chip can jump over another to the nearest square, if it is free, and the chip that was jumped over is r... | 360. "Nine" sequentially jumps over \(13,14,6\), \(4,3,1,2,7,15,17,16,11\). Then 12 jumps over 8, 10 jumps over 5 and 12, and 9 jumps over 10. | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,276 | |
361. Moving Chips. Divide a sheet of paper into 6 squares and place a stack of 15 chips numbered $1,2,3, \ldots, 15$ in square $A$ (see figure), 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 | 41,277 |
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 | 41,280 |
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 | 41,284 |
375. Incorrect Magic Square. The figure shown here is a correct magic square, composed of numbers from 1 to 16 inclusive. The sum of the numbers in any row, any column, and on either of the two main diagonals is 34. Suppose now that you are forbidden to use the numbers 2 and 15, but instead you can repeat any two numbe... | 375. Taking the numbers 7 and 10 instead of 2 and 15, you can form a square as shown in the figure. You can practically form a magic-like square from any 16 numbers if you manage to arrange them in such a way that the differences between any two adjacent numbers horizontally and the differences between any two adjacent... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,291 |
376. The Magic Square Misunderstanding. Before you is a magic square of the fifth order. I have found that the vast majority of people, not deeply familiar with the theory of magic squares, are convinced that in squares of the fifth order, the number 13 must invariably stand in the center. One reader, who had been amus... | 376. If you make 9 squares, coinciding with the square depicted in our drawing, and then form a larger square from them, you will find magic squares of the fifth order with any number in the center on it. This square is called a Nasik square (named by the late Mr. Frost after Nasik - a place in India where he lived) an... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 41,292 |
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 | 41,293 |
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 | 41,294 |
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 | 41,308 |
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 | 41,312 |
398. Wine Mixture. One vessel is filled with wine to $\frac{1}{3}$, and another vessel of equal capacity - to $\frac{1}{4}$. Each of these vessels was topped up with water, and all their contents were mixed in a jug. Half of the resulting mixture was then poured back into one of the two vessels.
In what ratio did the ... | 398. The mixture contains \(\frac{7}{24}\) of wine and \(\frac{17}{24}\) of water. | \frac{7}{24} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,313 |
399. Stolen Balm. Three thieves stole a vase containing 24 ounces of balm from a gentleman. Hurriedly making their escape, they met a glassware seller in the forest, from whom they bought three containers. Finding a secluded spot, the thieves decided to divide the loot, but then discovered that the capacities of their ... | 399. Here is one of several solutions:
\begin{tabular}{lrrrr}
Capacity of vessels in ounces & 24 & 13 & 11 & 5 \\
Contents in the vessel: & & & & \\
before pouring & 24 & 0 & 0 & 0 \\
after 1st pouring & 0 & 8 & 11 & 5 \\
after 2nd pouring & 16 & 8 & 0 & 0 \\
after 3rd pouring & 16 & 0 & 8 & 0 \\
after 4th pouring & ... | 8&8&8&0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,314 |
400. Milk Delivery. One morning, a milkman was transporting two 80-liter barrels of milk to his shop when he encountered two women who begged him to sell them 2 liters of milk each right away. Mrs. Green had a 5-liter jug, and Mrs. Brown had a 4-liter jug, while the milkman had nothing to measure the milk with.
How di... | 400. The simplest solution to the problem is as follows (the capacity of the vessels is indicated above, the initial amount of contents below, and in each subsequent line - the amount of contents after each operation):
\begin{tabular}{llll}
80 liters & 80 liters & 5 liters & 4 liters \\
80 & 80 & 0 & 0 \\
75 & 80 & 5 ... | 80&76&2&2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,315 |
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 | 41,328 |
415. Путешествие миссис Симпер. На рисунке изображена упрощенная схема маршрута, по которому моя приятельница миссис Симпер собирается путешествовать следующей осенью. Можно заметить, что на схеме представлено 20 городов, соединенных между собой железнодорожными линиями. Миссис Симпер живет в городе \(H\) и хочет посет... | 415. Существует 60 маршрутов, следуя по которым миссис Симпер могла бы посетить каждый город по одному и только по одному разу, закончив путь в \(H\), если считать различными маршруты, отличающиеся только направлением. Однако если леди должна избежать тоннелей между \(N\) и \(O\), а также между \(S\) и \(R\), то можно ... | HISTLKBCMNUQRGFPODEAH | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,330 |
422. "Classics". We often see children playing the ancient and universally popular game of "classics". In one of its variations, the figure shown here is drawn on the ground. We want to know if it is possible to draw this figure with one continuous line. It turns out that it is possible.
 and finish it at \(B\), or vice versa. Otherwise, it is impossible to draw the required figure with one continuous line.
OTBETЫ
![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-301.jpg?height=271&width=765&top_l... | A\toBorB\toA | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,337 |
424. Planting Trees. One person planted 13 trees as shown in the diagram. As a result, he ended up with 8 rows, each containing 4 trees. However, he did not like how the second tree in the horizontal row was planted. He expressed his feelings about this rather vaguely, saying that "it is slacking off there and generall... | 424. The figure shows an elegant way of planting trees in 9 rows with 4 trees in each.
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,339 |
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 | 41,348 |
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 | 41,349 |
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 | 41,351 |
437. Missionaries and Cannibals. There is one unusual story about three missionaries and three cannibals who had to cross a river in a small boat that could only hold two people at a time. Having heard of the cannibals' tastes, the missionaries could not afford to be in the minority on either bank of the river. Only on... | 437. Let's denote three missionaries as Ммм, and three cannibals as Ккк; capital letters denote a missionary and a cannibal who can row. Then, \(\mathrm{K}\) crosses to \(\mathrm{K}\), \(\mathrm{K}\) returns with the boat; \(\mathrm{K}\) crosses; \(\mathrm{K}\) returns; \(\mathrm{M}\mathrm{m}\) cross; \(\mathrm{M}\math... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,352 |
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 | 41,353 |
441. The Broken Ruler. Here is an interesting puzzle that resembles (although in reality, it is significantly different from) one of the classic problems of Bache about a weight cut into pieces, with which it is possible to determine the weight of any load from 1 pound to the total weight of all pieces. In our case, a ... | 441. Let 8 divisions break a 33-centimeter ruler into 9 parts of lengths \(1,3,1,9,2,7,2,6,2\) cm. Then with their help, any whole number of centimeters from 1 to 33 cm can be measured. Of course, the divisions themselves are located at distances of \(1,4,5,14,16,23,25\) and 31 cm from one of the ends of the ruler. Ano... | 1,4,5,14,16,23,25,31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,356 |
442. Six Cottages. A 27 km road surrounds an abandoned and uninhabited area. Along it are 6 cottages (see figure) arranged in such a way that some of them are 1, 2, 3, and so on up to 26 km apart from each other. For example, Brown can be 1 km from Stiggins, Jones can be 2 km from Rogers, Wilson can be 3 km from Jones,... | 442. If cottages are arranged in a circle at intervals of 1, 1, 4, 4, 3, 14 km, then for any integer number of kilometers from 1 to 26 inclusive, there will be two cottages that are that distance apart.
[This problem is clearly a variation of the previous one. Just as before, Dudeney could have increased the length of... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,357 |
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 | 41,358 |
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 | 41,360 |
446. The "Effervescent" Puzzle. In how many ways can the letters in the word EFFERVESCES \({ }^{*}\) be arranged in a line so that no two \(\mathrm{E}\) are adjacent? Of course, we do not distinguish between identical letters like \(F F\), since swapping them does not result in a new arrangement.
By changing the position of the numbers (but not... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,364 |
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 | 41,365 |
451. Chained Prisoners. Once upon a time, there were 9 very dangerous prisoners who had to be carefully watched. Every weekday, they were taken out to work, chained together as shown in the picture, which, incidentally, was drawn by one of the guards. No two people were ever chained together more than once in the same ... | 451. In the following solution, each prisoner is chained with each of the others exactly once.
\[
\begin{array}{ccccccc}
1-2-3 & 2-6-8 & 6-1-7 & 1-4-8 & 7-2-9 & 4-3-1 & 4-5-6 \\
5-9-1 & 9-4-2 & 2-5-7 & 3-6-4 & 5-8-2 & 7-8-9 & 3-7-4 \\
\\
8-3-5 & 6-9-3 & 8-1-5 & 9-7-6
\end{array}
\]
If the reader is looking for a diff... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,366 |
453. Archery Competition. Three archers, each with six arrows, hit the target shown in the image. Hitting the "bullseye" is worth 40 points, and each subsequent ring from the center is worth 39, 24, 23, 17, and 16 points, respectively. The results were as follows: Miss Dora
Talbot - 120 points, Reggie Watson - 110 poi... | 453. Mrs. Finch scored 4 times 17 and 2 times 16, for a total of 100 points; Reggie Watson hit 2 times 23 and 4 times 16, for a total of 110; Miss Dora Talbot scored once 40 and 5 times 16, for a total of 120 points. She could have scored her 120 points in various ways if it were not stated that someone's arrow hit the... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,368 | |
456. Seven children. Four boys and three girls sit in a row in a random order.
What is the probability that the two children at the ends of the row will be girls?
\section*{PUZZLES} | 456. Children can sit in 5040 different ways, of which in 720 cases girls will end up at both ends. Therefore, the desired probability is \(\frac{720}{5040}\), or \(\frac{1}{7}\). Of course, this can also be expressed differently, saying that there is 1 chance in 6 that girls will end up at the ends. | \frac{1}{7} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,371 |
457. Tic-Tac-Toe. Every child knows how to play this ancient game. A square is divided into 9 cells. Each player, in turn, places their mark (an X or an O) in a free cell, trying to align three of their marks in a straight line. The player who manages to do this wins. If two good players are playing, each game between ... | 457. Let's renumber the cells as shown in the figure. Case \(A\): Mr. O (the first player) can start the game in three ways: from the center 5, or from any corner \(-1,3,7\) or 9, or from any side \(-2,4,6,8\). Let's analyze these starts one by one. If Mr. O starts from the center, then Mr. X has the option to go to a ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 41,372 |
458. The Horseshoe Game. Here is a small game akin to tic-tac-toe. Two players participate. One player has two white chips, the other has two black chips. Taking turns, each player places a chip on a free circle (see the diagram),
, called out by the first player, is ... | 459. The first player is best off naming 2 or 3, since in these cases only one outcome of the dice throw will lead to his defeat. If he names 1, then an unfavorable outcome would be rolling a 3 or 6. If he names 2, then only rolling a 5 would be unfavorable. If he names 3, then only a 4 would be unfavorable. If he name... | 2or3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,374 |
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