Stumped the 4th grade math problem for the master s degree, and stumped the 4th grade math problem f

If you press the law of "-" backwards, you will get 4

The simplest rules, don't think about the complicated. It is that these three groups of numbers are all arranged from smallest to largest.

This is categorized by tone.

1, 3, 5, 7, 8 are the first tone, and 5, 9 are the third.

2, 4, 6 is the fourth tone.

Question from a quick search. The difference between the sum of odd and even digits is 4, and I am ashamed to think about this question for almost 20 minutes. I just wanted to show my face, but it turned out that there were already two people in front of me.

Landlord, is there a mistake, there are 5 in the first group, and there are 5 in the third group, all of which are divided according to the same law, how can there be different groups with the same number? If it is allowed to exist, then 9 can also be grouped together.

Du Nima stupid) force) ah 5 read a sound?? The ??? taught by your foreign language teacher Nyima

1. The number in the second row 2=3-1 3=5-1 6=7-1 should be followed by 8-1=7 ......

2. The number of the third row 5=3+2 9=5+4 should be followed by 6+7=13 ......3. The number 8=1+2+5 in the first row should be followed by 3+4+9=16 ......4. Fill in the numbers regularly.

I think it's said in tone, one, four, three! There is no "5" in the first group of the original question

Divide 128 by 33, and regard the divisor 33 as (30) test quotient, quotient (4), quotient bias (large), quotient adjustment (small), change quotient (3), and the remainder is (29).

Have fun.

Divide 128 by 33, and regard the divisor 33 as (30) test quotient, quotient (4), quotient bias (large), quotient adjustment (small), change quotient (3), and the remainder is (29).

At this stage, there are many such test quotient questions, in order to learn three digits divided by two digits in the future to lay the foundation, you can find a few more such questions to do, the divisor is two digits, you can use the "rounding" method to treat the divisor as a whole ten test quotient.

This mainly helps students learn how to test the quotient.

Divide 128 by 33, and regard the divisor 33 as (30) test quotient, quotient (4), quotient bias (large), quotient adjustment (small), change quotient (3), and the remainder is (29).

Divide 128 by 33, and regard the divisor 33 as (30) test quotient, quotient (4), quotient bias (large), quotient adjustment (small), change quotient (3), and the remainder is (29). Hope to adopt.

Divide 128 by 33, and regard the divisor 33 as (30) test quotient, quotient (4), quotient bias (large), quotient adjustment (small), change quotient (3), and the remainder is (29). Hehe.

When Xiao Ming did a division problem, he mistakenly took the divisor 48 as 84, and the quotient obtained was 37 and 12What is the correct quotient?

Divide 128 by 33, regard the divisor 33 as 30 test quotient, quotient 4, quotient is larger, adjust the quotient smaller, change quotient 3, and the remainder is.

Divide 28 by 33, and regard the divisor 33 as (32) test quotient, quotient (4), quotient bias (1), quotient adjustment (5), change quotient (3), and the remainder is (29).

Regarded as 32, the quotient is 4, the quotient is larger, adjust the quotient to smaller, change the quotient to 3, and the remainder is 29.

30, 4, Big, Small, 3, 29

It seems to be a trial division.

Think of 33 as 30 test quotient, quotient 4, large change quotient 3 remainder 29

30, 4, Big, Small, 3, 29

OK test quotient is estimate, guess.

We can use the "what-if" method to calculate. If all 98 coats are considered to be double-breasted coats, there are a total of 98 8 784 buttons, and it is known that the buttons of the double-breasted coat are 182 more than the single-breasted coats, then the total number of buttons is 784 minus the difference between the entire double-breasted and single-breasted buttons by 182, and 602 is obtained.

Divide 602 by the sum of the buttons of each double-breasted coat and each single-breasted coat, i.e.: 602 (8+6) 43 pieces, 43 pieces is the number of single-breasted coats.

Column: [98 8-182] (8+6).

43 (pieces) 98-43 = 55 (pieces).

A: 43 single-breasted coats and 55 double-breasted coats.

There are more in a double row than a single row: 8-6 = 2.

If there are as many double-rows as there are 49 single-rows, it should be more: 49*2=98.

182 more, actually. More than 49 are described as double-rowed.

With a single row for a double row, it is more: 8 + 6 = 14 So, the number of changes is: 84 14 = 6 pieces.

That is, the double-row ones: 49 + 6 = 55 pieces. There are 49-6 = 43 pieces in a single row.

（182+98*6）÷（8+6）

55 pcs. 98-55 = 43 pieces.

So there are 55 double-breasted coats and 43 single-breasted coats.

There's something wrong with that.

Each double-row coat has 2 more buttons than a single-row coat, 182 2 = 91, so a double-row coat has 91 more than a single-row coat, and there are 98 pieces in total, which is not an integer.

Double-row coats are 182 more than single-row coats 182 (8-6) = 91 (pieces).

Single-row coat 98-91 = 7 (pieces).

8-6 is a double-breasted coat with more buttons than a single-breasted coat; And the condition gives a total of 182 more buttons, then 182 2 = 91 is the number of pieces more than a double-breasted coat than a single-breasted coat; Using 98-91=7 is the number of single-breasted coats.

182 (8-6)=91 is more double buttons than single buttons.

98-91) 2 is the number of single-button coats.

98-91) 2+91 is the number of double-button coats.

This is a chicken-rabbit cage problem.

Suppose all the clothes are double-breasted, 98*8=784 (pieces), assuming that all the clothes are single-breasted, 98*6=588 (pieces), double-breasted is 784-588=196 (pieces), assuming that double-breasted is more than single-breasted, 196-182=14 (pieces), assuming that all clothes are double-breasted, then single-breasted is 784-182=602 (pieces), so single-breasted 602 14=43 (pieces).

Double-breasted 98-43 = 57 (pieces).

182 2 = 91 = double-breasted coat.

98-91=7=Single-breasted coat.

Let the original number be x, then the wrong number is 10x, then 10x-x=, so x=, x=, that is, this decimal is .

If the original number is a decimal place, then the number after the missing decimal point is 9 times larger than the original;

If the original number is two decimal places, then the number after the missing decimal point is 99 9 11 times larger than the original;

If the original number is three decimal places, then the number after the missing decimal point is 999 9 111 times larger than the original;

Cannot be dispelled by 11 and 111.

So, the number turned out to be.

is a decimal place.

So what is missing is also a decimal place.

So it's 10 times the original, so it is.

Hello, you have to make it clear that you forgot about a few decimal places.

Otherwise, you can't do it.

- - I Khan. There is such a difficult question in the fourth grade.

The number with the decimal point missing turned out to be.

It should be 6 + 9 + 8 = 23 fish.

The reason is as follows: the common feature of these 6, 9, and 8 fish is that they all have bodies, at least half of them, so the 6, 9, and 8 fish are taken from different fish, and the number of 23 fish is given.

I would like to repeat the answer upstairs again, it doesn't make much sense. What I want to say now is what is behind this incident.

I don't think there's any point in asking why, to put it bluntly, it's a play on words, and the words of the game are a fish, and the author of the question thinks that the lack of a head, a lack of a tail and a body can also be counted as a fish, and its purpose is just to mess with our brains. It shows the divergent and broad nature of his thinking. It was as if he himself felt that he had an advantage.

But in my opinion, it's really not advisable.

It shows that China's education is still divorced from reality, and it shows that there are many places that run counter to quality education, which is sad, lamentable, and helpless.

Wishing you a happy new year! All wishes come true! Good luck!

The upstairs is right! Current**.

No head, no tail, and half of the body are all one. So 6 + 9 + 8 = 23, this answer is fine.

0 6 has no head, 9 has no tail, 8 is only half, it is 0.

6+9+8=23?

The head of 9 is 0, the tail of 6 is 0, and half of 8 is, isn't it 0.

The answer is 18 hours, and the analysis is as follows:

After fundamental analysis, from A to B is downstream, and from B to A is countercurrent, and the equation is listed:

v boat + v water) * 3 = (v boat - v water) * v boat = 5v water from it, which will be brought into the equation on the left, and the total distance obtained is s = 18v water, so the barrel goes down the water, and it takes t v water = 18 hours from a to b.

Suppose the speed of the ship is v and the water velocity is v ab and the distance is s, then s (v+v)=3

s/(v-v)=

So s (v) = 18 hours hehe.

Get to the point! The barrel itself is not accelerating! Let the speed of the ship be x and the speed of the water be y, we can get (x+y)*3=1, (x-y)*solution y=1 18, so it takes 18 hours.

Of course, because the decimal point of an addition is shifted one place to the left, the addition is reduced by a factor of 10, and the result is less wrong than the correct result. That is, since the addition is reduced by a factor of 10, it is doubled less than the real addition. i.e. one additive is, so the other is.

Column: Answer: One addition is the other.

, because one of the numbers is shifted to the left, so that one of the numbers becomes a multiple of the original, that is, the number is reduced by a multiple of the original number, that is, the original number that is wrong, and the other number is.

(A decimal point of an addition is shifted one place to the left, which is equivalent to a 10-fold expansion of an addition, i.e., the result is a 9-fold undercount of the addition.)

The correct addition is the one that is correct and the one that is wrong is overcome.

This is the added number that is misunderstood.

This is the second plus.

Find the number of the difference between the wrong additions, which represents the multiple relationship between the wrong additions, and indicates that one is, ! The two additions are and !!

Let one addition have a value of x and another y

then the incorrect addition is.

x+y=Solve equations.

The method is to divide this by one of these numbers, and the other number is

Method 2: Let one addition have a value of x and another y

then the incorrect addition is.

x+y=

...It's all tough, it's all a decimal...

It's really popular.

One plus is, the other is.

One addition is for the other.

x+y=

Simultaneous equations yield:

So x= y=

Put, since it is shifted one place to the left, so the final sum should be a multiple of one of the numbers, and it is a multiple of the difference of that value, divide by this is one of the numbers, and the other number is.

The difference between the "right sum" and the "wrong sum".

The purpose of the division is because the decimal point is moved to the left and is reduced by a factor of 10.

The original two additions are each and ! Hehe.

Let the two numbers be a and bThen there is a system of equations: a+b=; a+b/10=。The calculation shows that a=; b=。

And, the sum of the two numbers to make the difference, get, is the multiple of the wrong number, divided by get, then get another number.

**No, you still need to find an expert, right?

There are only stupid teachers, no stupid students!

This ,,, is normal... I can read long English texts**, but I can't do elementary school English questions... I'm not a teacher ...

This kind of thing is normal, and the IQ of the person who asks the question is generally higher than that of the fourth grade.

The old teacher was stumped by the fourth-grade math problem and asked for help**Looking for authoritative analysis?

Is it normal for the teacher to not be able to do the questions?

The answer is x, because I don't know.