How many zeros are there in all the natural numbers from 1 to 1,000,000?

111111 0s.

Analysis: There are 9 zeros and 2 zeros in 100.

So: 1 to 100: There are 11 zeros.

Similarly, from 101 to 200, there are 11 zeros;

201 to 300, there are 11 zeros;

801 to 900, with 11 zeros;

901 to 1000 with 12 zeros.

So, from 1 to 1000, there are 111 zeros.

1001 to 2000, there are 111 zeros;

2001 to 3000, there were 111 zeros;

8001 to 9000, there are 111 zeros

9001 to 10000 with 112 zeros

So: 1 to 10000, there are 1111 zeros

So: from 1 to 1000000, there are 111111 zeros.

Consider from 000000

to 999999

These 1,000,000 numbers total.

Bit. where the number ......The number of occurrences to 9 is equal, all are.

Therefore, from 1 to 999999 the numbers 1 to 9 appear for 600,000*9. Again, from 1 to 999999, there are a total number of digits.

Piece. So from 1 to 999999, the number 0 appears 5888889. Add 6 out of 1,000,000 for a total of 0488895 from 1 to 1,000,000.

From 0000 to 9999, a total of 10,000 numbers are rounded.

Among the numbers that do not contain 1 at all, they are:

9*9*9*9 = 6561

Then there are 10000 - 6561 = 3439 with 1.

From 1 to 10,000, add 10,000 next to the collapse or collapse itself.

Numbers with 1 are 3439 + 1 = 3440.

Natural numbers are all integers, and in all natural numbers from 10 to 170, the number with 0 is an integer ten, and there are two zeros in 100, so there are a total of 17+1=18 zeros.

Out of all the natural numbers of 10 170, there are a total of 27 zeros.

If there are 10 numbers from 1 to 100 and 10 numbers with 1, then from 1 to 10000 there are 100 numbers of 10, then 100x10

1000 counts. There are 1,000 numbers containing 1.

There are l single digits, 18 two-digit numbers, 244 three-digit numbers, and 2952 four-digit numbers, for a total of 3216.

How many natural numbers 1-1000 contain the number 0? How many zeros are there?

0 at the end: multiples of 10, multiples of 100 and multiples of 1000, and as long as it contains 0, so only multiples of 10 are counted: multiples of 10 have 1000 10 = 100;

0 in the middle: Only 3 digits have such a number.

Among them, there are 101 to 109 in the hundredth place, a total of 9 (100 has been calculated and cannot be counted repeatedly);

In the same way, there are 9 from 2 to 9 in the hundred;

Then the number 0 in the middle has 9 9 = 81;

So there are 100 + 81 = 181 numbers with 0.

The number of zeros contained is also calculated from the end and the middle:

There is 1 0 at the end: there are 100 multiples of 10, and there are 100 0s;

There are 2 zeros at the end: there are 10 multiples of 100, but the number of 0s has been calculated once in multiples of 10, so it cannot be counted repeatedly, counting 10;

There are 3 zeros at the end: only 1000, but the 0s on the single digit and the ten digits have been calculated earlier, and only 1 0 can be counted;

0s in the middle: the same calculation as above, there are 81 zeros;

So there are 100 + 10 + 1 + 81 = 192 0s

1-9 No.

10 out of 90-9.

In 100-999, only the ten digit 0 has 9*9=81, and the single digit 0 has 9*9=81

There is 1 x 1*2 of the two 0's

There are 3 out of 1000.

Adds up to 9 + 81 + 81 + 2 + 3 = 176

Single digits: 1, 11, 21 ,......91

A total of 10 100 = 1000.

Ten places: 10, 11 ,......19

A total of 10 100 = 1000.

Hundred: 100, 101 ,......9199

1000 in total.

Thousand: 1000, 1001 ,......1999

1000 in total.

10,000

There is 1 in total, so the number of 1s is 4*1000+1=4001

In 1-99, there are 10 with a single digit of 1, and 10 with a 10 digit of ten.

The same goes for it. In 100-199, there are 10 with 1 single digit, 10 with 10 10 digits, and 100 with 100 digits with 100 digits.

200-999: 80 with 1 digit, 80 with 10 digits, and 0 with 1 100 digits.

1000: There is only one 1

So the number of 1 in the natural numbers from 1 to 1000 is:

10 + 10 + 10 + 10 + 100 + 80 + 80 + 1 = 301.

Solution: 1 10 x 1000 = 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100

The thousand digits contain 1:1

Total: 100 + 100 + 100 + 1 = 301 There are 301 numbers in total

In the single digit of 1:

One digit: 1 Two digits: 11, 21....91 of 9.

Three-digit numbers: 101, 111....191,201,211...291,..901...991 of 90.

1 in the 10th digit digit: 10....19 of 10 three-digit numbers: 110....119，210...219...910...919 of 90

In 100:100, 101 ... 199, a total of 100 in 1:1000 thousands

A total of 301 1

112, "1" is one, "10 19" is ten, "100 199" is a hundred, and "1000" is one, for a total of 112.

111111 0s.

Analysis: There are 9 zeros and 2 zeros in 100.

So: 1 to 100: There are 11 zeros.

Similarly, from 101 to 200, there are 11 zeros;

201 to 300, there are 11 zeros in Zaohetan;

801 to 900, with 11 zeros;

901 to 1000 with 12 zeros.

So, from 1 to 1000, there are 111 zeros.

1001 to 2000, there are 111 zeros;

2001 to 3000, there were 111 zeros;

8001 to 9000, there are 111 zeros

9001 to 10000 with 112 zeros

So: shoot 1 to 10000, a total of 1111 0s, so: 1 to land 1000000, a total of 111111 0s.

Consider from 000000

to 999999

These 1,000,000 numbers total.

Bit. where the number ......The number of occurrences to 9 is equal, all are.

Therefore, from 1 to 999999, the number 1 to 9 is out of the aystool, and now there are 600,000 * 9. Again, from 1 to 999999, there are a total number of digits.

Piece. From 1 to 999999, the number 0 appears 5888889. With the addition of 6 limbs out of 1,000,000, there are 0488895 total from 1 to 1,000,000.