How many Olympiad questions can you answer?

First: the front and back of the three cakes are marked as A1, A2, B1, B2, C1, C2 The order of frying is: A1 and B1 together, A2 and C1 together after one minute, B2 and C2 together after one minute, and it takes three minutes to complete.

Second: Suppose the school ends at 12 o'clock at noon, then another 36 hours will be the past day and 12 hours, and it should be 0 o'clock the night after tomorrow, so there will be no sun no matter whether it rains or the sky is sunny!

Third: the rising tide lifts all boats, and the sea water will not reach the fourth level!

Hope it helps!

Question 1 Answer 3 minutes The three cakes are numbered 1, 2, 3 1 and 2 fry one side at the same time for 1 minute, 2 and 3 at the same time for 1 minute, at this time No. 2 has been fried both sides and back, and the remaining No. 1 and No. 3 have one side that has not been fried, and it takes another 1 minute, a total of 3 minutes.

Answer to question 2 No, because it is 12 o'clock when I speak, and it will be 12 o'clock in the evening in 36 hours, and the sun will definitely not rise.

Answer to Question 3 Never, because a rising tide lifts all boats.

Problem 1: For 3 minutes, 3 cakes ABC, A and B are fried first, and C is fried when A is reversed. At this point, the a-frying is finished. Then B and C fry at the same time.

Question 2: Definitely not. In another 36 hours, it will be 12 p.m.

Question 3: No, the sea will rise as well.

Minute. 2.Thirty-six hours later is evening. No.

3.A rising tide lifts all boats. No.

1. 3 minutes.

a b c three loaves.

Fry for the first minute: a positive, b positive.

In the second minute, fry a reverse and c positive.

Fry B anti C reverse in the third minute.

2. No, because 36 hours later it will be night.

Three, how long will it be.

The first 3 minutes.

The second one didn't understand.

The third, if the boat is okay, never, because the rising tide lifts the boat.

Handsome Ninja King, hello :

Add 1 to its numerator, and it equals 1, indicating that the denominator is 1 more than the numerator.

Add 1 to its denominator, it becomes 24 25, and the denominator is more than the numerator: 1 1 2, so expand the numerator and denominator of 24 25 by 2 times at the same time: 48 50, and then subtract 1 from the denominator, so.

The original number is: 48 49

It turns out that the difference between the numerator and the denominator is 1, and after adding its denominator to 1, the denominator is 1+1=2 larger than the numerator

24 25 is the fraction of the fraction, the fraction of the fraction is 2, and the fraction of the fraction without the fraction is (24 2) (25 2) = 48 50

The original score was 48 (50-1) = 48 49

First time: A to B Ship B doubled.

The second time: B to A and A are doubled.

The third time: A to B and B are doubled.

Fourth: B to A Ship A doubled.

Reverse thrust ship A doubles 48, so A originally had 24, then B has 24 less and 48, and B is originally 72That is, at the end of the third time, A 24, B 72

Ship B doubled with 72, so B originally had 36, A had 36 less, and 24, and A was originally 60That is, at the end of the second time, A 60, B 36

Ship A doubled with 60, so A would have had 30, B had 30 less, and B had 36, and B had 66That is, at the end of the first time, A 30, B 66

Ship B doubled to 66, so B originally had 33 A had 33 and 30 less, and A was originally 63i.e. originally A 63, B 33

The process is as follows At the beginning, there were ships 63 in port A and ships 33 in port B

Set up two ports, A and B, each originally had small boats x and y.

A and B for the first time, x-y 2y for the second time, 2(x-y), 2y-(x-y)=3y-x

Thirdly, 2(x-y)-(3y-x)=3x-5y 6y-2xFourth, 6x-10y 6y-2x-(3x-5y)=-5x+11y

6x-10y=48,5x+11y=48,x=63,y=33

Draw a statistical table based on the data on the first floor to calculate. See it more clearly.

If a number is divisible by 9, the sum of the numbers in each bit is also divisible by 9.

There are 5-digit numbers from largest to smallest, and there are ABCDs.

abcde-edcba=(a-e) (b-d-1) 9 (d+10-b-1) (e+10-a), these 5 numbers correspond to abced, but the order is unknown.

Because assuming that a is the largest, a=9, the remaining four numbers are 9-e b-d-1 9+d-b e+1

Because it is also 5 digits from small to large, e is not 0

Assuming e=1, then there are two numbers 8 2 (9-e and e+1), in order to be divisible by 9, so the remaining numbers are 7,98721-12789=85932, which is not true.

In the same way, e=2, then 7 3 6 97632-23679=73953, which is not true.

e=3, then 6 4 5 96543-34569=61974, which is not true.

e=4, then 5 5 4 95544-44559=50985, which is not true.

e=5, while the assumption that e is the smallest 9-e=4 is less than e, is not true.

To sum up, there is no such thing as a five-digit magic number.

There is no such number.

Let a b c d e, the difference = (a*10000+b*1000+c*100+d*10+e)-(e*10000+d*1000+c*100+b*10+a).

a-e)*9999+(b-d)*990

The difference must be a multiple of 9. So the sum of each digit is also a multiple of 9.

Difference = (a-e) * 10000 + (b-d) * 1000 - (b-d) * 10 - (a-e).

a-e)*10000+(b-d)*1000-(b-d+1)*10+(10-a+e)

a-e)*10000+(b-d-1)*1000+9*100+(9-b+d)*10+(10-a+e)

So the hundred digits of the difference must be 9, so a=9

If b=d, then the ten digits of the difference are also 9, so b=c=d=9, then the difference can only be 89991, it is not a magic number, so b must not be equal to d

Difference = (9-e)*10000+[(b-d-1)*1000+9*100+(9-b+d)*10+(e+1)

If e is greater than 4, then the highest digit 9-e is less than 5, indicating that the smallest number is less than 5, so e is less than 5.

When e=0, the number of 10,000 digits is 9-e=9, and the single digit is e+1=1, according to the sum of the numbers is a multiple of 9, the number of tens is 9-1=8, but 99810 is not a magic number.

When e=1, the number of 10,000 digits is 9-e=8, and the single digit is e+1=2, according to the sum of the digits is a multiple of 9, the number of tens is 9-2=7, but 98721 is not a magic number.

When e=2, the number of 10,000 digits is 9-e=7, and the single digit is e+1=3, according to the sum of the digits is a multiple of 9, the number of tens is 9-3=6, but 97632 is not a magic number.

When e=3, the number of 10,000 digits is 9-e=6, the single digit is e+1=4, and according to the sum of the digits is a multiple of 9, the number of tens is 9-4=5, but 96543 is not a magic number.

When e=4, the number of 10,000 digits is 9-e=5, and the single digit is e+1=5, according to the sum of the numbers is a multiple of 9, the number of tens is 9-5=4, but 95544 is not a magic number.

So no matter what the value of e is, it is not a magic number.

Because the quotient is 3 and 2/11, take the decimal as 2 11, 2 11 is a circular decimal number, yes, according to this law, the sum of 2000 numbers after the decimal point is 9000.

The quotient is 3 and 2/11, take the decimal as 2 11, and 2 11 into a circular decimal is: , take two numbers as a group, there is 2000 2 = 1000 (group), the sum of the two numbers is 1 + 8 = 9, and the sum is 1000 9 = 9000

The bottom surface of the upper part is projected on the bottom surface of the lower part, which is actually an inscribed circle of the bottom surface of the lower part.

So the ratio of the base area of the two parts is: up:down = :4

Cone volume = 1 3 base area * height.

Prismatic volume = base area * height.

The height of the two parts is the same, so the volume ratio is up:down=1 3:4, so the volume of the upper part is 12

There are 4 students per class and 14 students remaining. There are 5 in each class, which is exact, which means that there are exactly 14 classes, with a total of 5*14=70 balls.

Profit and loss issues. The total difference is 14

The difference between classes is 5-4=1

Number of classes 14 1=14 classes.

Number of footballs 14 5=70.

Using the pigeon energy theorem to solve, there are 14 classes and 70 balls.

Let x be the number of classes and y be the number of footballs.

Solve the system of equations 4x+14=y

5x=y