Olympiad practice every day, every day to practice thinking is the Olympiad a few star questions

Updated on educate 2024-04-11
14 answers
  1. Anonymous users2024-02-07

    1. Because the last digit is 8, the single digits of these three consecutive even numbers are 2, 4, and 6 respectively

    So these three numbers are 12, 14, 16, and the smallest number is 12

    2. Yes, out of 4 people, one person shook hands three times. Of the ten people, one person held nine people.

  2. Anonymous users2024-02-06

    1.Single-digit time 8

    Three consecutive even numbers are only 2, 4, and the mantissa of 6 is 8

    According to the first digit is 2, we can quickly get 12 14 16 = 26882Yes. Think of it this way.

    A and B shook hands, then A shook once, B also shook once, and the two shook together 2 times.

    In this way, after everyone has shaken hands, their total handshake must be a multiple of 2, which is an even number.

    Regardless of the number of people who shook hands evenly, the number of times they shook hands always added up to an even number.

    And the number of people who have shaken hands odd times is different, if this number is an odd number, then the odd number is equal to the odd number, that is to say, their total number of handshakes is an odd number, so that the sum of all the handshakes is also an odd number, and this is impossible. So the number of people who have shaken hands odd times must be even.

  3. Anonymous users2024-02-05

    1. Since the last digit of the product is 8, and the last digit that affects the last digit is the single digit of these three numbers, so the single digit is , and because the product of these three consecutive even numbers is more than two thousand, so these three numbers are

    2. Not necessarily, for example, if three people shake hands with each other and shake hands three times, the number of people is an odd number.

  4. Anonymous users2024-02-04

    The answer is a five-star question. Practicing thinking every day is a way to practice thinking every day, which can help children improve their thinking ability, expand their thinking horizons, enhance their problem-solving ability, and improve their learning efficiency, which is an effective way for children to learn Olympiad mathematics. Five-star questions refer to difficult questions, which usually require more thinking, more analysis, and more reasoning to solve, so five-star questions can help children improve their thinking skills and cultivate their problem-solving skills.

  5. Anonymous users2024-02-03

    a b A B.

    1. The place of the first head-on encounter is 2800 kilometers away from Village A, and after the encounter, the two continue to walk to another village and return immediately, and meet head-on at 2400 kilometers away from Village B. At the time of the first encounter, A walked a distance of 2800 and B walked x-2800, and the two together were x, and A walked 5600-x more than B.

    The second time they met, A walked a total distance of x+2400 and B 2x-2400, and the two together were 3x, and A walked 4800-x more than B. The result is: x to 3x = (5600-x) to 4800-x, and the solution is x = 6000.

    2. The place of the first head-on encounter is 2800 kilometers away from Village A, and after the encounter, the two continue to walk, A travels to another village and immediately returns until they return to Village A, and B still does not reach Village A, and A walks to 2400 kilometers away from Village B and meets B.

    At the first encounter, A walks 2800 miles and B is X-2800. At the time of the second encounter, A walked a total distance of 3x-2400 and B walked x-2400. It is not realistic to get the distance traveled from the first encounter to the second encounter B to -400, i.e., this is not possible.

    3. The place of the first head-on encounter is 2800 kilometers away from Village A, after the encounter, the two continue to walk, B travels to another village and immediately returns until returning to Village B, A still does not reach Village B, and B walks to 2400 kilometers away from Village B and meets A.

    When they first met, A walked a distance of 2800 and B walked X-2800, and the two walked x together, and B walked X-5600 more than A. In the second encounter, A walked a total distance of x-2400, B walked 2x+2400, and the two walked 3x together, and B walked x+4800 more than A. Get:

    x ratio 3x = x - 5600 to x + 4800 gives x = 10800

  6. Anonymous users2024-02-02

    The distance between the two villages is: 2800x3-2400=6000 (km).

  7. Anonymous users2024-02-01

    Solution: The distance between the two villages is more than x kilometers.

    2800 (x-2800)=(x+2400) (2x-2400) to get x -6000x=0

    Solve x=0 (undesirable, discarded) or x=6000

    A: The distance between the two villages is 6,000 kilometers.

  8. Anonymous users2024-01-31

    1. Set x questions correctly and answer (19-x) questions incorrectly.

    5x-2(19-x)=81

    7x-38=81

    7x=119

    x=17 2=48 (person).

    1) Class: 48 + 8 = 56 (people).

    2) Class: 48-8=40 (person).

    3. The second room: 90 1 3-8=22 (people).

    The first room: 90 2 3 + 8 = 68 (people).

    4 = angle) 5, set. Jack bauer.

    X years old, uncle.

    3x years old. 3x-x=20

    2x=20x=10

    3 10 = 30 (years).

    5 (5-2) = 30 (kg).

    Corner: 70 sheets.

    8 points: 30 sheets.

    I don't know the specific formula, I made up the ......)

    8. There are 27 students.

    108 saplings.

    33 = 63 (yuan).

    63 7 + 33 3 = 20 (yuan).

    20-11=9 (yuan).

    10. The first pile: 178 2 + 6 = 95 (tons).

    Second pile: 178 2-6 = 83 (tons).

    11. Class A: 104 2-2+12=62 (person) Class B: 104 2+2-12=42 (person).

    62-34=4 (person).

    Note: Some questions don't need to be calculated, they are all made up, and it should be okay.

  9. Anonymous users2024-01-30

    Let the two two-digit numbers on the left be.

    10a+a,10b+b,(10a+a)(10b+b)=121b;

    The four digits on the right are.

    1000m+100m+10n+n=11(100m+n).

    121ab=11(100m+n)

    ab=9m+(m+n)/11.

    Obviously, 0 m+n to the left of 0 is a multiple of 11, in the case of:

    m=2,n=9;m=3,n=8;m=4,n=7;m=5,n=6;m=6,n=5;m=7,n=4;m=8,n=3.

    Check one by one, and know that there is only one set of suitables: m=3, n=8

    At this point, the left = ab = 4 7 = 28;

    Right = 9m+(m+n) 11=9 3+(3+8) 11=28

    Therefore, the original equation is:

    44 77 = 3388 or 77 44 = 3388.

  10. Anonymous users2024-01-29

    He could ask a neighbor to borrow a cow so that there were 20 of them.

    Eldest son: 20 1 2 = 10

    Second son: 20 1 4 = 5

    Three sons: 20 1 5 = 4

    10+5+4=19, which is just in line with the requirements.

    Finally, he returned the borrowed cattle to his neighbors.

  11. Anonymous users2024-01-28

    The least common multiple of 20 is 20 and 20 is half plus a quarter plus a fifth is exactly equal to 19, which is a clever solution that uses a kind of thought.

  12. Anonymous users2024-01-27

    Big brother: 19 2 = , rounded = 10

    Second brother: 19 4 = , rounded = 5

    Little brother: 19 5 = , rounded = 4

    Well. Just @@

  13. Anonymous users2024-01-26

    This question is actually a text question. Half of the total, a quarter of the total, and a fifth of the total. This total does not refer to 19, but to say that the three brothers add up to exactly 19 when they get their hands. Let the total be x. Namely.

    1 2 + 1 4 + 1 5) x = 19, so x = 20, so the eldest son has 10 heads, the second son gets 5 heads, and the younger son gets 4 heads.

  14. Anonymous users2024-01-25

    Mathematics is an important basic subject, to lay a good foundation of mathematics, we should start from primary school, the knowledge of mathematics in primary school seems to change a lot, but from the requirements of quality education, we should focus on cultivating students' sense of innovation and practical ability. In recent years, with the full implementation of quality education in primary schools, teachers, students and parents have increasingly felt the importance of letting students who are interested in mathematics learn a little Olympiad. This book has changed the problems of many previous Olympiad books with many descriptions, difficult topics, and low interest in learning, and has reflected the characteristics of several aspects in the process of compilation

    Synchronized with the textbook] Improve and deepen on the basis of mastering the basic knowledge of the course. [Easy for self-study] The example questions give detailed ideas and solution processes to help students master the methods. [Hierarchical] The topics are from easy to difficult, the level is clear, and the connection with the reality of life is noted.

    Thinking] Enlightening thinking, multiple solutions to one question, and there are ideas in the answers. [Open-ended] Some of the questions are open to conditions, solutions, and results, giving full play to students' imagination and creativity. [Interesting] The topics in the lower grades focus on observation, drawing and measuring by hand, and have the characteristics of animation.

    In order not to increase the excessive burden on students, each lecture in this book arranges three days of self-study and practice, the first day of self-study "special analysis" and "basic refinement", and do two imitation training questions; On the second day, do two consolidation exercises; On the third day, do two extended improvement questions, so that you can learn and practice every day.

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