(1) If one number is randomly selected from the 100 integers 1-100, the probability that the number can be divided by 8 is zero (2) If you toss an even coin twice at will, the probability that at least one coin will face up is zero (3) If you throw two dice at will, the probability that the sum of points is greater than 8 is 0 (4) At present, there are 7 playing cards, 2 of which are "King". After washing, randomly select 2 of them, and the probability of drawing at least 1 trump card is 0

(1) If one number is randomly selected from the 100 integers 1-100, the probability that the number can be divided by 8 is zero (2) If you toss an even coin twice at will, the probability that at least one coin will face up is zero (3) If you throw two dice at will, the probability that the sum of points is greater than 8 is 0 (4) At present, there are 7 playing cards, 2 of which are "King". After washing, randomly select 2 of them, and the probability of drawing at least 1 trump card is 0

(1) If one number is randomly selected from the 100 integers of 1-100, the probability that the number can be divided by 8 is 12 / 100 = 3 / 25 (2) toss an even coin twice at will, and the probability that at least one time the coin faces up is 3 / 4 (3) toss two dice at will, and the probability that the sum of the number of points is greater than 8 is 10 / 36 = 5 / 18 (4)

What is the probability that any number from 1 to 100 can be divided by 2 and 3

The number that can be divided by 2 and 3, that is, the number that can be divided by 6, [100 / 6] = 16]
P (number divisible by 6) = 16 / 100 = 0.16

Write the numbers 1, 2, 3, 4 and 5 on the five cards, then mix them and arrange them into a line. The probability that the number can be divided by 2 or 5 is zero___ ．

According to the meaning of the question, we know that this question is a classical probability type. All the events in the experiment are arranged in a row randomly by five cards, with a total of A55 = 120 results. The events that meet the conditions are arranged with the last three digits of 2, 4 and 5, with a total of c31a44 = 72. According to the probability formula of the classical probability type, we get P = 72120 = 35, so the answer is: 35

Take any two numbers from the set {1,2,3,..., 100}, so that their sum can be divided by 4. How many kinds of methods (regardless of the order) are there?

All elements can be divided into four categories: multiples of 4, numbers divided by 4 to 1, numbers divided by 4 to 2, and numbers divided by 4 to 3. There are 25 elements in each category. Therefore, multiples of 4 have the following situations: (1) both numbers are multiples of 4. Choose 2 from 25. There are 300 kinds of C (25,2) = 2

From 1, 2 How many of these 100 numbers (regardless of the order) can you take any two numbers so that the sum can be divisible by 4

1 to 100 can be divided into intermediate cases, which are 4n-3,4n-2,4n-1,4n (n = 1,2... 25)
Since (4n-3) + (4N-1), 2 (4n-2) and 2 * 4N can be divided by 4, there should be C (25,1) * C (25,1) + C (25,2) + C (25,2) = 25 * 25 + 25 * 24 / (2 * 1) + 25 * 24 / (2 * 1) = 1225

In the positive integer less than 20, take three different numbers, so that their sum can be divided by 3, then the number of different methods is______ ．

In 1-19, there are 6 numbers divided by 3, 7 by 1, and 6 by 2. There are four kinds of numbers and ways that can be divided by 3, which are {0, 0, 0}, {1, 1, 1}, {2, 2}, {0, 1, 2}, so there are C36 + C37 + C36 + 6 × 7 × 6 = 327

There is a number whose ending number is 2 or 4 or 6 or 8, which can be divided by 3 and is the square of 2 (the square is a positive integer). How much is this number?
Please prove if you have, and if you don't, please prove if you don't!

Suppose there is such a number x, then x = 2 ^ n, X can be divisible by 3, so there is an integer K0, such that x = 3 * K0, then 2 ^ n = 3 * K0 = > 2 * 2 ^ (n-1) = 3 * K0 = > the left side is an even number, and the right side must be an even number, that is, K0 is an even number, that is, K0 = 2 * K1, K1 is an integer, so 2 ^ (n-1) = 3 * K1, just like the front side, do this all the time

Find the number of all positive integers within 100 that can be divided by 3 or 4 and each digit is 6

There are 10 numbers with 6 bits: 6, 16, 26, 36, 46, 56, 66, 76, 86, 96
Among them, 26, 46, 86 are unqualified, and the remaining 7 meet the conditions

In this paper, we prove that there must be two numbers in any four positive integers whose difference is divisible by 3

Using the principle of drawer, the remainder divided by 3 must be 0,1,2
There must be two repeated remainders in the four numbers. If the two repeated remainders are subtracted, the difference can be divided by three

A positive integer can be divided by three, by four, by five, and by six______ ．

3. The least common multiple of 4, 5 and 6 is 60, so the answer is 60