A math problem that thinks of himself as a math master

Updated on educate 2024-03-15
23 answers
  1. Anonymous users2024-02-06

    By 96-57 = 39, 148-96 = 52, [39, 52] = 13, that is, the greatest common divisor of 39,52 is 13

    There are three gemstone needles A, B, and C, 64 pieces of No. 1 are the smallest, No. 64 is the largest, and all 64 pieces are on A at the beginning, 1) 1 to B, Shift 1 time, 1) 2 to C, 1 to C Shift 2 times, 2) 3 to B, 1 to A, 2 to B, 1 to B, Shift 4 times, 3) 4 to C, 1 to C, 2 to A, 1 to A, 3 to C, 1 to B, 2 to C, 1 to C 8 times.

    4) After 5 moves, it takes 16 times, and after 64 pieces are moved, it takes 1+2+2+2+2+. 2 to the 63rd power.

    Let n=1+2+4+8+. 2^63,2n=2+4+8+..2^642n-n=n=2^64-1.

    This number, if shifted once a second, would have to move by 585 billion years.

  2. Anonymous users2024-02-05

    Regardless of the credibility of this legend, if you consider moving 64 pieces of gold from one needle to another, always in the order of the top and the bottom. How many moves does this take? A recursive approach is needed here.

    Suppose there are n pieces and the number of moves is f(n).Obviously f(1) = 1, f(2) = 3, f(3) = 7, and f(k+1) = 2*f(k)+1. After that, it is not difficult to prove that when f(n)=2, f(64)= 2 64-1=18446744073709551615 If it happens every second, how long does it take?

    There are 31,536,000 seconds in 365 days in a common year, 31622400 seconds in 366 days in a leap year, an average of 31556952 seconds per year, calculated, 18446744073709551615 31556952 = years This indicates that it would take more than 584.5 billion years to remove these gold pieces, while the Earth has only existed for 4.5 billion years, and the life expectancy of the solar system is said to be tens of billions of years. It has really been 584.5 billion years, not to mention the solar system and the Milky Way, at least all life on the earth, including the Van Pagoda and temples, has long been wiped out.

  3. Anonymous users2024-02-04

    There is a similar "Hanoi Tower Game", you can play it, there are 7 in total, how many times do you have to use 7, and then calculate according to the dimensional equation, how many times will 64 take to complete! and board games on the same principle, 1, 2, 4, 8, 16, 32, 64, 128...

  4. Anonymous users2024-02-03

    If the distance from the school to Baitashan via Yantan Bridge is x kilometers, and the distance from the school to Baitashan via Yintan Bridge is y kilometers.

    x/40-y/60=15/60

    x/60=y/40

    The solution is x=18 and y=12

    The distance from the school to Baitashan via Yantan Bridge is 18 km, and the distance from the school to Baitashan via Yintan Bridge is 12 km.

  5. Anonymous users2024-02-02

    From the school to Baitashan via Yantan Bridge is x kilometers, and from the school to Baitashan via Yintan Bridge is y kilometers, x 40-y 60=15 60

    x/60=y/40

    x=18, y=12

  6. Anonymous users2024-02-01

    One road is long x the other is y

    x/60=y/40

    x/60=y/

    x = 18 km y = 12 km.

  7. Anonymous users2024-01-31

    The two lines are 18 km and 12 km long, respectively.

  8. Anonymous users2024-01-30

    If two roads are respectively x,y kilometers long, then the equation can be listed:

    x/40-15=y/60, x/60=y40.

    The solution is x=540, y=360

  9. Anonymous users2024-01-29

    I've simplified the title, it's a right-angled triangle, the hypotenuse is 1 meter longer than one right-angled side, and the other right-angled side is 5 meters long, you know?

    Suppose the hypotenuse is c, and the right-angled edges are a and b, respectively

    So a 2 + b 2 = c 2

    a+1=cb=5 and substituting the three formulas into one formula to get it.

    a^2+25=c^2=(a+1)^2

    a^2+25=a^2+2a+1

    a=12, i.e. the height of the flagpole is 12 meters.

  10. Anonymous users2024-01-28

    If the flagpole is x meters high, the rope length is x + 1 meter, and there is.

    x^2+5^2=(x+1)^2

    Get x=12

    So, the flagpole is 12 meters high.

  11. Anonymous users2024-01-27

    Let the flagpole be high x, by the Pythagorean theorem: x 2+5 2=(x+1) 2x=12

    So, the flagpole is 12 meters high.

  12. Anonymous users2024-01-26

    There is no picture???

    Set the height of the flagpole h from the inscription:

    h+1)^2=h^2+25

    Solution: h=12

    Do you understand? Feel free to ask.

    Please adopt.

  13. Anonymous users2024-01-25

    Set the height of the flagpole to x

    ac=x-1,bc=5,ab=x

    According to the Pythagorean theorem, yes.

    AC squared + BC squared = AB squared.

    x-1 squared + 5 squared = x squared.

    The solution is x=12

  14. Anonymous users2024-01-24

    If the flagpole is long x, then the rope (x+1).

    Using the Pythagorean theorem (x+1) 2=x 2+5 2

    The solution is x=12

  15. Anonymous users2024-01-23

    Set the flagpole height to xThen the rope length is x 5 2 (x 1) so x=12

  16. Anonymous users2024-01-22

    1 If the length of the three sides of the triangle is 5, 12, and 13, then the area of the triangle is, 5 +12 =13

    So it's a right triangle, so area = 5 * 12 2 = 30

    2 In the triangle ABC, the angle C=90° AC=10 BC=24 then the height on the hypotenuse AB is, AB =AC+BC

    So ab=26

    Area = ac*bc 2 = height 2 on ab*ab

    High on ab = 10 * 24 26 = 120 13

    3 One steamer left the port at a speed of 16 km h and sailed northeast, and the other ship left the port at the same time and sailed southeast at a speed of 12 km h, and they were separated half an hour after leaving the port, km

    Distance = [(16*1 2) +12*1 2) ] = 10 km.

    4 know that the length of the three edges of the same vertex of a cuboid is 3, 4, 12 The length of the longest wooden stick that can be contained in this cuboid is (3 + 4 + 12 )=13

    5. In the RT triangle ABC, the angle c=90° if c=10 a; b=3;4 then ab=

    a²+b²=10²

    b=4a/3

    a²+16a²/9=100

    a²=36a=6b=8

    ab=486 in the RT triangle abc The hypotenuse ab=2 is known then ab +bc +ac =

    ab²+bc²+ac²=ab²+ab²=4+4=8

  17. Anonymous users2024-01-21

    1+5 to explain the situation to you, 1 here is not the number 1 represents the age of the child, you can understand that the age of the child is x, then the adult is 5x, his answer does not set x for you, take it to understand that 1 is the age of the child, 5 is the age of the adult, 5 times the child, right. Take it and get 6, and 6 is the age of 6 children, and it is also the sum of the ages of adults and children. And then there was 36 6

    Get the age of the little one 6 years old. In fact, it is to get an x with a value of 6

  18. Anonymous users2024-01-20

    The mother's age is 5 times that of Xiaowen.

    The age of the mother and daughter is 6 times that of Xiaowen.

    That is, 1+5=6

    Mother and daughter age and is 36

    So Xiaowen is equal to 36 6 = 6

    I hope it helps!

  19. Anonymous users2024-01-19

    ACE= DAC (equal internal wrong angles) and AEC= BAD (equal isotope angles) from AD CE

    Divide the bac equally by ad to get bad= dac

    So ace= AEC, i.e. ace= e

  20. Anonymous users2024-01-18

    Hello, did you make a mistake in your question?

    Looks like it should be"then ace= e" instead"‖"Right? There is no concept of parallelism between corners.

    It's easy to prove equality.

    1.Because of ad ce, e= bad (isotopic angle), ace = diac (inner misalignment angle).

    2.ad divides bac, then bad = dac3Replace the above with the above, you can get ace = e doom, in short, I think it is misprinted, it is an equal sign and not a parallel sign.

  21. Anonymous users2024-01-17

    8,24, divide by 8 to get 9, decompose the prime factor, you get.

    Divisible by three, the sum of the numbers on each digit should be divisible by three, 1+2+3+4=10, so add at least 2 and subtract at least 1

  22. Anonymous users2024-01-16

    8 24 2 1

    PS topic is not rigorous、、It should be an integer multiple、、、

  23. Anonymous users2024-01-15

    One, 1, right.

    2. Right. II, 1, B, III

    The factors of 15 are: 1, 3, 5, 15

    Multiples of 7 (within 40): 7, 14, 21, 28, 35, 4, 60 factors are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

    Multiples of 3 (within 60) are, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60

    Then the numbers that are both factors of 60 and multiples of 3 are 3, 6, 12, 15, 30, and 60

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