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Single digit, 1+2+......9=45

Two digits, from 10 to 99

The ten digits appear 10 times from 1 to 9, i.e. 10 to 19 ,......90 to 99, so a total of 10*(1+2+......9)=450

If 0 is not counted, then 1 goes from 11, 21, to 91, 9 times.

then the others are also 9 times.

So it's 9*(1+2+......9)=405

So the two-digit number is 450 + 405 = 855

Three digits. hundreds, 1 from 100 to 199, 100.

So the others are also 100 pcs.

So it's 100*(1+2+......9) = 4500 decimal digits, 1, is 110, 111 ,......119，210，……919 appears 10*9=90 times.

So the sum of the ten digits is 90*(1+2+......9) = 4050 digits, 1, from 101 to 991, is 90.

So the single digits are a total of 90 (1+2+......9) = 4050, so the same goes for the three-digit number 4500 + 2 * 4050 = 12600. Four digits, the thousand digits are 1000*(1+2+......9) = 45000 hundreds, ten digits and single digits are 900*(1+2+......9) = 40500 so a total of 45000 + 40500 * 3 = 166500 so total.

The answer is 18000, method 1: the number from 1 to 9 is 45, the number from 10 to 19 is 55, the number from 90 to 99 is 135, the number from 1 to 99 is 900, the number from 100 to 199 is 1000, and the number from 900 to 999 is 1800, they are in a series of equal differences, so the number from 1 to 999 is 14500 for the number to 1999, The sum of the numbers from 2000 to 2999 is 15500 and so on, with a tolerance of 1000 and a series of 100 terms with 4500, the sum of 10 terms is 166500, so the sum of the numbers from 1 to 9999 is 180000

Method 2: Fill all numbers into four digits, such as 1 to make up 0001, 16 to make up 0016, so it seems to be the arrangement of ten numbers from 0 to 9.

A total of 10,000 numbers, 40,000 numbers, 0-9 each appear 4,000 times 45 times 4,000 times 4,000 = 180,000

9999 + 1 = 10000 digits sum = 37, excluding zeros.

9990 + 10 = 10000 digits sum = 28, including a zero.

9900 + 100 = 10000 digits sum = 19, including two zeros.

9000 + 1000 = 10000 digits sum = 10, including three zeros.

9009 + 991 = 10000 digits sum = 37 is calculated without zero.

9909 + 91 = 10000 digits = 37 is calculated without zero.

9099 + 901 = 10000 digits = 37 is calculated without zero.

9890 + 110 = 10000 digits = 28, calculated as including a zero.

The sum of 8600 + 1400 digits = 19, calculated as containing two zeros.

This leads to the inclusion of three zeros. There are four groups.

Contains two zeros. There are 45 groups.

Contains a zero. There are 450 groups.

Contains no zeros. There are 4500

Add 5000 as the sum of the middle numbers = 5

Positive integers 1, 2, 3 ......The sum of all numbers of integers, 9998, 9999 is an extended formula: 9*5+(1+2*9)*9*5+9*5*10(1+3*9)+9*5*100(1+4*9)+9*5*1000(1+5*9).

Make up all numbers into four digits, e.g. 1 makes up 0001 and 16 makes up 0016

In this way, it looks like a ten-digit arrangement of 0-9.

A total of 10,000 numbers, 40,000 numbers, 0-9 appear 4,000 times each.

There are a total of 9999 single digits, 1 9 each 1000 times, 0 999 times, and the sum of the single digits can be 45000

There are 9999 tens of digits, 1 9 1000 times each, and 0 999 times, and the sum of the tens is 45000

The number of hundreds and thousands is the same as above, and the sum is 45000*4=180000

Answer: 180000

Explanation: 1+9998=9999 numbers and 9+9+9+9=362+9997=9999 numbers and 9+9+9+9=369998 2 4999 A total of 4999 groups.

1.Solution: Bringing x=0 into the original equation yields: k crack search + 3k-4 = 0k + 4) (k-1) = 0

k1=-4 k2=1

2.You can use a multiplication formula (perfect square formula or square difference formula) to decompose the factors to get two unary linear equations.

2p)/2]²=2p²-1

p²=2p²-1

The value of 1=p p is p1=1 p2=-23Solution: -x+2 Hunger 6=k x

x²+2√6x-k=0

x²-2√6x+k=0

There is only one rotten source shouting a public point, 0

2√6)²-4k=0

24-4k=0k=6

1.What is the number that is 50% more than 180?

2.For a project, a certain engineering team worked for 7 days to complete 20% of it, according to this calculation, how many days will it take to complete the project?

28 days. 3.The teacher corrected the essays of the two classes, 30% of the total number was approved in the first lesson, 40% of the total in the second lesson, 24 books in the first lesson, and how many books were approved in the second lesson?

32 copies. 4.There are 24% of the students in the playground kicking the shuttlecock, 36% of the students are playing with the pocket, the students who play the pocket are 6 more than the students who play the shuttlecock, how many students are playing the pocket on the playground now?

18 people. 5.In a brine with a salt content of 20%, how much less salt is than water?

After the price of a product is reduced by 20%, the price is increased by 20%, and the current price is higher than the original price? Or is it lower than the original price?

Therefore, the current price is lower than the original price.

1 100x（1+50%）=150

5x7 = 35 days.

80x40%-32

6÷12％x36%=16

5 has a water content of 75

Since the content is 20%, let the salt be 20 and the water be 80, so the salt water is 100, and the salt is less than the water (80 20) 80 = 75%.

6 is lower than the original price, which is 96

It's all a matter of proportions and unary equations, take a good look at the book and do the math yourself......It's not hard.

1,11 min. 200x12 (200+20)2,190pcs.

3. Truck: 284km passenger car, 568km 852 (1+2) 3 and 1 4 hours 980 square meters.

6x8÷10

The original number is Let the original number be x 10x-x= x=

1.200 12 (200+20)=minutes) Answer: It takes about 11 minutes to reach the end of the line.

2.A: You can do 190 now.

3.852 3 1 = 284 (km).

852 3 2 = 568 (km).

A: 568 km for buses and 284 km for trucks.

5. s=6×8÷2=24

h=24×2÷10=

6.100x+10y-(10x+y)=10x+y=original number.

On [0,1) is the increment function.

then 0<=a0

So f(x) at (-1,0) is a subtractive function.

Domain. -1a^2-4

a 2-a-2<0,-10,a 2-4>0 multiplier function. a-20

a>2,a<-1

So 2a 2-4

a^2+a-6<0

34-a^2

a^2+a-6>0

a<-3,a>2

So 2 so. √3

a, b, c are all integers, and a 2 + b 3 = c 4, find the minimum solution of c: a 2 + b 3 = c 4

b^3=c^4-a^2

b 2 * b = (c 2 + a) * (c 2-a) because a, b, c are all integers.

So: c 2 + a = b 2 (1) or c 2 + a = b (3).

c^2-a=b （2） c^2-a= b^2 （4）

1) + (2) or (3) + (4).

Equally: 2c 2 = b 2 + b

Multiply the two sides by 4 to get 1 plus 1 to get

8c^2+1=4b^2+4b +1

Recipe, obtain: 8c 2 + 1 = (2b + 1) 2

Since b is an integer, the minimum value of (2b+1) 2 is 1, i.e., the minimum value of 8c 2+1 is 1

So, 8c 2+1=1

Solution: c=0

c minimum is 0

I know what you mean, you don't understand in breaking down 2 formulas, right?

Indeed, integers also include negative numbers, so C 2-A and C 2+A cannot tell who is bigger and who is smaller, because A may be ...... negative number

But you can think of it this way, if one of them is negative, then c can't have a minimum, and c can't be infinitely smaller, so c can only be positive, and the same can be carried together, and a and b are also positive.