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5. There were 5 times more strawberry bushes on the first bed than on the second. When 22 bushes were transplanted from the first bed to the second, the number of strawberry bushes on each bed became the same. How many bushes were there on each bed? | ## 5. First method of solving.
Let's build a graphical model of the problem's condition
1) $22: 2=11$ (bushes) - this is $1/5$ of all the bushes (that were on the second bed);
2) $11 \ti... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Between the numbers $1,2,3,4,5,6,7,8,9$ insert arithmetic operation signs and parentheses so that the resulting expression has a value of 100. | 1. a) $1+2+3+4+5+6+7+8 \times 9=100$;
b) $1 \times 2 \times 3+4+5+6+7+8 \times 9=100$;
c) $1+2 \times 3+4 \times 5-6+7+8 \times 9=100$;
d) $1 \times 2 \times 3 \times 4+(5+6-7)+8 \times 9=100$
e) $(1 \times 2+3) \times 4 \times 5+6-7-8+9=100$;
f) $(1+2+3) \times(4+5+6)-7+8+9=100$;
g) $((1+2): 3+4+5-6) \times 7+8 \... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Let 2 cups and 2 pitchers weigh as much as 14 plates, 1 pitcher weighs as much as 1 cup and 1 plate. How many plates will balance a pitcher
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。 | 4. First method of solution (graphical).
Reduce the number of items on each scale by half.
 $22725: 9=2525$ (rubles) — the amount the students paid for one month.
To determine how much each student paid monthly, we need to know the number of students in the class. This is unknown in the problem. However, from the problem's condition, it follows that this is a natural number that is a divisor of 2525.
... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Write the smallest four-digit number in which all digits are different. | 1. According to the problem, to record a four-digit number, four different digits should be used. They can be chosen from $0,1,2,3,4,5,6,7,8,9$.
The smallest digit in the thousands place can be $1: 1^{* * *}$.
The smallest digit in the hundreds place can be $0: 10^{* *}$.
Since we have already used 1 and 0, the digi... | 1023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest number that gives a remainder of 1 when divided by 2, and a remainder of 2 when divided by 3. | 3. First, let's write down the numbers that give a remainder of 1 when divided by 2:
$$
3,5,7,9,11 \ldots
$$
Next, let's write down the numbers that give a remainder of 2 when divided by 3:
$$
5,8,11,14,17 \ldots
$$
We will choose from the obtained numbers the one that satisfies both conditions and is the smallest.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. A motorcyclist traveled 980 km over three days. In the first two days, he traveled 725 km, and on the second day, he traveled 123 km more than on the third day. How many kilometers did he travel each of these three days? | 6. Let's represent the condition of the problem in a drawing
980 km
1) $980-725=255$ (km) - traveled on the third day.
2) $255+123=378$ (km) - traveled on the second day.
3) $725-378=347$... | 347 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. How many tiles with dimensions $15 \times 15$ cm are needed to tile a wall that is 3 m 6 dm long and 27 dm wide $?$ | 1. The first method of solving.
1) $15 \times 15=225$ (cm$^2$ - the area of one tile;
2) $360 \times 270=97200\left(\mathrm{~cm}^{2}\right)$ - the area of the walls;
3) $97200: 225=432$ (tiles).
The second method of solving (Rukhov Arseny, student of School No. 25).
1) $360: 15=24$ (tiles) - will fit in one row along... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given the sequence of numbers $0,1,2,6,16,44,120, \ldots$ Extend it and write down the next two numbers. | 4. All numbers, starting from the third, in the given sequence are formed according to the rule:
$$
\begin{gathered}
a_{n}=\left(a_{n-1}+a_{n-2}\right) \times 2 \\
(0+1) \times 2=2 \\
(1+2) \times 2=6 \\
(2+6) \times 2=16 \\
(16+6) \times 2=44 \\
(44+16) \times 2=120 \\
(44+120) \times 2=328
\end{gathered}
$$
Thus, w... | 328 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. How many days have passed from March 19, 1990 to March 23, 1996 inclusive | 5. 6) \(365 \times 6 + 2 = 2192\) (days) - have passed from 19.03.1990 to 19.03.1996
2) \(23 - 19 = 4\) (days) - have passed from 19.03.1990 to 23.03.1996
3) \(2192 + 4 = 2196\) (days) - have passed from 19.03.1990 to 23.03.1996\).
Answer: a total of 2196 days have passed. | 2196 | Other | math-word-problem | Yes | Yes | olympiads | false |
6. It is known that the perimeter of one rectangle is greater than the perimeter of another rectangle. Compare the areas of these rectangles.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. The solution to this problem can be obtained by students as a result of conducting a computational experiment with various rectangles, for example:
a) $a=3 \text{~cm}, b=4 \text{~cm}$;
$$
\begin{aligned}
& a=4 \text{~cm}, b=2 \text{~cm} ; \\
& p_{2}=12 \text{~cm} ; \\
& S_{2}=8 \text{~cm}^{2}
\end{aligned}
$$
$$
\... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Which of the following time intervals is the largest? a) 1500 minutes b) 10 hours c) 1 day. | 1. To compare time intervals, it is necessary to express them in units of the same denomination, for example, in hours:
$1500: 60=25$ (hours);
1 day $=24$ hours.
Thus, the greatest time interval is 1500 minutes.
Answer: the greatest time interval is 1500 minutes. | 1500 | Other | MCQ | Yes | Yes | olympiads | false |
3. In the box, there are geometric shapes: triangles, squares, and circles. In total, there are 24 shapes. There are 7 times as many triangles as squares. How many of each shape could be in the box? | 3. The first method of solving. Let's represent the condition of the problem in the form of a drawing.
squares
triangles
, we c... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. By how much is the largest five-digit number greater than the smallest five-digit number | 6. The largest five-digit number is 99999, and the smallest is 10000.
$$
99999-10000=89999
$$
Answer: greater by 89999. | 89999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of all three-digit numbers that can be written using the digits $1,2,3$ such that all digits in each number are different. | 1. Let's first determine the three-digit numbers that can be formed using the digits $1,2,3$ such that all digits in each number are different.
Thus, we have the following numbers that sati... | 1332 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Anton was given a scale, and he started weighing his toys. A car was balanced by a ball and two cubes, and a car with a cube - two balls. How many cubes balance a car? (All of Anton's balls are the same, and the cubes are the same too.) | 4. First method of solving. Since the machine is balanced by the ball and two cubes, the machine with a cube will be balanced by the ball and three cubes. From the second condition, we have that 2 balls balance a ball and 3 cubes, meaning one ball by mass is equal to 3 cubes. Thus, the machine can be balanced by 5 cube... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. A chocolate bar consists of $5 \times 8$ square pieces. The bar is broken along straight lines, separating the pieces until 40 individual pieces are obtained. How many times will the bar need to be broken? | 6. To get 40 equal pieces from a chocolate bar, you first need to divide it into 8 strips along its length. For this, 7 breaks are required. Then, each of the 8 strips needs to be divided into 5 pieces. For this, 4 breaks are needed in each strip. In total, the chocolate bar will need to be broken 28 times $(7 \times 4... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Two numbers were first multiplied, and then the larger number was divided by the smaller one, resulting in equal outcomes. What are these numbers? How many such pairs of numbers exist? | 2. Let $a$ and $b$ be the two numbers mentioned in the problem, with $a > b$. According to the problem,
$$
a \times b = a : b
$$
or
$$
a b^{2} - a = 0
$$
Factoring out the common factor, we get
$$
a \left(b^{2} - 1\right) = 0
$$
From this, we can conclude that $b = 1$, and $a$ can be any number.
Students might r... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Katya is 6 times younger than her great-grandfather; if a 0 is placed between the digits of her age, the age of her great-grandfather is obtained. How old is she | 1. The first way to solve the problem. From the problem statement, it follows that Katya's age is a two-digit number, and her great-grandfather's age is a three-digit number. The number of units in Katya's age, when multiplied by 6, gives a number in which the number of units will be the same as in Katya's age. Let's f... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit numbers do not contain the digit 8? | 2. First method of solving. There are a total of 900 three-digit numbers. To find out how many of them do not contain the digit 8, we need to subtract from 900 the number of three-digit numbers that contain the digit 8. Let's determine the number of these numbers. From 800 to 899, there are one hundred such numbers. Ot... | 648 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Apples were divided into two unequal piles. When half of the apples from the first pile were moved to the second pile, and then half of the apples from the second pile were moved back to the first pile, there were 18 apples in the first pile and 8 in the second. How many apples were in each pile originally? | 3. The first method of solving.
1) $8 \times 2=16$ (beans) - was in the second pile after half was moved from the first pile;
2) $18-8=10$ (beans) - was in the first pile after half was moved from it;
3) $10 \times 2=20$ (beans) - was originally in the first pile;
4) $16-10=6$ (beans) - was originally in the second pil... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given a triangle with side lengths of 7 cm, 12 cm, and 9 cm respectively. Explain how to construct a segment connecting a vertex and the opposite side with a length of 9 cm so that the perimeters of the two resulting triangles are equal. | 6. First method of solution. Since in the obtained triangles one side will be common ( $A D$ ), in order for the perimeters to be equal, it is necessary that the side of 9 cm be divided into parts, the difference in lengths of which would equal the difference of the other two sides ( $12-7=5 \text{ cm}$ ). Based on thi... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. How many times will a three-digit number increase if the same number is appended to it? | 1. Let's write down a random three-digit number (531) and append the same number to it, resulting in -531531. Let's find their quotient:
$$
531 \times 531: 531=1001
$$
The pattern found is also valid for other three-digit numbers.
Answer: The three-digit number will increase by 1001 times. | 1001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Carlson has a runny nose. He uses square handkerchiefs measuring 25 cm $\times 25 \mathrm{~cm}$. Over eight days, Carlson used $3 \mathrm{~m}^{2}$ of fabric. How many handkerchiefs did Carlson use per day? | 2. First method of solving.
1) $25 \times 25=625\left(\mathrm{~cm}^{2}\right)-$ area of one handkerchief;
2) $3 \times 10000=30000\left(\mathrm{~cm}^{2}\right)$ - contained in $3 \mathrm{~m}^{2}$;
3) $30000: 625=48$ (handkerchiefs) - Carlson used over 8 days
4) $48: 8=6$ (handkerchiefs) - Carlson used per day.
Second ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. If a schoolboy bought 11 notebooks, he would have 8 rubles left, but for 15 notebooks, he would be short of 12 rubles and 24 kopecks. How much money did the schoolboy have? | 3. Let's represent the condition of the problem in a drawing:
1) $15-11=4$ (notebooks) - the difference in the number of notebooks purchased;
2) $800+1224=2024$ (kop.) - the cost of 4 note... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a box, there are 7 blue and 5 red balls. What is the minimum number of balls that need to be taken out to ensure that among them there are at least 2 blue and 1 red? | 5. The worst-case scenario is if we draw 7 balls and they all turn out to be blue. To get another red ball, we need to draw one more ball. In this case, there will definitely be one red ball and 2 blue balls among the balls.
Answer: 8 balls. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The boy caught a fish. When asked how much the caught fish weighed, he said: “I think the tail weighs 1 kg, and the head weighs as much as the tail and half the body, and the body weighs as much as the head and the tail together.” How much does the fish weigh? | 2. From the problem statement, it is known that the tail ( $X$ ) weighs 1 kg. The weight of the head is equal to the weight of the tail plus an additional $1 / 2$ of the body weight (T):
$$
\Gamma=\mathrm{X}+\mathrm{T} / 2 \text { or } 2 \Gamma=2 \mathrm{X}+\mathrm{T} .
$$
Since the tail weighs 1 kg, then $2 \Gamma=2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. This figure consists of 6 identical squares. Its perimeter is 84 cm. Find the area of this figure.
 | 3. 1 ) $84: 14=6$ (cm) - the length of the side of the square.
2) $6 \times 6=36$ (cm$^2$) - the area of one square.
3) $36 \times 6=216$ (cm$^2$) - the area of the entire figure.
Answer: 216 sq. cm. | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. How many numbers less than 96 are divisible by 2 and 3? | 5. We need to select numbers from 1 to 95 that are divisible by 2 and 3, which means they are divisible by 6. Such numbers are: $6,12,18$, $24,30,36,42,48,54,60,66,72,78,84,90$. There are 15 such numbers in total.
The number of such numbers could also be determined as follows:
$$
95: 6=15 \text { (remainder 5). }
$$
... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. If from each of two numbers you subtract half of the smaller one, the remainder of the larger one is three times the remainder of the smaller one. How many times larger is the larger number than the smaller one? | 1. The first way to solve the problem is to represent the condition of the problem in the form of a drawing.
From the drawing, it is clear that the first number is twice as large as the sec... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Three teams scored 285 points in the Olympiad. If the team from School No. 24 had scored 8 points less, the team from School No. 46 had scored 12 points less, and the team from School No. 12 had scored 7 points less, then all of them would have scored equally. How many points did the teams from Schools No. 24 and No... | 2. First method of solving. Let's make a brief record of the problem in the form of a drawing
1) $7+12+8=27$ (points) - this is how many fewer points all schools scored;
2) $285-27=258$ (p... | 187 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There were 2 minutes left before the electric train departed when the motorist was 2 km from the station. For the first minute, he drove at a speed of 30 km/h. At what speed should he drive during the second minute to catch the train? | 5. The first method of solving.
1) $2 \times 1000=2000$ (m) - need to travel;
2) $30 \times 1000=30000(\mathrm{m} / \mathrm{h})$ - speed of the car;
3) $30000: 60=500$ (m/s) - speed in the first minute;
4) $2000-500=1500$ (m) - left to travel in 1 minute;
5) $1500 \times 60=90000(\mathrm{m} / \mathbf{h})=90(\mathrm{km}... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. One side of the square was increased by 5 times, and the other side was reduced by 2 times, resulting in a rectangle with an area of $160 \mathrm{~cm}^{2}$. What is the side length of the square | 6. First method of solution.
1) $160: 5=32\left(\mathrm{~cm}^{2}\right)$ - area of half the square;
2) $32 \times 2=64\left(\mathrm{~cm}^{2}\right)$ - area of the original square;
3) $64: ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate in the most rational way
$$
12 \times 171+29 \times 9+171 \times 13+29 \times 16
$$ | $$
\begin{aligned}
& 12 \times 171 + 29 \times 9 + 171 \times 13 + 29 \times 16 = \\
& = (171 \times 12 + 171 \times 13) + 29 \times 9 + 29 \times 16 = \\
& = 171 \times (12 + 13) + 29 \times (9 + 16) = \\
& = 171 \times 25 + 29 \times 25 = (171 + 29) \times 25 = \\
& = 200 \times 25 = 5000
\end{aligned}
$$
Answer: 50... | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. From two points $A$ and $B$, the distance between which is 300 km, two cars set off simultaneously. The speed of the car that set off from $A$ is 40 km/h. Determine the speed of the second car, given that after two hours the distance between the cars was 100 km. | 3. Since the problem does not specify the direction they were traveling, we need to consider two cases: moving in the same direction and moving in different directions.
First case. Moving in the same direction. Let the first car travel from $A$ to $B$ at a speed of 40 km/h.
Answer: 9 sheep. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Arrange the order of operations in the expression
$$
1891-(1600: a+8040: a) \times c .
$$
and calculate its value when $a=40$ and $c=4$. Show how the expression can be modified without changing its numerical value. | $$
\begin{aligned}
& 1891-\left(1600: 40+8040^{5}: 40\right)^{4} \times 4= \\
& =1891-(40+201) \times 4= \\
& =1891-241 \times 4=1891-964=927 .
\end{aligned}
$$
Based on the properties of arithmetic operations, we can write:
$$
\begin{aligned}
& 1891-(1600: a+8040: a) \times c= \\
& =1891-(8040: a+1600: a) \times c= ... | 927 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In this century, 200 years will be marked since the birth of the famous Russian mathematician, a native of Kaluga Governorate, Pafnuty Lvovich Chebyshev. Among the numbers that record his year of birth, the sum of the digits in the hundreds and thousands place is three times the sum of the digits in the units and te... | 3. Since in the 21st century the 200th anniversary will be marked, the first digit in the number will be -1, and the second will be 8, and the year of birth has the form $18 a c$.
The sum of the digits in the hundreds and thousands place is 9, and it is three times the sum of the digits in the units and tens place ($9... | 1821 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The perimeter of quadrilateral $A B C D$ is 100 cm. The length of side $A B$ is 41 cm, side $B C$ is 18 cm shorter than side $A B$, but 6 cm longer than side $C D$. Find the length of side $A D$. | 1. The first method of solving.
1) $41-18=23$ cm - the length of side $B C$;
2) $23-6=17$ cm - the length of side $C D$;
3) $41+23+17=61$ cm - the sum of sides $A B, B C$, and $C D$;
4) $100-61=19$ cm - the length of side $A D$.
The second method of solving. Another approach can be used to solve this problem. First, d... | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. First, three, and then four people shook hands with each other. How many handshakes were there? Find the pattern in counting the number of handshakes and determine their number for 7 people. | 6. When solving this problem, we should use incomplete induction. Let's analyze the first situation and try to find a pattern in determining the number of handshakes based on the number of participants. In the first case, when there were three people, the following options are possible:
$$
\text { 1-2, 1-3, 2-3. }
$$
... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. From 16 m of fabric, 4 men's and 2 children's coats were sewn. How many meters of fabric are needed to sew one men's and one children's coat, if from 18 m of the same fabric, 2 men's and 6 children's coats can be sewn | 7. Let's make a brief record of the problem's condition:
$$
\begin{aligned}
& 4 \text { m, } 2 \text { d }-16 \text { m; } \\
& 2 \text { m, } 6 \text { d }-18 \text { m. }
\end{aligned}
$$
To ensure that the brief record contains the same quantities, we will double the second order:
$$
\begin{aligned}
& 4 \text { m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The teacher cut a square sheet of paper with a side of 5 cm into two rectangles. The perimeter of one of these rectangles is $16 \mathrm{cm}$. What is the perimeter of the other? | 1. The first method of solving.
1) $16: 2=8$ cm - the semi-perimeter of the first rectangle;
2) $8-5=3$ cm - the width of the first rectangle;
3) $5-3=2$ cm - the width of the second rectangle;
4) $(2+5) \times 2=14 \text{cm} -$ the perimeter of the second rectangle.
The second method of solving. Before solving by the... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The sum of three numbers is 1281. From the first number, 329 was subtracted, and to the second number, 401 was added. What needs to be done with the third number so that the sum does not change? Will the solution to the problem change if the word "sum" is replaced with "difference"? | 2. First method of solving. Find out by how much the sum has increased ( $401-329=72$ ). To keep the sum the same, it is necessary to subtract 72 from the third number.
Second method of solving. Let $a, b, c$ be the first, second, and third numbers. According to the condition, their sum is 1281
$$
a+b+c=1281
$$
From... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two trains leave from two cities at the same time. The first one travels at 40 km/h, while the second one travels at 48 km/h. How far apart will these trains be from each other after 8 hours, if they are moving in the same direction and the distance between the cities is 892 km? | 3. The trains are moving in the same direction, so they can move in the direction of $A B$ or $B A$. Let's consider each of these cases.
The trains are moving in the direction of $A B$.
1) $40 \times 8=320$ km - the first train traveled;
2) $48 \times 8=384$ km - the second train traveled;
3) $384-320=64$ km - by thi... | 956 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In A.S. Pushkin's fairy tale "The Priest and His Workman Balda," it is said: "Balda sat on the mare and rode a verst so that the dust rose in a column." Determine how many tens of meters are contained in the length of the path that Balda rode on the mare. $(1$ verst $=500$ sazhen, 1 sazhen $=3$ arshins, 1 arshin $=$... | 1. 2) $3 \times 500=1500$ arshins - in one verst;
2) $71 \times 1500=106500$ cm - in one verst;
3) $106500: 100=1065$ m - in one verst;
4) $1065: 10=106$ tens (remainder 5) — the number of tens of meters.
Answer: in one verst there are 106 tens of meters. | 106 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. At the mathematics olympiad in two rounds, it is necessary to solve 14 problems. For each correctly solved problem, 7 points are given, and for each incorrectly solved problem, 12 points are deducted. How many problems did the student solve correctly if he scored 60 points? | 3. Let's determine the number of problems solved by the student, given that he received 7 points for each problem: $60: 7=8$ (remainder 4). Since 12 points were deducted for each incorrectly solved problem, the number of correctly solved problems was more than 8. Further, by trial and error, we can establish that there... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. One side of a rectangle is $5 \mathrm{~cm}$, and the other is the smallest two-digit number that is divisible by 3. For the second rectangle, one side is equal to the shorter side of the first. The area of one rectangle is $25 \mathrm{~cm}^{2}$ greater than the area of the other. Determine the unknown side of the se... | 6. First, let's determine the unknown side of the rectangle. For this, we will find the smallest two-digit number that is divisible by 3, which is 12. According to the problem, one side of the second rectangle is $5 \mathrm{~cm}$. Let's determine the area of the first rectangle $(12 \times 5=60)$. Since the problem doe... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. To number the pages in a mathematics textbook, 390 digits were required. How many pages are in the mathematics textbook? | 7. For writing the first nine pages, 9 digits are needed $(1,2,3,4,5,6,7,8,9)$.
For numbering pages from 10 to 99, 90 two-digit numbers are needed, which use $90 \times 2=180$ digits.
In the next step, we determine how many digits were used to write three-digit numbers $(390-180-9=201)$. Now we find out how many page... | 166 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The pages of materials for conducting a math club session are numbered. The last page is numbered 128. How many digits were needed to number all the pages? | 2. For numbering the first nine pages, 9 numbers are needed.
For numbering pages from 10 to 99, 90 two-digit numbers are needed, which use $90 \times 2=180$ digits.
Let's find out how many three-digit numbers were used for numbering the pages. There will be $29(128-99=29)$. Determine how many digits were used to writ... | 276 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. If the thought number is multiplied by 6, and then 382 is added to the product, the result will be the largest three-digit number written with two identical even numbers and one odd number. Find the thought number. | 3. Let's determine the largest three-digit number written with two identical even digits and one odd digit — this is 988. Subtract 382 from the obtained number, we get 606. This is the sixfold of the thought number. Dividing 606 by 6, we determine the thought number -101.
Answer: 101. | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A sheet of cardboard measuring 48 cm in length and 36 cm in width needs to be cut into cards measuring 16 cm in length and 12 cm in width. How can this be done to obtain the maximum number of cards? | 1. In the first step, it is necessary to determine how many cards will fit if the length of each is $16 (48: 16=3)$. Then determine how many cards will fit in the width (36:12 = 3).
Finally, find the number of cards $(3 \times 3=9)$. Then conduct similar reasoning for the case where the cards are laid out by width ( $... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The sum of three numbers is 121,526. One addend is the largest five-digit number, all digits of which are even; the second is the smallest four-digit number, all digits of which are odd. What is the third number? | 2. We will sequentially consider all the conditions of the problem. Given that the first addend is a five-digit number, it follows that five digits will be used for its notation (XXXXX). Since it must be the largest and all its numbers are even, this will be the number 88888. The second addend is a four-digit number (X... | 31527 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (Ancient Indian problem 3-4th century AD). Of the four sacrificial donors, the second gave twice as many coins as the first, the third three times as many as the second, the fourth four times as many as the third, and all together gave 132 coins. How many did the first give? | 4. The first method of solving. By analyzing the condition of the problem, one can conclude that the number of coins given by the first person will not be more than 10. We will then refine this number through trial and error. Let's try the number 5.
$$
5+5 \times 2+5 \times 2 \times 3+5 \times 2 \times 3 \times 4=5+10... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Nikolai was walking to a math olympiad, covering 100 meters per minute. After 36 minutes, he stepped into School No. 15. Then he remembered that he had left the pen at home, with which he had won the school olympiad, and ran back home at a speed of 3 meters per second. How much time did he spend on the return trip? | 5. 6) $100 \times 36=3600$ m - the distance from home to school;
2) $3 \times 60=180$ m - the number of meters he ran per minute on the way back;
3) $3600: 180=20$ min - the time he spent on the return trip.
Alternatively, the second step could have determined the number of seconds spent on the return trip $(3600: 3=1... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Two boats departed from piers $A$ and $C$. The speed of the boat traveling from $A$ is 7 km per hour, and from $C$ is 3 km per hour. Determine after how much time the boats will meet, if the distance between the piers is 20 km, and the speed of the river current from $A$ to $C$ is 2 km per hour. | 4. Since the problem does not specify the direction in which the boats are moving, we need to consider various scenarios: they are moving towards each other; they are moving in the same direction downstream; they are moving in the same direction upstream; they are moving in opposite directions. Given that the boats mus... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (An old entertaining problem). Divide 46 rubles into 8 parts so that each part is 50 kopecks (50 cents) more than the previous one. | 5. First method of solution. Let the first part be $x$, then the second is $x+50$, the third is $-x+50 \times 2$, the fourth is $-x+50 \times 3$, the fifth is $x+50 \times 4$, the sixth is $-x+50 \times 5$, the seventh is $-x+50 \times 6$, and the eighth is $x+50 \times 7$. We will calculate the sum of all parts.
$$
\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A wire 104 centimeters long was bent into a rectangular frame. The length and width of this frame are whole numbers of centimeters. In how many ways can the frame be obtained? | 6. First, let's determine what the sum of the length and width is, that is, we will calculate the semi-perimeter. For this, we divide 104 by 2 and get 52. Let's represent the number 52 as the sum of two addends:
$$
52=1+51=2+50=3+49=\ldots=25+27=26+26 \text {. }
$$
We get 26 ways.
Answer: 26 ways. | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. From points $A$ and $B$, which are 300 km apart, two cars set off simultaneously. The speed of the car that left from $A$ is 40 km/h. Determine the speed of the second car, given that after two hours the distance between them was 100 km. | 2. The problem statement does not specify the direction in which the cars are moving, so all possible scenarios should be considered: moving in opposite directions, towards each other, both in the direction of $A B$, and both in the direction of $B A$.
Assume the cars are moving in opposite directions. This scenario, ... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The great commander, Marshal of the Soviet Union, Georgy Konstantinovich Zhukov, was born in the village of Strelkovka, Kaluga Governorate. He lived for 78 years. In the 20th century, he lived 70 years longer than in the 19th century. In what year was G.K. Zhukov born? | 3. Let's make a brief record of the problem statement
1) $78-70=8$ years - doubled the time of life in the 19th century;
2) $8: 2=4$ years - G.K. Zhukov lived in the 19th century;
3) $1900... | 1896 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If a schoolboy bought 11 pens, he would have 8 rubles left, but for 15 pens, he would be short of 12 rubles 24 kopecks. How much money did the schoolboy have? | 4. 5) $15-11=4$ pens - bought more the second time;
2) $8+12=20$ rubles - amount to four pens;
3) $20: 4=5$ rubles - cost of one pen;
4) $5 \times 11+8=63$ rubles - the schoolboy had.
Answer: the schoolboy had 63 rubles. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A string 60 dm long was cut into three pieces. The length of one of them is 1 cm more than another and 1 cm less than the third. Find the length of each piece. | 2. First method of solving. Let's make a drawing according to the problem statement.
1) $600-3=597$ cm - the length of the string if all parts were as small as the smallest one
2) $597: 3... | 199 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. From a class of 20 people, a team of three students is formed to participate in mathematics, Russian language, and informatics olympiads. In this class, all students excel academically. How many ways are there to form the team of olympiad participants, if each member of the team participates in one olympiad? | 6. We can select a participant for the mathematics olympiad in twenty ways, for the Russian language olympiad in nineteen ways, and for informatics in eighteen ways. In total, this results in 6840 ways $(20 \times 19 \times 18=6840)$.
Answer: 6840 ways to form the team. | 6840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit numbers less than 500 do not contain 1? | 2. Let's determine which digit can represent the hundreds place of $-2.3$ or 4. Consider numbers of the form $2 x x$. There will be a total of one hundred such numbers. The digit 1 can be in the tens or units place. If 1 is in the tens place ($21 x$), we get the numbers: $210, 211, 212, 213, 214, 215, 216, 217, 218, 21... | 243 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Two friends collected 300 stamps. One collected 5 per week, the other - 3. The second boy collected for 20 weeks longer than the first. How many days did each boy collect stamps? | 2. 3) $20 \times 3=60 \text{m}$ - the second boy collected in 20 weeks;
2) $300-60=240 \text{m}$ - the boys collected during the time the first boy was collecting;
3) $5+3=8$ m - they collected per day;
4) $240: 8=30$ days - the first boy collected;
5) $30+20=50$ days - the second boy collected.
Answer: the first boy ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Every 24 hours, the clock runs ahead by 3 minutes. They were set accurately. After what shortest time will the clock hands show the exact time again? | 4. To make the clock show the exact time, it is necessary for it to run ahead by 12 hours. Let's find out how many minutes are in 12 hours $\operatorname{cax}(12$ hours $=60$ min. $\times 12=720$ min.). Determine the number of days $-720: 3=240$.
Answer: 240 days. | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. In this century, the 200th anniversary of the birth of the famous Russian mathematician, a native of the Kaluga province P.L. Chebyshev, will be celebrated. In the number that records his year of birth, the sum of the numbers in the hundreds and thousands place is three times the sum of the numbers in the tens and u... | 6. Since in the 21st century the 200th anniversary of P.L. Chebyshev's birth will be celebrated, he was born in the 19th century - 18th. The sum of the numbers in the thousands and hundreds place is $9(1+8=9)$. The sum of the numbers in the tens and units place is three times less than 9. Therefore, it is $3(a+b=3)$. T... | 1821 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. How many numbers less than 2011 are divisible by 117 and 2? | 7. When solving this problem, one can reason as follows. First, list the numbers less than 2011 and divisible by 117 (2), then among them, select those that are divisible by 2 (117), and determine how many there are.
Since the number is divisible by both 2 and 117, it will also be divisible by their product $2 \times ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A rectangular section of road with a length of 800 m and a width of 50 m was paved with asphalt. For every 100 m$^{2}$ of road, 3 tons 800 kg of asphalt were used. How many tons of asphalt were used? | 1. 2) $800 \times 50=40000 \mathrm{~m}^{2}$ - the area of the road;
2) $40000: 100=400$ times - the road contains $100 \mathrm{m}^{2}$ segments;
3) $3800 \times 400=1520000 \mathrm{~kg}=1520 \mathrm{~T}$ - the amount of asphalt used.
Answer: 1520 tons of asphalt were used. | 1520 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. What digits stand in the natural number sequence at the thirteenth and one hundred twentieth positions?
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。 | 2. Let's write down the first fifteen numbers:
$$
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
$$
To fill the first nine positions, 9 digits are required, and then two digits are needed to write each two-digit number. Starting from the tenth position, two-digit numbers are written. For each of them, two places are allocated. ... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
4. A two-digit number was increased by 3, and it turned out that the sum is divisible by 3. When 7 was added to this same two-digit number, the resulting sum was divisible by 7. If 4 is subtracted from this two-digit number, the resulting difference is divisible by four. Find this two-digit number. | 4. Since the sum of a two-digit number and three is divisible by three and the second addend is also divisible by three, the unknown two-digit number is divisible by three. Similarly, we can establish that the two-digit number is also divisible by seven and four. Thus, it is divisible by $84(3 \times 4 \times 7=84)$. A... | 84 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. At $17-00$ the speed of the racing car was 30 km/h. Every 5 minutes thereafter, the speed increased by 6 km/h. Determine the distance traveled by the car from $17-00$ to $20-00$ of the same day. | 5. Let's write down the sequence of numbers expressing the speed at which the car was traveling on the highway: $30,36,42,48,54 \ldots 240$. To determine the distance, we need to calculate the sum:
$$
\begin{aligned}
30: 12 & +36: 12+42: 12+\ldots .+234: 12+240: 12= \\
& =(30+36+42+. .+234+240): 12
\end{aligned}
$$
I... | 405 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A cube with a side of 8 dm is cut into cubes with sides of 4 cm. How many cubes will be obtained? | 2. Let's find out how many 4 cm cubes fit into one side of the larger cube $(80: 4=20)$. Then, in the first layer of cubes, there will be 400 $(20 \times 20=400)$. We have 20 such layers. In total, there will be 8000 cubes $(400 \times 20=8000)$.
$. We need to find the greatest number of candies, so Maria will have 33 candies. Sasha will have 68 candies $(35... | 136 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Students from one of the schools are tasked with making 455 Christmas tree ornaments for kindergartens in the city. All students should make the same number of ornaments. More than 10, but fewer than 70 people participated in making the ornaments. How many students made the ornaments, and how many did each of them m... | 4. Let's represent the number 455 as the product of several numbers. Since the number ends in 5, it is divisible by $5 (455: 5=91)$. The number 91 is also divisible by $7 (91: 7=13)$. Thus, the number 455 can be represented as the product of the following numbers: $455=5 \times 7 \times 13$.
Now let's represent the nu... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. How many different products that are divisible by 10 (order does not matter) can be formed from the numbers $2,3,5,7,9$? Numbers in the product do not repeat! | 7. For a number to be divisible by 10, it must be divisible by 2 and 5.
Thus, the product must necessarily include 2 and 5. We get the first product $-2 \times 5$.
Now, let's add one more factor to this product:
$$
2 \times 5 \times 3, 2 \times 5 \times 7, 2 \times 5 \times 9
$$
Add one more factor to the obtained ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. From 2 kg 600 grams of copper, three items were made: a spoon, a candlestick, and a teapot. Each item is three times heavier than the previous one. What is the weight of the teapot? | 1. Let's make a brief record of the problem statement.
1) $1+3+9=13$ (parts) - corresponds to 2 kg 600 grams;
2) $2600: 13=200$ (grams) - weight of the spoon;
3) $200 \times 3=600$ (grams... | 1800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the fourth grades of School No. 15, 100 students are studying. 37 of them have enrolled in the football section, 40 - in the swimming section, 15 people have enrolled in both sections. How many students have not enrolled in any section? | 4. 5) $40-15=25$ (students) - enrolled only in the swimming section;
2) $37-15=22$ (students) - enrolled only in the football section;
3) $25+22+15=62$ (students) - enrolled in the football and tennis sections;
4) $100-62=38$ (students) - did not enroll in any section.
Answer: 38 students did not enroll in any section... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In a village, two houses $A$ and $B$ stand along a straight road, 50 meters apart from each other. At what point on the road should a well be built so that the sum of the distances from the well to the houses is minimized? | $\triangle$ If we act fairly, we should build the well in the middle between the houses: then the distance to each house will be 25 meters, and in total 50. However, from the point of view of the problem's condition, any point $X$ on the road segment between $A$ and $B$ is equally good, since the segments $A X$ and $X ... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In village $A$, there are 100 schoolchildren, and in village $B$, there are 50 schoolchildren. The distance between the villages is 3 kilometers. At what point on the road from $A$ to $B$ should a school be built to minimize the total distance traveled by all schoolchildren? | $\triangle$ It seems fair to build the school closer to $A$, as there are more students there. But how much closer? Sometimes it is suggested to build the school one kilometer from $A$ and two kilometers from $B$ (twice as many students, so the school is twice as close). However, from the perspective of the given probl... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. How many degrees does the minute hand turn in a minute? The hour hand? | $\Delta$ In one hour, the minute hand makes a full revolution, and in half an hour, it turns by $180^{\circ}$ (a straight angle). Half an hour is 30 minutes, so in one minute, it turns by $1 / 30$ of the angle in $180^{\circ}$, which is $180 / 30=6^{\circ}$.
The hour hand moves $1 / 12$ of a circle in one hour, meanin... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
22. Five rays emanating from a single point divide the plane into five equal angles. Find the measure of these angles. | $\Delta$ If the sum of these five angles is a full circle, that is, two straight angles, which is $2 \times 180^{\circ}=360^{\circ}$. Each of them is one fifth of $360^{\circ}$, that is, $360 / 5=72^{\circ}$. $\triangleleft$
This problem and its solution require some comments. The reason is that different geometry tex... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
132. In a triangle, all sides are equal. Find its angles. | $\Delta$ An equilateral triangle (in which all sides are equal) is a special case of an isosceles triangle, and any side can be taken as the base. Therefore, all angles in it are equal. Since the angles sum up to $180^{\circ}$ (from the previous problem), each of them is $60^{\circ} . \triangleleft$ | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
141. What is the sum of the exterior angles at three vertices of a triangle? | $\Delta$ Together with the three interior angles (which sum up to $180^{\circ}$), this results in three straight angles, that is, $540^{\circ}$, so the sum of the exterior angles is $540-180=360^{\circ} . \triangleleft$ | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
142. What is the sum of the exterior angles at the four vertices of a convex quadrilateral? | $\Delta$ To find the sum of these angles (they are shown on the diagram with single arcs, but this does not mean they are equal!), consider the corresponding interior angles. Together with the 4 interior angles (which sum up to $360^{\circ}$), we get 4 straight angles, that is, $720^{\circ}$, so the sum of the exterior... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
143. What is the sum of the exterior angles at the five vertices of a convex pentagon? | $\checkmark$ Together with the 5 interior angles (which sum up to $540^{\circ}$), we get 5 straight angles, that is, $900^{\circ}$, so the sum of the exterior angles is $900-540=360^{\circ} . \triangleleft$
Similarly, one can verify that the sum of the exterior angles of a convex hexagon, heptagon, and generally any $... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
161. Find the angles of a right isosceles triangle. | $>$ The angles at the base of an isosceles triangle are equal, and together they sum up to $90^{\circ}$, so each of them is $45^{\circ} . \triangleleft$
Several of the following problems provide criteria for the congruence of right triangles. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
207. What angle does the diagonal of a square form with its side? | $\Delta$ This is an acute angle in a right isosceles triangle, which equals $45^{\circ} . \triangleleft$
## A few more problems | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
265. Inside the square $A B C D$, an equilateral triangle $A B M$ is constructed, and vertex $M$ is connected to points $C$ and $D$. Find the angles of triangle $C M D$. | $\Delta$ The angles of an equilateral triangle are $60^{\circ}$, so the angle at vertex $A$ of the isosceles triangle $A M D$ is $90-60=30^{\circ}$. Therefore, the angles at its base are $(180-30) / 2=75^{\circ}$. Thus, the angle $C D M$ is $90-75=15^{\circ}$ (as well as $\angle M C D$ ). $\triangleleft$ | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
315. Two of the angles of a trapezoid are $60^{\circ}$ and $130^{\circ}$. What can be said about its other angles? | $\Delta$ The angles at the lateral side of the trapezoid add up to $180^{\circ}$ (as interior alternate angles). Therefore, there are also angles $180^{\circ}-60^{\circ}=120^{\circ}$ and $180^{\circ}-130^{\circ}=50^{\circ}$ in the trapezoid. Thus, all four angles have been found. $\triangleleft$ | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
390. A circle is divided into five equal arcs by five points. Find the measure of each of them. | $\triangleright$ The total of all arcs is $360^{\circ}$, so each of them is $360^{\circ} / 5=72^{\circ} . \triangleleft$
(The Inscribed Angle Theorem) Let $A, B$ and $C$ be points on a circle with center $O$. Then the angle $B A C$ is equal to half the measure of the arc $B C$, that is, half the angle $B O C$.
The an... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
449. Two points $A$ and $B$ are 8 units apart. How many lines exist that are 5 units away from point $A$ and 3 units away from point $B$? | $\triangleright$ The line $l$ is at a distance $d$ from the point $A$ when it is tangent to the circle of radius $d$ centered at $A$. Therefore, the problem can be restated as follows: given two circles with radii 3 and 5, and the distance between their centers is 8. How many common tangents exist?
We have already sol... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
500. Find the area of a rectangle with sides 3 and 5. | $\triangleright$ Such a rectangle can be cut into 15 unit squares, so its area is equal to the sum of the areas of these squares, that is, $15 . \triangleleft$ | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
501. How many square centimeters are in a square meter | $\triangleright$ A square with a side length of 1 meter can be divided into $100 \times 100$ squares with a side length of 1 centimeter, so in one square meter, there are $100 \cdot 100=10000$ square centimeters. $\triangleleft$ | 10000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
521. By what factor should the side of a square be increased to make its area four times larger? | $\triangleright$ The area of a rectangle with sides $a$ and $b$ is $a b$, so the area of a square with side $a$ is $a \cdot a=a^{2}$. By doubling the side of the square, we increase its area by four times: $(2 a)^{2}=2^{2} a^{2}=4 a^{2} . \triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
577. Draw an isosceles right triangle on graph paper with the legs equal to the side of a cell, and check the Pythagorean theorem for it: the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. | $\triangleright$ The squares built on the legs have a unit area. The square built on the hypotenuse is divided into four triangles, each of which makes up half of a grid square, and thus has an area of 2 - as required. $\triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
578. Check the Pythagorean theorem for a right triangle with legs 1 and 2. | $\triangleright$ The squares constructed on the legs have areas of 1 and 4. It remains to find the area of the square constructed on the hypotenuse. This can be done in two ways. First, it can be cut into 4 triangles and a central square. The triangles are equal to the original (legs 1 and 2) and have an area of 1. The... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
662. Point $A$ was moved 13 units to the right and resulted in a point with a coordinate of 8. What was the coordinate of point $A$? | $\triangleright$ Let $A$ have the coordinate $a$. Then $a+13=8$, and $a=8-13=(-5) . \triangleleft$
 | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
663. The grasshopper jumped from the point with coordinate 8 to the point with coordinate 17.5. Then it made another jump of the same length (in the same direction). At the point with what coordinate did it end up? | $\triangleright$ The length of the jump is $17.5-8=9.5$; another jump will give $17.5+9.5=27 . \triangleleft$ | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
664. Given two points $A(-3)$ and $B(7)$ (the coordinates are given in parentheses). Find the coordinate of the midpoint of segment $A B$. | $\triangle$ The distance $A B$ is $7-(-3)=10$, half of the distance is 5. Shifting $A$ 5 units to the right (or $B$ 5 units to the left), we get the point with coordinate 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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