Arrange the consecutive odd numbers 1, 3, 5, 7, 9 into the following number table

It can't be 2012, because the middle of the cross is the average of five numbers, 2012 is not divisible by 5, 2015 can, and the five numbers are 393,,401,403,405,413

Let the middle number be x, then.

x-10x-2 x x+2

The addition of x+10 is 5x, and with 2012 5, we get x=, which is not an integer, so it cannot be equal to 2012

2015 5=403, is an integer, so it can be equal to 2015

Observe first, and you will soon see the pattern.

Let the middle number be x, then the number above is x-10, the number below is x+10, the number on the left is x-2, and the number on the right is x+2

Column equation: x+(x-10)+(x+10)+x-2)+(x+2)=2012

5x=2012

x= (can't).

5x=2015

x=403 (can).

These numbers are 393,401,403,205,413, respectively

5*(2i-1)

5*(2i+1)-2 5*(2i+1) 5*（2i+1）+25*(2i+3)

The sum is 5*5*(2i+1) = 25*(2i+1) and must be divisible by 25, so neither 2012 nor 2015 meets the requirements.

1. Because the general term of the number series is 2n-1, when 2n-1=2013, n=1007, that is, the number 2013 is the 1007th term in the increasing arrangement.

It is known that the number table is 8 columns, and the number of rows is calculated: 1007 8 = 125 and 7 remains, so to include the number 2013, this number table has at least 126 rows, and 2011 is ranked in the 7th column of the 126th row of this number table.

2. The numbers in column b of row A are arranged in increasing order in item 8*(a-1)+b, and its value is: 2*(8*(a-1)+b)-1, that is: 16a+2b-17, and the values of column b+1 in row a, column b in row a+1, and column b+1 in row a+1 are:

16a+2b-15,16a+2b-1,16a+2b+1.The sum of the four numbers is: 4*(16a+2b-8).

Let 4*(16a+2b-8)=1416, obtain: 16a+2b-8=354, or: 16a+2b-17=345, 16a+2b-15=347, 16a+2b-1=361, 16a+2b+1=363

That is, when the sum of the 4 large peimeng numbers that are framed is 1416, the 4 numbers framed are 345,347,361,363,9,1, a total of (2013-1) 2+1=1007 numbers, 1007 8=125... 7, 7 columns in 125 rows.

2. The upper left corner of the four numbers in the frame is x, and the other three numbers are x+2, x+16, and x+18

x+x+2+x+16+x+18=1416

4x=1380

x=345 These 4 numbers are ,1, and the positive odd numbers 1,3,5,7,,2013,.According to a certain rule, it is arranged into the following number table.

1.In this table, how many numbers are there? What rows and columns is 2013 in? (e.g. 57 in row 4, column 5).

(1) Let the sum of the numbers in the cross box be s

s=5+15+25+13+17

There is also a median number a

s=（a-10)+(a+10)+(a-2)+(a+2)+a=a+a+a+a+a

5a(2) The rule still exists after the cross box is moved, and s=5a still holds.

3) s cannot be equal to 2012 for reasons 2, 1All additions are odd, and the addition of 5 odd numbers cannot be even.

2.According to the formula s=2012

2012 5= The median cannot be a decimal.

s can be equal to 2015

a=2015/5=43 43-10=33 43+10=53 43-2=41 43+2=45

The five numbers are 33, 31, 43, 45, 53

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**Rule 1: Let the odd number in the middle of the cross box be a, then the sum of the five odd numbers in the box is expressed as [5a] by the algebraic formula containing a (the upper and lower numbers, the sum of the left and right numbers is equal to twice the middle number, plus the middle number for a total of five times).

Conclusion: This shows that the sum of the five odd numbers that can be crossed must be an odd multiple of the natural number p, which is (no matter which odd number a is in the middle, it is always a multiple of three, so the sum is 5a is also a multiple of three).

**Rule 2: The odd numbers that fall in the middle of the cross box and are in the second column are 15, 27, 39....The number of this column can be expressed as 12m+3 (m is a positive integer), and similarly, the odd number that falls in the middle of the cross box and is in the fourth column can be expressed as [12m+7] (the odd number in the fourth column is 4 and 12m+3+4 is larger than the odd number in the second column, respectively).

Application rules: 1) If the sum of the five odd numbers in the cross box is 6035, then the odd number in the middle of the cross box is (6035 5), and this odd number falls in the column from left to right. 【1207=12*100+7】

2) Could the sum of the five odd numbers in the crossbar be 485? Tell me your reasoning. [485 5=97=12*8+1, so it's impossible].

Rule 1: 5a, 5;

Rule 2: 12m+7,1)6035 5=1207,1207=12 100+7 is the fourth column.

2) 485 5 = 97 = 12 8 + 1, because only the number centered on 12 m m m m + 9 can form a cross box, so it cannot be 485.

am+7

The fourth column 5=97 97=12*8+1, in the first column, it is impossible to be a median.

(1) 1207 4 (2) It cannot be 485, there is a law that the sum of countless additions is 5a, and the odd number in the middle is 97, and it can be found that 97 is in the first place, and it cannot be in the middle of the cross box.

1)1027；101

2) Impossible, 485 5 = 97 = 12 * 8 + 1, that is to say, if 485 can, the number in the middle of it is 97, and 97 is the first of the thirteenth line, contradictory, so it can't.

(1) Solution: (5+13+15+17+25) 15=5, so the sum of the five numbers in the cross box is 5 times the middle number 15;

2) Solution: The sum of these five numbers is equal to 2015

Let the number in the middle of the cross box be x, according to the title: (x-10)+(x-2)+x+(x+2)+(x+10)=2015,5x=2015,x=403

Therefore, these five numbers are 393, 401, 403, 405, 413 Give points!!

(1) 7 + 21 + 23 + 25 + 39 = 115 = 23 * 5 (5 times in the middle).

In the middle is the number x (7, 21, 23, 25, 39), which is expressed algebraically and the sum of the five numbers in the cross box is 5x.

2) Move the cross box up and down, left and right, and the frame can be framed to another five numbers, and these five numbers still have this rule.

3) 2012 5=not divisible, so it cannot be equal to 20122015 5=403 divisible, so can be equal to 2015The five numbers are: 387, 401, 403, 405, 419

(1) What is the relationship between the 5 digits in the cross box and the middle digit 23 The sum of the 5 digits = 5 * 23

2) Let the middle number be a, and use the algebraic expression to represent the 5 numbers in the cross box and =5a(3) Promise to move the cross box up, down, left and right, you can frame the other 5 numbers, these 5 numbers still have the rule in (2)? Yes (4) Can the sum of the 5 numbers in the cross box be equal to 2010? Please write 5 numbers, please can't, please explain the reason.

Let the middle number of the cross box be x

then 5x=2010x=402

No, because these numbers are all odd numbers, they can't be equal to 402

(1) Calculate the sum of the five numbers in the cross box, and get 7 + 21 + 23 + 25 + 39 = 115, and 115 = 23 5, and the sum of the five numbers in the sound cross box is 5 times the middle number 23.

2) If the middle number is a, the sum of the five numbers in the cross box is expressed by the formula 5a.

3) By calculation, no matter what kind of five numbers are in the box, these five numbers still have this pattern.

4) If it can be equal to 2010, then according to the above law, there is 5a=2010, and the solution is a=402, thinking that 402 is an even number, so the sum of the five numbers in the cross box cannot be equal to 2010

The middle number multiplied by 5 is equal to the sum of the 5 numbers.

Can't be equal to 2010The largest middle number in the box is 4545 times 5 equals 225

The answer is in group 32 number 4.

The specific algorithm is as follows:

Make sure 999 is the first number.

2n-1=999

n = 500 then determined in the group.

1+2+3+……n>=500 (1)

n(n+1)>=1000 (2)

n minimum take 32

In this case, the left side of (2) is equal to 1056

Observe 1 1 1-1=0*0

4 13 15 17 19 13-4=3*35 21 23 25 27 29 21-5=4*4…… 32 ？？x-32=31*31x=993 is the first digit of the 32nd group, then 999 is the fourth digit.

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