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What are the classifications of computers?

① According to the form of information and processing methods can be classified: 1, electronic digital computer: all information expressed in binary numbers. 2, electronic analog computer: the internal form of a continuous change of analog voltage, the basic operational components for operational amplifiers. 3, hybrid electronic computer: both digital and analog, design

It is known that there is a two digit number whose ten digit number is less than one digit number by 2. Three times of the product of the ten digit number and the one digit number is just equal to the two digit number

Let X be the number of one digit, then the number of ten digits is (X-2). According to the meaning of the question, we can get 3x (X-2) = 10 (X-2) + X. after sorting out, we can get 3x2-17x + 20 = 0, that is, (x-4) (3x-5) = 0. We can get X1 = 4, X2 = 53 (not suitable for the meaning of the question, rounding off), then X-2 = 4-2 = 2. A: these two digits are 24

For a two digit number, the number on the tenth digit is 3 times smaller than the number on the single digit. The sum of the number on the tenth digit and the number on the single digit is equal to 4 / 1 of the two digits

Let X be the number of one digit and y be the number of ten digits
Because "the number in ten is three less than the number in one"
So y + 3 = X
Because "the sum of ten digits and one digit is equal to 4 / 1 of the two digits"
So y + x = (Y × 10 + x) / 4
Substitute x = y + 3 into
Y = 3
Substituting y = 3 into y + 3 = x, we get x = 6
So the double-digit number is 36

It is known that an n-digit number is an arrangement of such n-digits as 1,2,3,... N (a positive integer whose n is less than or equal to 9), and its first k digits form a positive integer which can be divisible by K (k = 1,2,3,. N). We call it a "smart number". Try to find all the six digit "smart numbers" and explain the reason

K value 1-6 for analysis, the preliminary conclusion is: the first digit is unlimited; the second digit is even; the first three digits add up to a multiple of three; three or four digits constitute a multiple of four; the fifth must be 5; the end must be even
OK, experiment
result:
one hundred and twenty-three thousand six hundred and fifty-four
three hundred and twenty-one thousand six hundred and fifty-four

Three times of the number on the single digit of a two digit number plus one is the number on the tenth digit. The sum of the number on the single digit and the number on the tenth digit is equal to 9. What is the two digit number?

Let X be x, x + 3x + 1 = 9, the solution is: x = 2, then the single digit is: 2 × 3 + 1 = 7, answer: the two digit is 72

Three times of the number on the single digit of a two digit number plus one is the number on the tenth digit. The sum of the number on the single digit and the number on the tenth digit is equal to 9. What is the two digit number?

Let X be x, x + 3x + 1 = 9, the solution is: x = 2, then the single digit is: 2 × 3 + 1 = 7, answer: the two digit is 72

There is a two digit positive integer whose sum of digits is n. if the digits on the ten digits are exchanged to form a new two digit number, find the sum of the new two digits and the original two digits
Who will? Who will answer

Suppose that the previous ten digit is a, the single digit is B, and a B is a positive integer greater than 0
a+b=n
Then (10a + b) + (10b + a) = 11n

N positive integers A1, A2 , an satisfies the following conditions: 1 = A1 ＜ A2 ＜ < an = 2009; and A1, A2 The properties of any n-1 different numbers in an
Arithmetical mean is a positive integer, find the maximum value of n
It's up to you to clean up

Let A1, A2 The number of n-1 left after removing AI in an
The arithmetic mean is a positive integer Bi (I = 1,2,...) n. Namely
bi= [(a1+a2+...+an)-ai]/(n-1)
Therefore, for any 1 ≤ I < J ≤ n, Bi BJ = (AJ AI) / (n-1)
Thus, n-1 ∣ (AJ AI)
Because b1-bn = (an-a1) / (n-1) = 2008 / (n-1) is a positive integer, n-1 ∣ 23 × 251
Because an-a1 = (an-an-1) + (an-1-an-2) + +（a2-a1）≥+(n-1)+(n-1)+…… （n-1）=（n-1）2
(n-1) 2 ≤ 2008, then n ≤ 45, combined with n-1 ∣ 23 × 251, ∣ n ≤ 9,
On the other hand, let A1 = 8 × 0 + 1, A2 = 8 × 1 + 1, A3 = 8 × 2 + 1 If A8 = 8 × 7 + 1, A9 = 8 × 251 + 1, then these nine numbers meet the requirements of the problem design. To sum up, the maximum value of n is 9
Ensure that there is no error in this solution!
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There are several numbers, the first is A1, the second is A2, the third is A3, and the nth is an, if A1 = - 1 / 3

The first number is denoted as A1, and the nth number is denoted as an, if A1 = - 1 / 2, from the second number, every number is equal to the reciprocal of the difference between 1 and the number in front of it, A1 = - 1 / 3a2 = 1 / [1 + 1 / 3] = 3 / 4 A3 = 1 / [1-3 / 4] = 4 A4 = 1 / [1-4] = - 1 / 3... It can be seen that every three are a cycle. 2003 / 3 = 667... 2, so A2003 = A2 = 2 / 3A