One column number: 3,6,8,8,4,2 From the third number, each number is a digit of the product of the first two numbers One column number: 3,6,8,8,4,2 From the third number, each number is a digit of the product of the first two numbers. Then the remainder of the number in this column divided by 3 is ()

One column number: 3,6,8,8,4,2 From the third number, each number is a digit of the product of the first two numbers One column number: 3,6,8,8,4,2 From the third number, each number is a digit of the product of the first two numbers. Then the remainder of the number in this column divided by 3 is ()

The remainder is 2. If you write a few more numbers, you can find the rule. The number in this column is 368842868, 8842868. After that, there will be 842868 cycle, so the number of the number 2008 should be (2008-3) / 6, and the remainder is 1, which is the first number in the 842868 cycle, so the number should be 8, and the 8 / 3 remainder is naturally 2

There is a column of numbers: 3, 6, 8, 8, 4, 2. Starting from the third number, each number is a digit of the product of the two preceding numbers, and the third digit of the column of numbers

Write more numbers to find out the rules. According to the requirements, this sequence should be like this: 3, 6, 8, 8, 4, 2, 8, 8, 8, 4, 2, 8, 6, 8, 8, 4, 2. It can be seen that after removing the first 3 and 6, this is the cycle of 8, 8, 4, 2, 8, 6. So the number 2012 is the number 2010 except 3 and 6. Dividing 2010 by the number of each group is more than 6, So it's the last number of each group, which is 6

There is a column of numbers 3,6,8,8,4,2 , starting from the third number, each number is a digit of the product of the first two numbers, then the number 2008 in this column

3X6=18,
6X8=48
,8X8=64,
8X4=32
,4X2=8
,8X2=16
, 6x8 = 48. By analogy, we will find that there are six numbers in the cycle section 6,8,8,4,2,8
Because the cycle section is 6,8,8,4,2,8, starting from the second number, so
2008-1=2007
2007÷6=334.3
The third number of cycle section 6,8,8,4,2,8 is 8
So the number 2008 in this column is 8

Set up equations to solve these problems (only one unknown number can be set)
1、 When they run to another point, they immediately return. A runs 6.5 meters per second, B runs 5.5 meters per second?
2、 There are 20 questions in a mathematics test paper. If you answer one question correctly, you will get 8 points. If you answer one wrong, you will get 5 points. If you don't answer one question, you will get 0 points. It is known that Xiao Liang took part in the test and did all the questions, but only got 134 points. How many questions did he answer correctly? How many questions did he answer wrong?

1. Let's set the X second for the first time
X（5.5+6.5）=100
12X=100
X=25/3
Suppose he answers question x wrong
20*8-X（8+5）=134
160-13X=134
13X=26
X=2

Set the unknowns according to the following conditions, and then list the equations!
Team a and team B carry out football match. It is stipulated that each team wins one game and gets 3 points, draws one game and loses one game and gets 0 points. Team a and team B have played a total of 10 games, and team a has kept an unbeaten record and got 22 points. How many Games has team a won?
Setting: 1
Column:
A communication company has two ways to pay for mobile phone calls: the first way is to pay 0.6 yuan per minute without paying the monthly rent; the second way is to pay 50 yuan per month and 0.2 yuan per minute. How many minutes of a month's call can make the two ways pay the same cost?
Setting: 1
Column:

First question
Suppose: a wins X Games and draws 10-x games because he is unbeaten
So 3 * x + (10-x) = 22
We get x = 6
Second question
Set up: call for X minutes to make the two payment methods cost the same
So there is 0.6 * x = 50 + 0.2 * X
X = 125

How to set the equation of two unknowns
Give an example

The sum of the two numbers is 25, and the difference between the two numbers is 5
Let these two numbers be x and Y respectively
X+Y=25
X-Y=5
The solution is x = 15, y = 10

Two equations with three unknowns
Is there n equations with N + 1 unknowns···
I did a Google search, and the results didn't give me any answers~

In ancient China, there were such mathematical problems. Such practical problems are called indefinite equations. That is the problem that you said has n + 1 elements, but only n equations
These problems in real life, because to life as the standard (for example, when calculating the number of people, the solution must be a positive integer), so you can get the unique solution
It is very interesting to learn about the history of ancient mathematics in our country. And you will be proud of those achievements in ancient China!

Two equations, three unknowns
3X+2Y-Z=5
2X-Y+3Z=1

If this equation is not a condition, then you should not know
10. Y, Z are positive integers, and then find out the specific value
Otherwise, we can only treat an unknown as a constant and solve ordinary quadratic equation with one variable

What is the meaning of using one unknown to represent another
According to the title of this problem, solve: 2x + 3Y = 7x=______ y=______
5x-2y=8 x=______ y=______
Two thirds x + three thirds y = 2 x=______ y=______
4X = quarter Y-2 x=______ y=______

2x+3y=7    x=(7-3y)/2    y=(7-2y)/35x-2y=8     x=(8+2y)/5   y= -(8-5x)/2x/2+y/3=2      x=(2-y/3)*2   ...

A mathematical problem (x is unknown)
It is known that the parabola y = xsquare - 2x + m intersects the axis X at point a (x1,0). B (x2,0) (x2 > x1)
(1) If the point P (- 1,2) is on the parabola y = xsquare - 2x + m, find the value of M
(2) Let the vertex of parabola y = x square - 2x + m be m. if the triangle AMB is a right triangle, find the value of M

Y = x ^ 2-2x + m, (1) substituting the point P (- 1,2) into the above formula, we get 2 = 1 + 2 + m, that is m = - 1 (2) y = x ^ 2-2x + M = (x-1) ^ 2 + M-1, then the point m is (1, m-1)