How many natural numbers from 1 to 1003 can't be divided by 7, 11 or 13?

How many natural numbers from 1 to 1003 can't be divided by 7, 11 or 13?

1003÷7=143… 2 & nbsp; the number of numbers divisible by 7 is 143. Similarly, the number divisible by 11 is 91; the number divisible by 13 is 77; the number divisible by 7 × 11 is 13; the number divisible by 7 × 13 is 11; the number divisible by 7 × 11 × 13 is 1; the number divisible by 11 × 13 is 1

How many of the 100 natural numbers between 1 and 100 are not divisible by 3 and 11

There are only 33, 66 and 99 divisible by 3 and 11. The others are not. There are 97. If 3 and 11 are changed to 3 or 11, there are 61

In the natural numbers from 1 to 1999, the numbers that can be divisible by 3 but not by 5 and 11 are

Among the natural numbers from 1 to 1999, the number of numbers divisible by 3 is 666 quotients of 1999 / 3,
Among these 666 numbers, there are 133 quotients of 1999 / (3 * 5) that cannot be divisible by 5
Among these 666 numbers, the number that cannot be divided by 11 has the quotient of 1999 / (3 * 11) of 60
So the number of natural numbers from 1 to 1999 that can be divisible by 3 but not by 5 and 11 is
666-133-60 = 473

From 0, 1, 2, 3 In the 2013 natural numbers of 2011 and 2012, take out several numbers, so that the sum of any two numbers taken out is an integral multiple of 50
How many can be taken out at most?

2013÷50=40..13
most
40 + 1 = 41 numbers

How many natural numbers between 1 and 1000 can be divided by 2, 3 and 5 at the same time?
one hundred and eleven

The least common multiple of 2,3,5 is 30
1000 / 30 = 33 + 10
So there are 33 numbers that can be divided by 2,3,5 at the same time

How many numbers can be selected from all the natural numbers from 1 to 99 so that the sum of any two of them cannot be divisible by 5?

41. Choose the number of 5K + 1, that is, divide 5 by 20 numbers with the remainder of 1, 1,6,11,..., 96, and then choose the number of 5K + 2, that is, divide 5 by 20 numbers with the remainder of 2, 2,7,12,..., 97, and then choose any number that can be divisible by 5 (that is, 5K type). For example, choose 5, a total of 41 numbers, that is, at most 41 numbers can be selected from all natural numbers from 1 to 99

Take some numbers from the 30 natural numbers from 1 to 30, so that the sum of any two numbers can not be divided by 7?

According to the stem analysis can be: at most 5 + 5 + 4 + 1 = 15 (number), answer: can take out at most 15 numbers, so that take out the number, the sum of any two different numbers are not a multiple of 7

What are the following numbers represented by scientific counting? 2 times 10 to the sixth power, 6.03 times 10 to the fifth power, 5.002 times 10 to the fourth power

2000000,603000,50020

It is known that the power of | n + 3 | of 2x is equal to eight. This is an equation of one variable, then the integer n is equal to?

The equation is a linear equation of one variable, indicating that the coefficient of X is equal to 1,
That is | n + 3 | = 1,
The solution is n = - 2 or n = - 4

8 to the 33rd power multiplied by a quarter to the 50th power
The best is simple algorithm!

8^33*(1/4)^50
=2^99*(1/2)^100
=(2*1/2)^99*1/2
=1^99*1/2
=1/2