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68. From Pickleminster to Quickville. Trains $A$ and $B$ depart from Pickleminster to Quickville at the same time as trains $C$ and $D$ depart from Quickville to Pickleminster. Train $A$ meets train $C$ 120 miles from Pickleminster and train $D$ 140 miles from Pickleminster. Train $B$ meets train $C$ 126 miles from Qui... | 68. There are two distances that satisfy the condition of the problem - 210 and 144 miles. The latter case is excluded, as the problem states that the trains move at speeds "not too different from the usual." (If we had taken the distance to be 144 miles, then \(A\) would have traveled 140 miles in the same time that \... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,984 |
69. Faulty Locomotive. We set off by railway from Angchester to Clinkerton. But an hour after the train started, a locomotive malfunction was discovered. We had to continue the journey at a speed that was $\frac{3}{5}$ of the original. As a result, we arrived in Clinkerton 2 hours late, and the driver said that if the ... | 69. The distance from Angchester to Clinkerton is 200 miles. The train traveled 50 miles at a speed of 50 miles/h and 150 miles at a speed of 30 miles/h. If the breakdown had occurred 50 miles further, the train would have traveled 100 miles at a speed of 50 miles/h and 100 miles at a speed of 30 miles/h. | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,985 |
70. The Runner's Puzzle. Two men are running around a circle in opposite directions. Brown, the better runner, gave Tomkins a start of $\frac{1}{8}$ of the distance, but overestimated his own strength: after running $\frac{1}{6}$ of the distance, he met Tomkins and realized that his own chances of success were very sli... | 70. When Brown had left behind only \(\frac{1}{6}\), or \(\frac{4}{24}\), of the entire distance, Tomkins had already traveled \(\frac{5}{6}\) minus \(\frac{1}{8}\), or \(\frac{17}{24}\), of the entire distance. Therefore, Tomkins' speed is \(\frac{17}{4}\) times greater than Brown's speed. Brown had \(\frac{5}{6}\) of... | 20\frac{1}{4} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,986 |
71. Two Ships. Two ships set out from one port to another, located 200 nautical miles away, and return back. The "Mary Jane" travels in one direction at a speed of 12 miles/h and on the return trip at a speed of 8 miles/h, spending a total of $41 \frac{2}{3}$ hours for the entire journey. The "Elizabeth Anne" travels a... | 71. The statement about the equality of average speeds is incorrect. In reality, the average speeds of the ships are not equal. The first ship takes \(\frac{1}{12}\) hours to travel one mile in one direction and \(\frac{1}{8}\) hours in the opposite direction. The half-sum of these fractions is \(\frac{5}{48}\). Theref... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,987 |
72. Determine the distance. Jones set out from $A$ to $B$ and on the way, 10 km from $A$, he met his friend Kenward, who had set out from $B$ at the same time as him. Upon reaching $B$, Jones immediately turned back. Kenward did the same upon reaching $A$. The friends met again, but this time 12 km from $B$. Of course,... | 72. The distance between two points is 18 km. The meeting points are 10 km and 12 km away from \(A\) and \(B\) respectively. Multiply 10 (the first distance) by 3 and subtract the second distance -12. What could be simpler? Try other distances to the meeting points (making sure the first distance is more than \(\frac{2... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,988 |
## 73. Man and Dog.
- Walking the dog, - a mathematician friend once told me, - gives me plenty of food for thought. Once, for example, my dog, after waiting for me to go out, looked to see which way I was going to head, and when I started down the path, he raced to the end of it. Then he returned to me, ran to the en... | 73. The dog ran at a speed of 16 km/h. The key to solving the problem lies in the following considerations. The distance the man had left to walk alongside the dog was 81 m, or \(3^{4}\) (the dog returned 4 times), and the length of the path was \(625 \mathrm{m}\), or \(5^{4}\). Therefore, the difference in speeds (exp... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,989 |
74. Baxter's Dog. Here is an interesting puzzle, complementing the previous one. Anderson left the hotel in San Remo at 9 o'clock and was on the road for a whole hour when Baxter set out after him along the same route. Baxter's dog ran out at the same time as its owner and kept running back and forth between him and An... | 74. It is quite obvious that Baxter will catch up with Anderson in one hour, by which time they will have each traveled 4 km in the same direction. Furthermore, the dog’s speed is 10 km/h; therefore, in this hour, it will have run 10 km! When this puzzle was presented to a French mathematics professor, he exclaimed, “M... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,990 |
75. Исследование пустыни. Девять участников экспедиции (каждый на автомашине) встречаются на восточной окраине пустыни. Они хотят исследовать ее внутренние районы, двигаясь все время на запад. Каждому автомобилю полного бака (содержащего 1 галлон бензина) хватает на 40 миль пути. Кроме того, он может взять с собой еще ... | 75. Девять исследователей \(A, B, C, D, E, F, G, H, J\) проезжают 40 миль, затратив на это по полному баку горючего. Затем \(A\) передает по 1 галлону остальным восьми участникам и поворачивает назад, причем у него остается 1 галлон на обратную дорогу. Остальные восемь участников едут еще 40 миль, затем \(B\) передает ... | 360 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,991 |
76. Mountain Exploration. Professor Walkinhom, a participant in the expedition, was given the task to survey the mountain from all sides at a certain height. He has to walk 100 miles around the mountain alone. The professor is capable of walking 20 miles a day, but he can only carry enough food for two days. For conven... | 76. Walkingholme deposits 5 rations at the 90-mile mark (see figure) and returns to base (5 days). Then he leaves 1 ration at the 85-mile mark and returns to the 90-mile mark (1 day). He leaves 1 ration at the 80-mile mark and returns to the 90-mile mark again (1 day). He transports 1 ration to the 80-mile mark, return... | 23.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,992 |
78. The Walking Passenger. The train is moving at a speed of 60 km/h. A passenger is walking from the end of the train to its beginning through the passageways between the cars at a speed of 3 km/h. What is the speed of the passenger relative to the railway track?
We are not going to engage in sophistry here, like Zen... | 78. Suppose a train travels for an hour and has an incredible length of 3 km. Then (see figure) in this time it will travel from \(B\) to \(C\) 60 km, while the passenger will move from \(A\) to \(C\), or 63 km. C
 of the distance, a ne... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,995 |
80. Two suitcases. A gentleman needed to get to a railway station located 4 km from his house. His luggage consisted of two equally heavy suitcases, which he could not carry alone. The gardener and the servant insisted that they should be entrusted with carrying the luggage. However, the gardener was too old, and the s... | 80. The servant must carry a suitcase \(1 \frac{1}{3}\) km and hand it over to the gentleman, who will carry the suitcase to the station. The gardener must carry another suitcase \(2 \frac{2}{3}\) km, and then hand it over to the servant, who will carry the suitcase to the station. Thus, each of them will carry one sui... | 2\frac{2}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,996 |
## 81. Escalator.
- Descending on the escalator, I counted 50 steps, - said Walker.
- I counted 75, - objected Trotman, - but I was going down three times faster than you.
If the escalator had stopped, how many steps could be counted on its visible part? It is assumed that both people moved at a constant speed and th... | 81. Let \(n\) be the number of steps on the escalator; the time it takes for one step to disappear at the bottom, we will take as a unit.
Trotman walks 75 steps in \(n-75\) units of time, or at a speed of 3 steps in \(\frac{n-75}{25}\) units of time. Therefore, Walker covers 1 step in \(\frac{n-75}{25}\) units of time... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,997 |
82. The Cart. "Three men," said Crackham, "Atkins, Brown, and Cranby, decided to go on a short trip. They have a distance of 40 km to cover. Atkins walks at a speed of 1 km/h, Brown at 2 km/h, and Cranby, with his cart pulled by a donkey, travels at 8 km/h. For some time, Cranby carries Atkins, then drops him off to wa... | 82. The journey lasted \(10 \frac{5}{41}\) hours. Atkins walked \(5 \frac{35}{41}\) km; Brown - \(13 \frac{27}{41}\) km, and Cranby's donkey ran a total of \(80 \frac{40}{41}\) km. I hope the donkey was given a good rest after such a feat. | 10\frac{5}{41} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,998 |
83. Four cyclists. Four identical circles represent four gravel tracks. Four cyclists start from the center at noon. Each moves along their own circle at speeds: the first - 6 km/h, the second - 9, the third - 12, and the fourth - 15 km/h. They agreed to ride until they all meet again at the center for the fourth time.... | 83. Cyclists \(A, B, C, D\) can ride one kilometer in \(\frac{1}{6}, \frac{1}{9}, \frac{1}{12}\) and \(\frac{1}{16}\) hours, respectively. Therefore, they complete a full lap in \(\frac{1}{18}, \frac{1}{27}\), and \(\frac{1}{45}\) hours and, thus, meet for the first time after \(\frac{1}{9}\) hours (or, equivalently, a... | 12:26:40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,999 |
84. Three cars. Three friends are driving in cars along a road in the same direction and at some point in time are positioned relative to each other as follows. Andrews is at some distance behind Brooks, and Carter is at a distance twice that from Andrews to Brooks, ahead of Brooks. Each driver is traveling at a consta... | 84. Brooks will catch up to Carter in \(6 \frac{2}{3}\) min. | 6\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,000 |
85. The Fly and the Cars. The length of the road is 300 km. Car $A$ starts at one end of the road at noon and moves at a constant speed of 50 km/h. At the same time, at the other end of the road, car $B$ starts at a constant speed of 100 km/h and a fly, flying at 150 km/h. Upon meeting car $A$, the fly turns and flies ... | 85. 1) The fly will meet \(B\) in 1 hour 48 minutes.
2) There is no need to determine the distance the fly will fly. This is too difficult a task. Instead, we can simply find the time when the cars could have collided, which is 2 hours. In reality, the fly flies (in kilometers):
\[
\frac{270}{1}+\frac{270}{10}+\frac{... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,001 |
86. Metro Stairs. Once, while exiting the "Kerli-street" metro station, we ran into a young athlete named Percy Longman. He stopped on the escalator and said:
- I always walk up the escalator. You know, extra training never hurts. This escalator is the longest on the line - almost a thousand steps. But here’s what’s i... | 86. The least common multiple of the numbers \(2,3,4,5,6\) and 7 is 420. Subtracting 1 from it, we get 419 - a possible number of steps. In addition, the conditions of the problem will be satisfied by numbers obtained by sequentially adding multiples of 420 to 419. Therefore, the number of steps in the escalator can be... | 839 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,002 |
87. Bus Ride. George went on a bus ride with his girlfriend, but after counting his limited resources, he realized they would have to walk back.
If the bus speed is 9 km/h, and the couple walks at 3 km/h, then how far can they ride so that the entire trip there and back takes 8 hours? | 87. Young people travel three times faster by bus than on foot; therefore, \(\frac{3}{4}\) of the total time they need to spend on the return trip and only \(\frac{1}{4}\) of the time to travel by bus. Thus, they will travel for 2 hours, covering a distance of 18 km, and walk for 6 hours. They will return exactly 8 hou... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,003 |
88. Transportation Puzzle. Twelve soldiers need to get to a point 20 km away from their location as quickly as possible. To do this, they stopped a small car.
- I drive at a speed of 20 km/h, - the driver said, - but I can only take four of you at a time. How fast do you walk?
- Each of us walks 4 km/h, - one of the s... | 88. The driver must transport four soldiers 12 km and drop them off 8 km from the destination. Then he must return 8 km and pick up another four soldiers (out of eight) who will be there by then, transport them 12 km and drop them off 4 km from the destination. Returning 8 km to pick up the remaining soldiers, who by t... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,004 |
93. A small train puzzle. The express from Basltown to Ironchester travels at a speed of 60 km/h, and the express from Ironchester to Basltown, which departs simultaneously with it, travels at a speed of 40 km/h.
How far apart will they be one hour before meeting?
I couldn't find these cities on any map or in any ref... | 93. To solve the problem, no algebraic calculations are required, nor do you need to know the distance between the cities. Let's send both trains back from the meeting point, wherever it may have occurred, at the same speeds. Then in one hour, the first train will travel 60 km, and the second 40 km. Therefore, the dist... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,009 |
95. The Pedestrian Problem. A man, walking out of town, looked back and noticed a friend who was walking in the same direction but 400 m behind him. Looking at each other, the friends walked straight ahead for another 200 m each. You might think they should have met, but no, there was still a distance of $400 \mathrm{m... | 95. The second man, seeing that his friend had turned around and was walking towards him, started to back away and thus walked 200 meters backwards. Of course, his behavior was rather eccentric, but he did act this way, and this is the only answer to the problem. As a result, the friends were able to move in a straight... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,011 |
96. Incorrect scales. When the pudding was placed on one pan of the scales, they showed 4 g more than $\frac{9}{11}$ of its true weight. When it was placed on the other pan, the scales showed 48 g more than in the first case. What is the true weight of the pudding?
 is 154. However, from the diagram in the problem, it is clear that... | 143 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,012 |
## 98. Weighing the Child.
- Last summer I witnessed a funny incident at a railway station, said my friend. - A small family stood in front of the automatic scales, which were calibrated for 200 pounds, trying in vain to solve the difficult problem of weighing the child. As soon as the parents left the child alone on ... | 98. It is important to note that the father, child, and dog together weighed 180 pounds, as shown in the figure. Further, the difference between 180 and 162 is 18, which matches twice the weight of the dog. Therefore, the dog weighs 9 pounds, and the child 30 pounds, since if you subtract \(70\%\) of this weight from 3... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,014 |
99. Fruits for Jam. For making jam, it was necessary to weigh fresh fruits. It turned out that apples, pears, and plums balance each other as shown in the figure.
Could you tell how many p... | 99. On the first scales, we see that an apple and 6 plums weigh the same as a pear, so on the second scales, we can replace the pear with an apple and 6 plums without disrupting the balance. Then we can remove 6 plums from each side and find that 4 apples weigh the same as 4 plums. Therefore, one apple weighs the same ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,015 |
100. Weighing Tea. A grocer needed to package 20 pounds of Chinese tea into two-pound bags, but his weights had gone missing. After a fruitless search, he found only a five-pound and a nine-pound weight.
How can the grocer complete his work as quickly as possible? Let's say right away that only 9 weighings are require... | 100. 1. By placing weights of 5 and 9 pounds on different scales, measure out 4 pounds. 2. Using the 4 pounds, measure out another 4 pounds. 3. Measure out 4 pounds for the third time. 4. Measure out 4 pounds for the fourth time, with the remainder also equaling 4 pounds. 5.-9. Using the scales, divide each portion of ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,016 |
101. A special number. What number is formed from five consecutive digits (not necessarily in order) such that the number formed by the first two digits, multiplied by the middle digit, gives the number formed by the last two digits. (For example, if we take the number 12896, then 12, multiplied by 8, gives 96. However... | 101. The desired number is 13452. The consecutive digits are \(1,2,3,4,5 ; 13 \times 4=52\).
[For a six-digit number, V. Milli proposed the following solution: \(947658.94 \times 7=658 .-M . G\). | 13452 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,017 |
102. Five cards. I have five cards with the digits $1,3,5,7$ and 9 on them. How can I arrange them in a row so that the product of the number formed by the first pair of cards and the number formed by the last pair of cards, minus the number on the middle card, equals a number composed of repetitions of the same digit?... | 102. The solutions are the numbers 39157 and 57139. In each case, the product of the numbers 39 and 57 minus 1 equals 2222. | 3915757139 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,018 |
103. Digits and Squares. What is the smallest square of an integer that ends with the longest sequence of identical digits?
If the longest sequence of identical digits were five, for example, the number 24677777 would fit (if it were the smallest square, but this is not true). Zero is not considered a valid digit. | 103. If the square of an integer ends in repeating digits, then these digits can only be 4, as in the case of \(144=12^{2}\). However, the number of such repeating digits cannot exceed three; therefore, the answer is the number \(1444=38^{2}\). | 1444 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,019 |
105. Repeating quartet of digits. If we multiply 64253 by 365, we get 23452345, where the first four digits repeat.
What is the largest number by which 365 must be multiplied to obtain a similar product containing eight digits, the first four of which repeat? | 105. Multiplying 273863 by 365, we get 99959 995. Notice that any eight-digit number, where the first four digits are repeated, is divisible by 73 (and by 137). Moreover, if such a number ends in 5 or 0, it is also divisible by 365 (or by 50005). Knowing these facts, we can immediately write down the answer. | 273863 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,021 |
106. Easy Division. By dividing the number 8101265822784 by 8, you will see that the answer can be obtained simply by moving the 8 from the beginning to the end of the number!
Could you find a number starting with 7 that can be divided by 7 in such a simple way? | 106. Let's divide 7101449275362318840579 by 7 using the "corner" method, as we were taught in school. When dividing 7 by 7, we get 1, the next digit 1 gives 0 in the quotient, then again 1, and so on, until we reach the end. Checking the quotient against the dividend, we see that it indeed results from moving the first... | 7101449275362318840579 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,022 |
107. Misunderstanding. An American reader asked me to find a number composed of any number of digits, for which division by 2 can be performed by moving the last digit to the front. Apparently, this problem arose for him after he became acquainted with a previously misstated problem. If it were required to move the fir... | 107. We can divide 857142 by 3 simply by moving the 2 from the end to the beginning, or divide 428571 by moving the 1. | 428571 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,023 |
108. Two Fours. I am constantly asked about the old puzzle "Four Fours." I published it in 1899, but later found out that it was first published in the first volume of the magazine Knowledge in 1881. Since then, various authors have referred to it. The puzzle is formulated as follows: "Find all possible numbers that ca... | 108. Here is how the number 64 can be expressed using two fours and arithmetic signs:
\[
\sqrt{(\sqrt{\sqrt{4}})^{4!}}=\sqrt{(\sqrt{2})^{24}}=\sqrt{2^{12}}=\sqrt{4096}=64
\]
[Interest in the "Four Fours" problem has periodically revived since its publication. I wrote about a relatively recent discussion of this probl... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,024 |
109. Two digits. Write any two-digit number (two different non-zero digits), and then express it using the same digits in reverse order (arithmetic operation symbols are allowed if necessary). For example, the number $45=5 \times 9$ would have worked if 9 on the right were 4, and the number $81=(1+8)^{2}$ could have be... | 109. What symbols to allow is a matter of taste, but I would personally prefer to do without any logs. Here are a few solutions:
\[
\begin{aligned}
25 & =5^{2} \\
36 & =6 \times 3! \\
64=\sqrt{4^{6}}, & \text { or }(\sqrt{4})^{6}
\end{aligned}
\] | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,025 |
110. Digital Coincidences. If I multiply two nines and add 9 and 9, I get 81 and 18 - two numbers consisting of the same digits. If I multiply and add 2 and 47, I get 94 and 49 - numbers with the same digits. If I multiply and add 3 and 24, I get 72 and 27 - two numbers consisting of the same digits.
Can you find two ... | 110. If we multiply 497 by 2, we get 994. If we add these two numbers, we get 499. The digits in both cases are the same. A similar result holds for 263 and 2. We get 526 and 265, respectively.
[G. Lindgren points out that by introducing nines after the first digit, one can obtain two answers with any desired number o... | 497\times2=994497+2=499;263\times2=526263+2=265 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,026 |
111. Palindromic Squares. Here is an interesting subject for investigation: to find squares of integers that can be read the same way forwards and backwards. Some of them are very easy to find. For example, the squares of the numbers $1, 11, 111$, and 1111 are $1, 121, 12321$, and 1234321, respectively. All the resulti... | 111. The square of the number 836, equal to 698896, contains an even number of digits, and it can be read the same way both from left to right and from right to left. Among all squares containing this even number of digits, the palindromic square is the smallest. | 698896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,027 |
112. Factorization. Into what factors does the number 1000000000001 decompose? This question is easy to answer if you know something about numbers of this particular form. It is no less easy to specify two factors if, instead of 11 zeros, you insert, for example, 101 zeros between the two ones.
There is one curious, s... | 112. If the number of zeros enclosed between two ones is equal to any number that is a multiple of 3 plus 2, then the two factors can always be written down immediately using the following curious rule: \(1001=11 \times 91 ; 1000001=101 \times 9901 ; 1000000001=\) \(1001 \times 999001 ; 1000000000001=10001 \times 99990... | 10001\times99990001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,028 |
113. Two factors. Find two integers, the difference between which is minimal, and their product is 1234567890. | 113. The number 1234567890 is factored as follows: \(2 \times 3 \times 3 \times 5 \times 3607 \times 3803\). If we multiply 3607 by 10 and 3803 by 9, we get two composite factors: 36070 and 34227, which when multiplied together give 1234567890 and have the smallest difference. | 3607034227 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,029 |
114. Division by 11. If the nine digits $1,2,3,4,5,6,7,8,9$ are written in a random order, for example 412539768, what is the probability that the resulting number is divisible by 11? The number I wrote, of course, is not divisible by 11, but if you swap the 1 and 8 in it, it will be divisible by 11. | 114. For a number to be divisible by 11, it is necessary that either four alternating digits sum to 17, and the remaining five sum to 28, or vice versa. For example, in the given number (482539761), the digits \(4,2,3\), 7,1 sum to 17, and \(8,5,9,6\) sum to 28. Furthermore, four digits can sum to 17 in nine different ... | \frac{115}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,030 |
115. Division by 37. I would like to know if the number 49129308213 is divisible by 37, and if not, what the remainder is. How can I do this without performing the division? It turns out that with a skilled approach, the answer to my question can be obtained in a few seconds. | 115. Let's write the numbers \(1,10,11\) from right to left under our number, as shown below:
\[
\begin{array}{rrrrrrrrrrr}
4 & 9 & 1 & 2 & 9 & 3 & 0 & 8 & 2 & 1 & 3 \\
10 & 1 & 11 & 10 & 1 & 11 & 10 & 1 & 11 & 10 & 1
\end{array}
\]
Now, let's multiply the numbers 1 and 10 at the bottom by the numbers above them and ... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,031 |
116. Once again about division by 37. Here is an interesting development of the previous puzzle. Nine digits $1,2,3,4,5,6,7,8,9$ are written in a random order, for example 412539768. What is the probability that the resulting number is divisible by 37 without a remainder? | 116. Let's denote our number as \(A B C A B C A B C\). If the sums of the digits denoted by the letters \(A, B\), and \(C\) are respectively:
\begin{tabular}{rrr}
A & B & C \\
18 & 19 & 8 \\
15 & 15 & 15 \\
12 & 11 & 22 \\
19 & 8 & 18 \\
22 & 12 & 11 \\
8 & 18 & 19 \\
11 & 22 & 12
\end{tabular}
then in the first thr... | \frac{1}{40} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,032 |
117. The task of ten digits. Arrange all ten digits $1,2,3,4,5,6,7,8,9,0$ in such an order that the resulting number is divisible by all numbers from 2 to 18. For example, if the digits are placed in the sequence 1274953680, the resulting number will be divisible by $2,3,4,5$ and so on up to 16, but will not be divisib... | 117. There are four solutions: 2438195760, 3785942160, 4753869120, 4876391520. The last digit must be zero. With any arrangement of the digits with an even digit before the zero, the number is divisible by \(2, 3, 4, 5, 6, 9, 10, 12, 15\), and 18. It remains to consider only \(7, 11, 13, 16\), and 17. (Divisibility by ... | 2438195760,3785942160,4753869120,4876391520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,033 |
118. Threes and sevens. What is the smallest number that is written using only the digits 3 and 7 and that both the number itself and the sum of its digits are divisible by 3 and 7? For example, 7733733 is divisible by 3 and 7 without a remainder, but the sum of its digits (33) is divisible by 3, not by 7, so it cannot... | 118. The smallest possible number is 3333377733. It is divisible by 3 and 7, and the sum of its digits (42) has the same property. The number must contain at least 3 sevens and 7 threes, with the sevens being moved as far to the right as possible. | 3333377733 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,034 |
119. Extracting Roots. Once, in a conversation with Professor Simon Greathead, a man of rather eccentric mind, I mentioned the extraction of cube roots.
- Amazing, - said the professor, - what ignorance people show in such a simple matter! It seems that in the extraction of roots, since the only roots were those extra... | 119. The sought numbers are 5832, 17576, and 19683. The sum of the digits of each of them, equal to 18, 26, and 27, respectively, coincides with the corresponding cubic root. | 5832,17576,19683 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,035 |
122. Digits and Cubes. Professor Rackbrain recently asked his young friends to find all five-digit squares for which the sum of the numbers formed by the first two and the last two digits is a perfect cube. For example, if we take the square of 141, which is 19881, and add 81 to 19, we get 100 - a number that, unfortun... | 122. There are three solutions: \(56169\left(237^{2}\right)\), where \(56+69=125\) \(\left(5^{3}\right) ; 63001\left(251^{2}\right)\), where \(63+01=64\left(4^{3}\right)\) and \(23104\left(152^{2}\right)\), where \(23+04=\) \(27\left(3^{3}\right)\). | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,038 |
123. In reverse order. What nine-digit number, when multiplied by 123456 789, gives a product where the nine least significant digits are 9,8,7,6,5,4,3, 2,1 (in that exact order)? | 123. The product of the numbers 989010989 and 123456789 is 122100120987654321, which is what we needed to find. | 989010989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,039 |
124. Progression. "If from nine digits, - said Professor Rackbrain, - you form three numbers $147, 258, 369$, you will find that each subsequent number differs from the previous one by 111 and that, consequently, an arithmetic progression has been formed."
Could you rearrange the nine digits in four different ways so ... | 124. The professor's answer was:
\begin{tabular}{lll}
297 & 564 & 831 \\
291 & 564 & 837 \\
237 & 564 & 891 \\
231 & 564 & 897
\end{tabular}
where the common difference of the progression is respectively \(267, 273, 327\), and 333. He pointed out that for each of the six permutations of the middle three digits, a cor... | 297,564,831;291,564,837;237,564,891;231,564,897 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,040 |
125. Formation of integers. Can the reader name the sum of all integers formed from the four digits $1,2,3,4$? In other words, it is required to calculate the sum of such numbers as $1234,1423,4312$, etc. Of course, one could write down all such numbers in a row and then add them up. However, it is more interesting to ... | 125. If you multiply 6666 by the sum of four given digits, you will get the correct answer. Since \(1,2,3,4\) add up to 10,
multiplying 6666 by 10 gives us the answer 66660. If we seek the sum of all combinations of four different digits, we get 16798320, or \(6666 \times 2520\). | 66660 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,041 |
126. Summation of numbers. Professor Rackbrain would like to know the sum of all numbers that can be formed from nine digits (0 is excluded), using each digit in each number only once.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 126. This problem can be solved in several ways. The answer, of course, is the same in all cases, equal to 201599999798400. The sum of nine digits is 45 and
\[
45 \times 8!=1814400
\]
Writing further
\[
18144
\]
18144
18144
18144 \( \qquad \)
nine times, adding them up and appending 00 at the end, we get the answ... | 201599999798400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 41,042 |
128. Digits and Squares. One of Professor Rackbrain's small Christmas puzzles reads as follows: what are the smallest and largest squares containing all ten digits from 0 to 9, with each digit appearing only once? | 128. The smallest square is 1026753849 ( \(32043^{2}\) ); the largest - \(9814072356\left(99066^{2}\right)\). | 10267538499814072356 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,044 |
129. Digital Squares. A very good puzzle is to find a number which, together with its square, contains each of the nine digits, excluding zero, exactly once. Thus, if the square of the number 378 were 152694, it would fit our criteria. But in fact, its square is 142884, which gives us two fours and three eights, while ... | 129. The problem has only two solutions: the numbers 567 (\(567^{2}=\) \(321489)\) and 854 (\(854^{2}=729316\)). When searching for a solution, only three-digit numbers should be considered, the sum of whose digits is 9, 18, and 27 or 8, 17, and 26. The smallest three-digit number whose square is a six-digit number is ... | 567^2=321489854^2=729316 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,045 |
130. Finding the Square. Six numbers are given: 4784887, 2494651, 8595087, 1385287, 9042451, 9406087. It is known that the sum of three of them is a perfect square. Which are these numbers?
The reader will probably not see any other way but the tedious method of trial and error, and yet there is a direct solution to t... | 130. The sums of the digits of the six given numbers are respectively
\[
\begin{array}{rrrrrr}
46 & 31 & 42 & 34 & 25 & 34 \\
1 & 4 & 6 & 7 & 7 & 7
\end{array}
\]
By repeatedly summing the digits of these sums, we eventually obtain single-digit numbers, which are listed in the second row.
These single-digit numbers ... | 2494651+1385287+9406087=13286025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,046 |
131. Juggling with numbers. Compose three simple arithmetic expressions from ten digits, using three of the four arithmetic operations - addition, subtraction, multiplication, and division. (In writing the expressions, only the symbols of the three chosen arithmetic operations are allowed.) Let us clarify this with an ... | 131. \(7+1=8 ; 9-6=3 ; 4 \times 5=20\). | 7+1=8;9-6=3;4\times5=20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,047 |
132. Equal Fractions. Can you form three ordinary fractions (say, something like $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ or $\frac{1}{9}$), using each of the nine digits exactly once? The fractions can be formed in one of the following ways:
$$
\text { either } \frac{a}{b}=\frac{c}{d}=\frac{e f}{g h j}, \quad \text { ... | 132. Let's provide five solutions to the problem:
\[
\begin{gathered}
\frac{2}{4}=\frac{3}{6}=\frac{79}{158} ; \quad \frac{3}{6}=\frac{7}{14}=\frac{29}{58} ; \quad \frac{3}{6}=\frac{9}{18}=\frac{27}{54} \\
\frac{2}{6}=\frac{3}{9}=\frac{58}{174} ; \quad \frac{.2}{1}=\frac{.6}{3}=\frac{97}{485}^{*}
\end{gathered}
\] | \frac{0.2}{1}=\frac{0.6}{3}=\frac{97}{485} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,048 |
133. Digits and Prime Numbers. Using each of the nine digits once and only once, form prime numbers (numbers that do not divide evenly by any integer except 1 and themselves) whose sum is the smallest.
For example, four prime numbers
61
+283
47
59
450
contain all nine digits once and only once, and their sum is ... | 133. The digits 4, 6, and 8 must be in the second place, since no prime number can end in these digits. The digits 2 and 5 can appear in the units place only if the prime number is a single digit, meaning there are no other digits. After this, the solution can be completed without much difficulty:
\[
\begin{array}{r}
... | 207 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,049 |
134. Once again about digital squares. From nine digits, in many different ways, one can form a square such that the numbers in the first and second rows, when added together, give the third row.
\footnotetext{
\({ }^{*}\) See note on page 33.
} | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,050 |
135. Nine digits. If 32547891 is multiplied by 6, using each of the nine digits once and only once, the product is 195287346 (also containing nine digits, each used once and only once). Could you find another number that has the same property when multiplied by 6? Remember that each of the nine digits must appear once ... | 135. The number 94857312, when multiplied by 6, gives 569143872, and all nine digits in each case are used once and only once.
[Two other solutions are known: \(89745321 \times 6=538471926\) and \(98745231 \times 6=592471386 .-M . \Gamma\). | 94857312\times6=569143872 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,051 |
136. Twenty-four. In one book it was written: “Write the number 24 using three identical digits, different from 8. (There are two solutions to this problem.)”
The book also provided two answers: $22+2=24$ and $3^{3}-3=24$. Readers familiar with the old puzzle “Four Fours” and similar puzzles might ask why there are on... | 136. It is not difficult to represent the number 24 using three fours, fives, eights, or nines:
\[
\begin{gathered}
(4+4-4)! \\
\left(5-\frac{5}{5}\right)! \\
\left(\sqrt{9}+\frac{9}{9}\right)!
\end{gathered}
\]
The number 24 can also be represented using three ones, sixes, and sevens. Indeed,
\[
\begin{gathered}
\l... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,052 |
137. Nine Barrels. In how many ways can nine barrels be arranged in three tiers so that the numbers written on the barrels to the right of any barrel or below it are greater than the number written on the barrel itself?
, written in decimal form. | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,057 |
142. Sum of Cubes. The numbers 407 and 370 match the sum of the cubes of their digits. For example, the cube of 4 is 64, the cube of 0 is 0, and the cube of 7 is 343. Adding 64, 0, and 343 gives 407. Similarly, the cube of 3 (27) added to the cube of 7 (343) results in 370.
Could you find a number, not containing zero... | 142. The desired number is 153. The cubes of the numbers 1, 5, and 3 are 1, 125, and 27, respectively, and their sum is 153.
[The author did not notice the fourth number: 371. If 1 is not counted, then 407, 370, 153, and 371 are the only four numbers that equal the sum of the cubes of their digits. For a more general ... | 153 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,058 |
152. Numbers instead of letters. One morning, Professor Rackbrain proposed the following rather difficult problem to his young friends. He wrote down the letters of the alphabet in the following order:
$$
A B C D \times E F G H I=A C G E F H I B D
$$
- Each letter, he said, represents its own digit from 1 to 9 (0 is ... | 152. \(6543 \times 98271=642987153\).
152. \(6543 \times 98271=642987153\).
(Note: The content is a mathematical equation, which is the same in both Chinese and English, so the translation is identical to the original text.) | 6543\times98271=642987153 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,068 |
153. The Merchant's Secret. A merchant, wishing to keep his accounts secret, chose a ten-letter word (all different) like ZACHERKNUTY, where each letter corresponds to a digit in the following order: $1,2,3,4,5,6,7,8,9,0$. For example, in the case of the given keyword, ZA means 12, CHER - 345, and so on. If the sum
$$... | 153. The only word (not a meaningless set of letters) that meets the given conditions is LAMP. The sum is deciphered as follows:
\[
\begin{array}{r}
36407 \\
+\quad 98521 \\
\hline 134928
\end{array}
\] | 134928 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,069 |
154. "Bee's wax". In a certain secret code, the word BEESWAX* represents a number. The police could not find the key to this code until they discovered the following note among the papers
$$
\begin{aligned}
& E A S E B S B S X \\
& B P W W K S E T Q \\
& \hline K P E P W E K K Q
\end{aligned}
$$
The detectives suspec... | 154. The key to the code has the form
\[
\begin{array}{cccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 \\
A & T & Q & B & K & X & S & W & E & P
\end{array}
\]
from which we get
\[
\begin{array}{r}
-917947476 \\
-408857923 \\
\hline 509089553
\end{array}
\]
and BEESWAX stands for the number 4997816. | 4997816 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,070 |
## 155. From "Wrong" to "Right".
- You can't make a "Right" out of two "Wrongs," someone said at breakfast.
- I'm not so sure about that,- countered Colonel Crackham - Here's an example (each letter represents a different digit, and all the encoded digits are non-zero):
$$
\begin{aligned}
& W R O N G \\
& \frac{W R O... | 155.
\[
\begin{array}{r}
+25938 \\
+25938 \\
\hline 51876
\end{array}
\] | 51876 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,071 |
156. Умножение букв. В этом маленьком примере на умножение пять букв соответствуют пяти различным цифрам. Каким именно? Среди цифр нет нуля.
$$
\begin{gathered}
\times S E A M \\
T M E A T S
\end{gathered}
$$ | 156.
\[
\begin{array}{r}
4973 \\
\times \quad 8 \\
\hline 39784
\end{array}
\] | 39784 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,072 |
157. Secret Code. Two conspirators had a secret code. Sometimes their correspondence contained simple arithmetic operations that looked completely innocent. However, in the code, each of the ten digits represented its own letter of the alphabet. Once, a sum was encountered which, after substituting the corresponding le... | 157.
598
\[
+507
\]
\[
\begin{array}{r}
8047 \\
\hline 9152
\end{array}
\] | 9152 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,073 |
158. Alphanumeric puzzle. This puzzle is not difficult to solve with the right approach:
$$
\begin{array}{llll}
A \times B=B, & B \times C=AC, & C \times D=BC, & D \times E=CH \\
E \times F=DK, & F \times H=CJ, & H \times J=KJ, & J \times K=E \\
& K \times L=L, & A \times L=L .
\end{array}
$$[^8]
Each letter represen... | 158. It is clear that \(A\) is 1, and \(B\) and \(C\) represent either 6 and 2, or 3 and 5. From the third equation, it is evident that they are 3 and 5, since \(D\) must equal 7. The letter \(E\) is 8, because in the product \(D \times E\) the value \(C=5\) appears. The rest is quite easy to complete, and we get the f... | A=1,B=3,C=5,D=7,E=8,F=9,H=6,J=4,K=2,L=0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,074 |
159. The Miller's Fee. Here is a very simple puzzle, although I have met people who pondered over it for several minutes.
A miller took $\frac{1}{10}$ of the total flour as payment for grinding. How much flour was obtained from the peasant's grain if he was left with one sack after paying the miller? | 159. From the peasant's grain, there should have been \(1 \frac{1}{9}\) bags of flour, which after paying \(\frac{1}{10}\) of all the flour would exactly amount to one bag. | 1\frac{1}{9} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,075 |
160. Hens and Eggs. Here is a new version of an old problem. Although it looks very complicated and confusing, it is extremely easy to solve with the right approach.
If one and a half hens lay one and a half eggs in one and a half days, then how many hens plus half a hen, laying eggs one and a half times faster, will ... | 160. Answer to the problem: half a chicken plus half a chicken, which is one chicken. If one and a half chickens lay one and a half eggs in one and a half days, then one chicken lays one egg in one and a half days. A chicken that lays better by one and a half times lays one and a half eggs in one and a half days, or on... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,076 |
162. Egg Sales. A woman brought eggs to the market and sold some of them. The next day, her hens worked hard, doubling the remaining eggs, and the owner sold as many as she did the previous day. On the third day, the new remainder was tripled, and the woman sold as many eggs as she did on the previous days. On the four... | 162. The smallest possible number of eggs is 103, and the woman sold 60 eggs daily. Any multiples of these numbers can be used as answers to the problem. For example, the woman could have brought 206 eggs and sold 120 daily, or brought 309 eggs and sold 180. Since the smallest number was required, the answer is unique. | 103 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,078 |
## 163. The Cat and the Mouse.
- One of these barrels has a mouse in it,- said the dog.
- Which one? - asked the cat.
- Well, the five hundredth.
- What do you mean by that? There are only five barrels here.
- The barrel I mean will be the five hundredth if you start counting forward and backward like this.
And the d... | 163. You just need to divide the given number by 8. If it divides evenly, with no remainder, the mouse is in the second barrel. If the remainder is 1, 2, 3, 4, or 5, the barrel number will match this remainder. If the remainder is greater than 5, subtract it from 10. The resulting difference is the barrel number. The n... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,079 |
164. Army formation. An army formation, consisting of a little more than 20 thousand people, includes 5 brigades. It is known that $\frac{1}{3}$ of the first brigade, $\frac{2}{7}$ of the second, $\frac{7}{12}$ of the third, $\frac{9}{13}$ of the fourth, and $\frac{15}{22}$ of the fifth brigade have equal numbers.
How... | 164. Five brigades consist of 5670, 6615, 3240, 2730, and 2772 people respectively. After converting all fractions to a common denominator (12012), the numerators will be 4004, 3432, 7007, 8316, and 8190 respectively. By combining all the different divisors contained in these numbers, we get 7567560, which when divided... | 21027 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,080 |
165. The Decisive Vote. The Congress of the United Society of Wandering Beggars (better known as the Union of Vagrants) convened to decide whether to declare a strike, demanding a shorter working day and increased alms. It was decided that during the vote, those members of the society who would vote in favor of the str... | 165. A total of 207 people voted. Initially, 115 voters were in favor and 92 against, with a majority of 23 votes, which is exactly one quarter of 92. But when 12 people, for whom there were no seats, joined the opposition, it turned out that 103 votes were in favor and 104 against. Thus, the opponents of the strike wo... | 207 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,081 |
166. Three Brothers. The military authorities had to decide which of the three sons of a certain merchant should be exempted from military service.
- I will tell you what they are capable of,-the father said.-Arthur and Benjamin can do the same work in 8 days that Arthur and Charles will take 9 days, and Benjamin and ... | 166. Arthur can complete the entire work in \(14 \frac{34}{49}\), Benjamin in \(17 \frac{23}{41}\), and Charles in \(23 \frac{7}{31}\) days. | Arthur:14\frac{34}{49},Benjamin:17\frac{23}{41},Charles:23\frac{7}{31} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,082 |
167. House Number. One person said that his friend's house is located on a long street (and on the side where the house is located, the houses are numbered in order: $1,2,3$ and so on) and that the sum of the house numbers from the beginning of the street to his friend's house is the same as the sum of the house number... | 167. The sum of the numbers of the houses located on one side of a given house will equal the sum of the numbers on the other side in the following cases: 1) if the number of the given house is 1 and there are no other houses; 2) if the number is 6 and there are a total of 8 houses; 3) if the number is 35, and there ar... | 204 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,083 |
168. Another house number puzzle. Brown lives on a street with more than 20 but fewer than 500 houses (all houses are numbered in sequence: $1,2,3$ and so on). Brown discovered that the sum of all numbers from the first to his own, inclusive, is half the sum of all numbers from the first to the last, inclusive.
What i... | 168. The house number of Brown is 84, and there are a total of 119 houses on the street. The sum of the numbers from 1 to 84 is 3570, and the sum of the numbers from 1 to 119 is 7140, which, as required, is exactly twice as much.
Let's write down the consecutive solutions (in integers) of the equation \(2 x-1=y^{2}\):... | 84 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,084 |
173. Offspring of a Cow. "Suppose," said my friend the farmer Hodge, "that my cow gives birth to a heifer at the age of two. Suppose also that she will give birth to a heifer every year and that each heifer, upon reaching the age of two, will follow her mother's example and give birth to a heifer annually, and so on. N... | 173. Let's write down the following sequence of numbers, first studied by Leonardo Fibonacci (born in 1175), who almost introduced the familiar Arabic numerals to Europe:
\[
0,1,1,2,3,5,8,13,21,34, \ldots, 46368
\]
Each subsequent number is the sum of the two preceding ones. The sum of all numbers from the first to t... | 121392 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 41,089 |
## 174. Sum Equals Product.
- Just think about it,- said one man to me,- there are two numbers whose sum equals their product; that is, you get the same result whether you add them or multiply them together. These are 2 and 2, since their sum and product are both 4.
He then made a gross mistake by saying:
- I discov... | 174. Taking any number, and then another, equal to 1 plus a fraction, where the numerator is 1 and the denominator is the number one less than the given one, we get a pair of numbers that give the same result when added and when multiplied. Here are a few examples: 3 and \(1 \frac{1}{2}, 4\) and \(1 \frac{1}{3}\), 5 an... | 1\frac{1}{987654320} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,090 |
175. Squares and cubes. Can you find two numbers, the difference of whose squares is a cube, and the difference of whose cubes is a square? What are the two smallest numbers with this property? | 175. The smallest possible solution has the form
\[
\begin{aligned}
& 10^{2}-6^{2}=100-36=64=4^{3} \\
& 10^{3}-6^{3}=1000-216=784=28^{2}
\end{aligned}
\] | 106 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,091 |
176. An interesting cube. What is the length of the edge of a cube (in meters) for which:
1) the total surface area and volume are expressed by the same number
2) the total surface area is equal to the square of the volume;
3 the square of the total surface area is equal to the volume? | 176. 1) \(6 \text{m}\); 2) approximately \(1.57 \text{m}\); 3) \(\frac{1}{36} \text{m}\). | 1)6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 41,092 |
177. "Common Divisor". Here is a puzzle that readers often ask me (of course, the specific numbers in it are different). A correspondent of a provincial newspaper reported that many teachers had undermined their health in vain attempts to solve it! Probably, he exaggerated a bit, because the question is actually simple... | 177. When dividing these numbers by the sought number, the remainders are the same. Therefore, if we subtract one number from the other as shown below, the difference will be divisible by the sought number without a remainder.
. By adding the numbers in the second column opposite the remainders equal to 1, we get \(... | 2231 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,094 |
179. The Rejected Cannon. This simple puzzle in the field of artillery technology you will probably solve without hesitation. It is so simple that a child can understand it. No knowledge of artillery is required to solve the puzzle. Nevertheless, some of my readers will have to ponder it for five minutes.
An inventor ... | 179. The experts were right. The cannon fires 60 shots in 59 minutes, if it really fires at a rate of 1 shot per minute. Time is counted from the moment of the first shot, so the second shot will be fired after the first minute, the third after the second, and so on. An analogy can be drawn. Suppose we marked 60 points... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,095 |
180. Twenty Questions. I recalled an old game that I often played in my youth. Someone thinks of something specific, for example, Big Ben, a knocker on a front door, the chime of a clock in the next room, the top button on a friend's jacket, or Mr. Baldwin's pipe. You have to determine what was thought of by asking no ... | 180. There are different ways to solve this puzzle, but the simplest one, I believe, is as follows. Suppose the six-digit number is 843712.
1) Is it divisible by 2 without a remainder? Yes.
2) Is the quotient divisible by 2 without a remainder? Yes.
3) Is the new quotient divisible by 2 without a remainder? Yes.
Yo... | 843712 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,096 |
181. Card Trick. Take a regular deck of cards (this time, let's consider all jacks, queens, and kings as tens). Glancing at the top card (let's say it's a seven), place it face down on the table, then, continuing to count out loud in order: "Eight, nine, ten..." and so on up to 12, lay out other cards from the deck on ... | 181. In each stack, the number of cards must equal 13 minus the value of the lowest one. Therefore, 13, multiplied by the number of stacks, minus the sum of the bottom cards and plus the number of remaining cards should equal the total number of cards in the deck, which is 52. Thus, the sum of the bottom cards equals 1... | 13\times( | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,097 |
182. Quarrelsome Children. A man married a widow, and each of them had children from their previous marriages. After 10 years, a battle broke out in which all the children (by then there were 12) participated. The mother ran to the father, shouting:
- Come quickly. Your children and my children are beating our childre... | 182. Each of the parents had 3 children from their first marriage, and 6 children were born from their second marriage. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,098 |
183. Apple Sharing. While the Crackhams were refueling their car in a picturesque village, 8 children heading to school stopped and watched them. In the basket, the children had 32 apples they were planning to sell. Aunt Gertrude, out of kindness, bought all the apples and said the children could divide them among them... | 183. Ned Smith and his sister Jane received 3 apples each. Tom and Kate Browns received 8 and 4 apples respectively. Bill and Ann Jones - 3 and 1 apple, and Jack and Mary Robinsons got 8 and 2 apples. In total, 32 apples were distributed. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,099 |
184. Buying a Rubber Band. Here is a puzzle that looks very much like some old puzzles, but requires a completely different approach. The author is unknown.
Four mothers (each with her daughter) went to the store to buy a rubber band. Each mother bought twice as many meters of rubber band as her daughter, and each of ... | 184. Mary's mother was Mrs. Jones. The purchases and expenses were distributed as follows.
Among the daughters:
Hilda bought 4 m for 16 cents,
Gladys bought 6 m for 36 cents,
Nora bought 9 m for 81 cents,
Mary bought 10 m for 1 dollar.
Among the mothers:
Mrs. Smith bought 8 m for 64 cents, Mrs. Brown bought 12 m... | Mary'motherwasMrs.Jones | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,100 |
185. Squares and triangular numbers. What is the third largest number (the smallest number is considered the first) that is both a triangular number ${ }^{*}$ and a square? Of course, the first two numbers with this property are 1 and 36. What is the next number? | 185. To find a number that is both a square and a triangular number, one must solve the Pell equation: \(8 x^{2}+1=y^{2}\). Successive values for \(x\) are \(1,6,35\), etc., and for \(y\) are \(3,17,99\), etc. The answer is the number 1225 \(\left(35^{2}\right)\), which has the required properties. | 1225 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,101 |
186. Exact Squares. Find four numbers, the sum of each pair of which, as well as their total sum, are exact squares. | 186. Of course, several solutions to this problem can be found, but apparently the smallest numbers will be:
\[
\begin{aligned}
a & =10430, \quad b=3970, \quad c=2114, \quad d=386 \\
a+b & =10430+3970=14400=120^{2} \\
a+c & =10430+2114=12544=112^{2} \\
a+d & =10430+386=10816=104^{2} \\
b+c & =3970+2114=6084=78^{2} \\
... | =10430,b=3970,=2114,=386 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,102 |
187. Elementary arithmetic. Here is one question similar to those that were so popular in Venice (and not only there) in the mid-16th century. Their appearance was largely due to Niccolò Fontana, better known as Tartaglia (the stammerer).
If a quarter of twenty were four, then what would a third of ten be? | 187. The answer is the number \(2 \frac{2}{3}\). To find it, a proportion needs to be set up: \(5: 4=3 \frac{1}{3}: 2 \frac{2}{3}\). | 2\frac{2}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 41,103 |
188. Перестановка цифр. Если мы хотим умножить 571428 на 5 и разделить на 4 , то для этого нам нужно лишь переставить 5 из начала в конец: число 714285 дает верный ответ.[^10]
Не сумели бы вы найти число, которое можно было бы умножить на 4 и разделить затем на 5 столь же просто: переставив первую цифру в конец?
Разу... | 188. Ответ имеет вид
\[
2173913043478260869565 .
\]
Данное число можно умножить на 4 и разделить затем на 5 , просто перенеся 2 из начала в конец. | 2173913043478260869565 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,104 |
193. A question about cubes. Professor Rackbrain one morning noticed that the cubes of consecutive numbers, starting from 1, can sum up to a perfect square. For example, the sum of the cubes of 1, 2, and 3 (i.e., \(1 + 8 + 27\)) is 36, or \(6^2\). The professor claimed that if taking consecutive numbers starting not fr... | 193. The cubes of all numbers from 14 to 25 inclusive (a total of 12) sum up to \(97344=312^{2}\). The next five cubes after the smallest answer are \(25,26,27,28\) and 29, the sum of which equals \(315^{2}\). | 312^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,109 |
194. Two Cubes. "Could you find," asked Professor Rackbrain, "two consecutive cubes, the difference between which is a perfect square? For example, $3^{3}=27$, and $2^{3}=8$, but their difference (19) is not a perfect square."
What is the smallest possible solution? | 194. \(7^{3}=343,8^{3}=512,512-343=169=13^{2}\). | 169=13^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,110 |
195. Разность кубов. Число 1234567 можно представить в виде разности квадратов, стоит только выписать два числа, 617284 и 617283 (половина данного числа плюс $\frac{1}{2}$ и минус $\frac{1}{2}$ соответственно), и взять разность их квадратов ${ }^{*}$. Найти же два куба, разность которых равнялась бы 1234567 , несколько... | 195. \(642^{3}=264609288 ; 641^{3}=263374721\) и разность между кубами равна 1234567. | 1234567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,111 |
196. Composite Squares. Can you find two three-digit squares (with no zeros) that, when written consecutively, form a six-digit number which itself is a square? For example, from 324 and $900\left(18^{2}\right.$ and $\left.30^{2}\right)$ we get $324900\left(570^{2}\right)$, but the number 900 contains two zeros, which ... | 196. The answer is the number 225625 (the squares of the numbers 15 and 25, written down one after the other), which equals the square of 475. | 225625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,112 |
198. Square Supplement. "What number," asked Colonel Crackham, "has the property that if it is added to the numbers 100 and 164 separately, a perfect square is obtained each time?" | 198. If we add 125 to 100 and 125 to 164, the resulting numbers are \(225=15^{2}\) and \(289=17^{2}\). | 125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,114 |
202. Nuts for Monkeys. One person brought a bag of nuts to the monkey enclosures. It turned out that if he divided these nuts equally among 11 monkeys in the first cage, there would be one extra nut; if he divided them among 13 monkeys in the second cage, there would be 8 nuts left; and finally, if he divided them amon... | 202. The smallest number of nuts is 2179. It is best to deal only with the first two cases at first and find out that 34 (or 34 plus any multiple of 143) satisfies the condition for 11 and 13 monkeys. Then, the smallest number of this kind that satisfies the condition for 17 monkeys should be found. | 2179 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 41,118 |
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