How many natural numbers can't be divisible by any number 3, 5 or 8

Divisible by 3: 1000 / 3 ≈ 333
Divisible by 5: 1000 / 5 = 200
Divisible by 8: 1000 / 8 = 125
Can be divided by 3, 5: 1000 / 15 ≈ 66 (pieces)
Divisible by 3 and 8: 1000 / 24 ≈ 41
Divisible by 5 and 8: 1000 / 40 = 25
Divisible by 3,5,8: 1000 / 120 ≈ 8
The number that cannot be divided by 3, 5 and 8 is 1000-333-200-125 + 66 + 41 + 25-8 = 468
100% right, believe me

There are -- natural numbers less than 100 that can be divided by 2 or 3, but not multiples of 5,

、
Note 0 exclusion
2, 4, 6 98. There are 49,
And then divisible by 3, but not divisible by 2 are 3, 9, 15, 21 A total of (99 △ 3 + 1) △ 2 = 17
That is, there are 49 + 17 = 66 less than 100 that can be divided by 2 or 3,
Then it is not a multiple of 5, that is to say, 5, 10, 15 95, these 19
So there are 66-19 = 47

What is the sum of all natural numbers less than 200 that can be divided by 3 but not multiples of 5

The number that can be divided by 3 is in the form of 3N, where when n = 5K, that is, the number of 15K can also be divided by 5
3N by 3N

Why is an integer a divisible by 7, 11 or 13
Will be an integer a from the last digit to the left, every three bits of a bar, called the first section, the second section The three digits in each section are called N1, N2,..., NK respectively, denoted by the (k-1) power of n = n1-n2 + n3-n4 +... + (- 1) * NK. If n can be divided by 7 or 11 or 13, then integer a can be divided by 7 or 11 or 13
Why?

Because 7 * 11 * 13 = 1001
Like 654321
654321=321321+（654-321）*1000
=321*1001+（654-321）*1000
Because 1001 is a multiple of 7,11,13, only 653-321 is a multiple of 7,11,13, and the original number is a multiple of 7,11,13

It is proved that the sum of 6 integers in any 11 integers can be divided by 6, but any 10 integers may not have this property

First, we prove that for any five natural numbers, the sum of three of them can be divisible by 3
It is proved that the remainder of any number divided by 3 can only be 0,1,2, which can be constructed as three drawers respectively: [0], [1], [2]
① If the remainder of the five natural numbers divided by 3 is distributed in the three drawers respectively, we can take one from each of the three drawers, and the sum will be divisible by 3
② If the five residuals are distributed in two drawers, there must be one drawer containing three residuals (drawer principle). The sum of the three residuals is either 0, 3 or 6, so the sum of the three corresponding natural numbers is a multiple of 3
③ If these five residuals are distributed in one of the drawers, it is obvious that the sum of three natural numbers can be divided by three
For any five natural numbers, the sum of three of them must be divisible by 3
Let the 11 integers be: A1, A2, A3 A11 and 6 = 2 × 3 (1) consider the case of divisible by 3 first
From the above, in 11 arbitrary integers, there must be:
Let a1 + A2 + a3 = B1;
Similarly, the remaining eight arbitrary integers, from above, must exist: 3 | A4 + A5 + A6;
Similarly, the other five arbitrary integers are: 3|a7 + A8 + A9, let: A7 + A8 + A9 = B3
② Consider B1, B2, B3 divisible by 2
According to the drawer principle, at least two of the three integers B1, B2 and B3 are identical odd or even. The sum of the two identical odd (or even) integers must be even
Then: 6|b1 + B2, namely: 6|a1 + A2 + a3 + A4 + A5 + A6
The sum of any 11 integers must be a multiple of 6
If it is 10 integers, B3 may not find that the sum of three integers is a multiple of 3 from the remaining four integers, so the sum of six integers of any 10 integers may not be divisible by 6

Characteristics of numbers divisible by 7 and 11

The common multiple of 7 and 11

My age is a double-digit number. This number plus 4 can be divided by 4; number plus 5 can be divided by 5; number plus 6 can be divided by 6. What's my age?

The number n is divided by 3, by 2, by 4, by 1, by 12. What is the remainder?

1.3A=4B+1-1=4B-1
2. Let a = a '+ 1,3 (a' + 1) = 4b-1 = > 3A '= 4b-4, because 4b-4 must be divisible by 4, so 3A' must be divisible by 12
3. Take the previous result into 3A + 2, and get 3 (a '+ 1) + 2 = 3A' + 5, so n is divided by 12 and the rest is 5

What are the characteristics of numbers divisible by 125

The last three digits are 125, 250, 375, 500, 625, 750, 875, 000

393a is a four digit number. Fill in the box with three numbers. The three four digits can be divided by 6, 11 and 8. What is the sum of the three numbers

The number is 3938,

How many natural numbers can't be divisible by any number 3, 5 or 8