Two people take turns to report the number, each time can only report 1 or 2, add up all the numbers reported by two people. (1) who reports the number after the sum is 20, who wins. Think about it: if you report the number first, in order to ensure that you win, how many should you report the first time? How to report next? (2) Who will win after 24, still let you first, can you guarantee to win? Why?

Two people take turns to report the number, each time can only report 1 or 2, add up all the numbers reported by two people. (1) who reports the number after the sum is 20, who wins. Think about it: if you report the number first, in order to ensure that you win, how many should you report the first time? How to report next? (2) Who will win after 24, still let you first, can you guarantee to win? Why?

(1) The person who reports first must report 2 at the first time, and 20-2 = 18 left. 18 is a multiple of 3. Therefore, the sum of the number reported each time and the number reported by another person is always 3, and the person who reports first at the last time is always the person who reports first, so as long as you do this, the person who reports first will win. (2) because 24 is a multiple of 3, if you still report first

Two people count in turn, each time can only report 1 or 2, add up all the numbers reported by two people, who will win if the sum is 10. If you are asked to count first, in order to ensure victory, how many times should you report for the first time? How to report next?

The person who reports the number first must report 1 at the first time, and the remaining 10-1 = 9. 9 is a multiple of 3. Therefore, the sum of the number reported each time and the number reported by another person is always 3, and the person who reports the number first at the last time is always the person who reports the number first, so as long as this is done, the person who reports the number first will win

Two people take turns to report the number, each time can only report 1 or 3, add up all the numbers reported by two people, who report the number after the sum is 31, who wins. Think about it, in order to ensure the victory, how should the first person report the number?

The person who reports the number first must report 3 at the first time, and 31-3 = 28 left. 28 is a multiple of 4. Therefore, the sum of the number reported each time and the number reported by another person is always 4, and the person who reports the number first at the last time is always the person who reports the number first, so as long as this is done, the person who reports the number first will win

Two people take turns to report the number, each time can only report 1 or 2, add up all the numbers reported by two people. (1) who reports the number after the sum is 20, who wins. Think about it: if you report the number first, in order to ensure that you win, how many should you report the first time? How to report next? (2) Who will win after 24, still let you first, can you guarantee to win? Why?

(1) The person who reports the number first must report 2 at the first time, and 20-2 = 18. 18 is a multiple of 3. Therefore, the sum of the number reported by the other person after each time is always 3, and the person who reports the number first at the last time is always the person who reports the number first, so as long as this is done, the person who reports the number first will win. (2) because 24 is a multiple of 3, no matter how many times you report first, only the number reported by the other party will start In the end, if the sum of the two numbers is 3, the other party will definitely win. Therefore, it is not guaranteed that the person who reports first will win, that is, the person who reports after the last time will win

A series calculation problem
If three different non-zero real numbers a, B and C form an arithmetic sequence, and a, B and C form an arithmetic sequence, then a / b =?

a. B and C are equal difference sequence, and a, B and C are equal ratio number
There are
2b=a+c
b^2=ac
Find a / b
So replace c with C
We get 2b-a = B ^ 2 / A
have to
（a-b）^2=0
a=b
therefore
a/b=1

An applied problem of sequence of numbers
1. There are a batch of computers with the original price of 5000 yuan, which are promoted in the computer stores of Party A and Party B. the promotion method of Party A's family is that the unit price of one computer is 4900 yuan, the unit price of two computers is 4800 yuan, and so on, but the minimum price of each computer can't be less than 3100 yuan, and the price of Party B's family is 80% of the original price. A vocational school needs to buy no more than 19 such computers, so which store costs less?

If A1 = 4900, d = - 100
Then A19 = 4900-1800 = 3100
So it's 310 yuan per set
And 80% is 5000 × 80% = 4000 yuan per set
So of course choose a

A practical problem of sequence of numbers
If you buy a house worth 2300W yuan by installment, you should pay 300W down on the day of purchase, and then pay 10W on this day of each month, plus the interest on the previous arrears, with the interest rate of 1%. How much should you pay in the 10th month? How much did you actually pay after paying off all the loans?

The first month repayment is 10W + (2300-300) * 1% = 30W, the second month repayment is 10W + (2300-300-10) * 1% = 29.9w, the third month repayment is 10W + (2300-300-10 * 2) * 1% = 29.8w Repayment in the nth month: 10W + [2300-300-10 * (n-1)] * 1%; repayment in the 10th month: 10W + [2300-300-10 * (10-1)] * 1% = 29.1w full

An applied problem about sequence of numbers
The total amount of timber in the forest is 100000 square meters. The forest grows at an annual growth rate of 25%, and X 10000 square meters of timber will be cut down at the end of each year. In order to realize that the total amount of timber in 10 years will be three times of the original, x (accurate to 0.1) can be calculated
one point nine
Please talk about the thinking and formulation process

The total amount of wood in the first year is A1 = 10
Ten years later, the total amount of wood is Sn = 30
Q is the common ratio, which is the annual growth
The amount of growth in the tenth year is based on the formula
N minus 1 power of an = A1 * q
Don't forget to subtract a fixed quantity x from this question
According to the sum formula of equal ratio sequence
SN=A1-AN*Q/1-Q
I think if you take the numbers one generation back, you can get the answer

At the end of 2001, the number of cars in a city was 300000. It is estimated that 6% of the car ownership at the end of the previous year will be scrapped every year, and the number of new cars will be the same every year. In order to protect the urban environment, it is required that the number of cars in the city should not exceed 600000, so how many new cars should be added every year?

Suppose the number of cars owned at the end of 2001 is B10000, then the number of cars owned at the end of 2001 is b20000, and the number of cars owned at the end of each year is B3 0000. If the number of new cars is x 10000, then B1 = 30. For n > 1, there is BN + 1 = BN × 0.94 + x = bn-1 × 0.942 + (1 + 0.94) x, so BN + 1 = B1 × 0.94n + X (1 + 0.94 + 0.942 +) +94n, when 30 − x0.06 ≥ 0, that is, X ≤ 1.8 when x ≤ 1.8, that is, X ≤ 1.8, when x ≤ 1.8, that is, X ≤ 1.8, when x ≤ 1.8, that is, X ≤ 1.8, that is, X ≤ 1.8, that is, x > 1.8, when 30 − x0.94n + 0.94n + 0.94n-0.94n0.06x = 0.06-0.060.060.06limn, you can be anywhere close to the x0.06limn → + ∞ BN = LiMn → + ∞ BN = LiMn → + ∞ = LiMn → + ∞ + [x0.06 + (0.06 + (30 - (30 - (30 − (30 − (30 − (30) (30 − (30 {(30 − 0.06) (30 − 0.06) 0.06) (30-0.= 1, 2, 3,) Then x0.06 ≤ 60, i.e. x ≤ 36000 vehicles. In conclusion, the number of new vehicles per year should not exceed 36000

How many different four digit numbers can the numbers 1, 2, 3, 4 make up?
Why can't we use p (top right 4, bottom right 4)? Isn't it four numbers out of four? Isn't it all permutation,

It's not clear. If there can't be duplicate numbers, it's 4 * 3 * 2 * 1
If it can be repeated, it's 4 * 4 * 4 * 4