Again, ask the math masters, ask the math masters

1.The next number should be 2

2.I can tell you how. First of all, you can use 1 for thousands, then 2 for hundreds, and the rest of the numbers are freely distributed until there is no other answer, and then use 3 for hundreds ......And so on, after all the numbers are done in hundreds, then use 2 to make thousands and 1 to make hundreds, and in the same way as 1 to make thousands, assign the remaining numbers, excluding 0, to get the answer.

There is also a way to calculate that 1 is more than a few in the thousand place and then multiply it by 9.

The first question will not, the second question, to use the permutations and combinations in high school mathematics, because it is an even number, so the last digit is one of the ones, because there is no repeated number, if the last digit is 0, the first three numbers are selected from the remaining nine numbers, there are sequential arrangements, 9*8*7 kinds, if the last digit is , there are 4*(9*8*7) permutations and combinations (but the first digit cannot be 0, so you have to subtract the permutations of the first digit 0 from it, 4*(8*7) So there are a total of 5*9*8*7-4* 8*7=2296

Find a pattern to fill in the numbers n years ago, the questions I did in elementary school were uninspired and wouldn't.

The second question is not difficult, first talk about the idea, otherwise talk about the process, you still can't talk about this type of question.

Topics like this are considered in terms of special elements, special positions, which include even numbers, especially 0, and special positions are thousands and units.

Because it is an even number, the single digit is even, and there are five cases: 0, 2, 4, 6, and 8.

Separate the 0 case from the other 4 cases.

When the single digit is 0, the other three digits are randomly ranked (the thousand digits cannot be 0), then there are a(9,3)=504 kinds.

When the single digit is other even digits, and the thousand digits cannot be 0, then there are 4*8*a(8,2)=1792 kinds.

Summing up the above, then there are 504 + 1792 = 2296 species.

After writing so much, I feel a little that if the civil service exam is this kind of question, then the civil servant is not a big handful on the street.

China's **ah won't say anything

The first question does not know, the second question 1When the mantissa is not 0, 9*8*7=5042When the mantissa is not 0, because 0 cannot be used as the first 4*8*8*7=1792

As AM DF, the extension of the CB is crossed with M, because AB=BF so AMB FEB

So be=bm

And de am

So ce me=cd ad=1000 1500=2 3me=2be

be+ce=bc=2800

So ce=2 3me=4 3be

i.e. be+4 3be=2800

be=2800*3/7

Mention the common factor x first, and then multiply the denominator of the answer by the root of two, plus one-fifth.

And then using the squared difference formula, the denominator becomes three-fourths minus one-twenty-fifth.

The radical formula runs away from the molecule, so be it...

Original = 10 and root number 3x-x = 160

1+10 and root number 3) x = 160x

Because be=eb=b, b and e

Exchangeable because of known conditions:

a^2=1/4(b

2+2be+e^2)=1/4(b

2+2BE) thus proven:

Sufficiency: If b 2 = e, then a 2 = 1 4 (e + 2b + e) = 1 2 (b + e) = a

So a 2 = a

Proven. Necessity: If a 2 = a, then 1 4 (b 2 + 2b + e) = a = 1 2 (b + e).

So b 2 + 2 b + e = 2 (b + e).

So b 2 = e

Proven. To sum up: the original proposition is true.

Note: This problem uses matrix operations in advanced algebra!

1.Let's take a cart of grain with x tons...

2x+4=7x*2/5

Just untie it.

You see, 7x is the total amount, multiply 2 5 is to be transported away in the morning, and in the morning to be transported is 2 cars + 4 tons.

In fact, it is 2 cars + 4 tons (2 5 in the morning) = = 7 cars * 2 5 (in the amount of the afternoon to express the morning).