1 has a few zeros all the way to the end of 10000

To be exact, there is only 1 0 at the end

But the landlord doesn't seem to mean that.

The specific analysis is as follows:

Let's start with a 0 at the end.

1—99 has 9 zeros

100—999 has 99 zeros

1000—9999 has 999 zeros

There are 4 zeros in 10000

There are a total of 1111 zeros

Consider the case where there are 5 at the end.

For every 5, there is a 0

1-10000 has 1000 5, resulting in 1000 0, so there is a total of 1111 + 1000 = 2111.

1 There are 2111 zeros in a row until the end of 10000

There is only 1 0 at the end

But the landlord doesn't seem to mean that.

The specific analysis is as follows:

Let's start with a 0 at the end.

1—99 has 9 zeros

100—999 has 99 zeros

1000—9999 has 999 zeros

There are 4 zeros in 10000

There are a total of 1111 zeros

There are 5 cases at the end.

For every 5, 5x2 produces a 0

1-10000 has 1000 5, resulting in 1000 0, so there is a total of 1111 + 1000 = 2111.

1 There are 2111 zeros in a row until the end of 10000

This 10,000 is too big, so it's not used to harm people.

The main thing is to see how many factors are 5 in this product.

The formula is like this: 10000 5 + 10000 25 + 10000 125 + 10000 625 + 10000 3125 = 2499.

The last formula is not divided by the whole number of parts.

Multiply 1 by 100 and there are 12 zeros at the end

There are 100 100s in 10000, that is, there are 1200 zeros

Add 12, I think it's 1212 0s, I don't know if it's right.

1-99, 9 pcs.

100-999, hundreds, that is, 9 of double 0s 9*2=18 into tens, a total of 9 groups of 9. 9*9=81

10000 4 pcs.

A total of 9 + 18 + 81 + 27 + 162 + 4 = 301

It seems that Zhou's pretty is right.

Two zeros. From the factor 10 to get 1 0, from the factors 2 and 5 multiply to get another 0, a total of two.

The product can be calculated and the result is obtained.

Original = 3628800. You see, there are exactly two zeros at the end of the product, and there is no one more than one.

If you expand the scale, lengthen the line, for example, from 1 to 20:

1×2×3×4×…×19×20。At this point, there are several zeros at the end of the product: now the answer becomes 4 zeros.

Among them, 1 0 is obtained from the factor 10, 1 0 is obtained from 20, 1 0 is obtained by multiplying 5 and 2, and 1 0 is obtained by multiplying 15 and 4, for a total of 4 zeros.

Multiplication ingenuitySuppose a times b equals c, i.e., denoted as ab = c or a·b = c.

In ancient China, multiplication calculations were performed using arithmetic chips. There are three levels of multiplication: the upper position is the multiplier, the median is the product, and the lower position is the multiplier.

First, the largest digit of the multiplier is multiplied by the multiplier, and after the multiplication, the calculation chip of this person is removed, and then the second digit is used to multiply, and the product of the two digits is added to the number on the corresponding digit, until the multiplication is finished.

249 pcs. Analysis: There are multiples of 5 1000 5=200.

Multiples of 25 have 1000 25 = 40.

Multiples of 125 have 1000 125 = 8.

Multiples of 625 are [1000 625] = 1.

Therefore, 200 + 40 + 8 + 1 = 249 power of 5 is the factor of the product.

So there are 249 zeros in total.

The product is the result of multiplying two numbers. For example, 3x4=12 is the product where 12 is the product.

The product (product) is the cumulative number or quantity or arithmetic multiplication of two numbers.

Sum refers to the new thing obtained by the addition of two or more things of the same attribute, and can also be narrowly understood as the result of the addition of two numbers.

Sum is a new thing obtained by adding things of the same attribute, such as 2 meters + 3 meters = 5 meters; 30 kg + 50 kg = 80 kg. However, things with different attributes and different units cannot be added or simply added by numbers, such as 5 meters of seconds + 10 seconds; 5 minutes + 1 hour.

The generation of sum: addition + addition = sum.

Words that represent summing: total, total, total, etc.

It's just how many 5s and how many zeros end in them. Since 000 999, the number of 5s at the end is 100, and the number of zeros at the end is 100-1=99 (minus 000) and subtract the number of 2 zeros and 3 zeros at the end, which is 89.

The number of the last two zeros is 9.

At the end there are 3 zeros and only 1000, 1.

So at the end there are 100 + 89 + 2 * 9 + 3 * 1 = 189 + 18 + 3 = 210.

From 1000 to 10000, there are 81 numbers with only two zeros at the end.

1100 to 1900 a total of 9.

2100 to 2900 for a total of 9.

3100 to 3900 a total of 9.

4100 to 4900 a total of 9.

5100 to 5900 for a total of 9.

6100 to 6900 a total of 9.

7100 to 7900 9 in total.

9 from 8100 to 8900.

9100 to 9900 9 pcs.

There are 9 9 = 81 in total.

From 1000 to 10,000 there are only two zeros at the end of the number, and there are 91.

Since it's from 1,000 to 10,000 and there's only two zeros at the end, it's actually 1001 9999, where the hundred digits are not 0s, and the last two digits are 0s, so you can look at the first two digits from 10 to 99, and the single digits are not zeros, so there are 90-9 = 81 numbers.

There are 81 numbers from one thousand to ten thousand followed by only two zeros, from 1100 to 9900.

If there is a number that is 0, then the product is 0, not 3 zeros It is incorrect to say that there are three zeros at the end of two non-zero factors, and there are at least three zeros at the end of the product.

First consider trying to write a number that meets the requirements, the specific number can be the last two digits, that is, the single digit and the 10-digit number are 0, and then the hundred digit number is not 0, and the value of the thousand digit value can be calculated again, in general, it is the principle of distributed multiplication counting.

This topic can be dissolved in this way.

Up to 10,000: ab00

a can be dusted to get 0 to 9, a total of ten numbers.

b cannot take 0, so it must be from 1 to 9, a total of nine numbers.

The combination is 90 numbers of blocks.

4500 pcs.

There are only two zeros at the end of the number between 1000 and 10000 that can be seen as a three- or four-digit number ending in "00", i.e. satisfying 1000 n 9999 and n mod 100 == 0.

First of all, we can know through a simple analysis: in 1000 9999, there are 90 brigades in hundreds, each number ends in "00", and then the number 4000 9000 in thousands is added, and there are 4500 in total.

So, from 1000 to 10000, there are 4500 numbers with only two zeros at the end.

Since it's from 1000 to 10000

And there are only two zeros at the end.

In fact, Yan Huai Lao is among the 1001 9999.

The hundred digits are not 0s, and the last two digits are both 0s

Then look at the first two digits 10 to 99.

A number where the single digit is not zero.

Therefore, there are 90-9 = 81 such ascends.

There are 10 5s in 100, 5 times even numbers ending in 0, and there are 10 in 10 at the end, and there are 100 in 100, 10 + 9 + 2 = 21.

From 1 to 10, multiply 10 consecutive integers:

How many zeros are at the end of a product?

The answer is two zeros. Among them, 1 0 is obtained from the factor 10, and 1 0 is obtained by multiplying the factors 2 and 5, making a total of two.

Exactly two zeros? Will there be a few more?

If you don't believe it, you can calculate the product and get the result.

Original = 3628800. You see, there are exactly two zeros at the end of the product, and there is no one more than one.

So, what if you expand the scale and lengthen the ranks? For example, multiply from 1 to 20:

1×2×3×4×…×19×20。How many zeros are at the end of the product?

Now the answer becomes 4 zeros. Among them, 1 0 is obtained from the factor 10, 1 0 is obtained from 20, 1 0 is obtained by multiplying 5 and 2, and 1 0 is obtained by multiplying 15 and 4, for a total of 4 zeros.

Exactly 4 zeros? Will there be a few more?

Rest assured, it's not much. To get a 0 at the end of the product, a prime factor 5 and a prime factor 2 need to be multiplied in pairs. In the prime factor of the product, 2 is more and 5 is less.

There is a prime factor of 5, which is multiplied by a 0 at the end of the family product. Multiply from 1 to 20, only there is a prime factor of 5 in each of them, and there can only be 4 zeros at the end of the product, and there is no more.

Multiply the scale a little more, from 1 to 30:

1×2×3×4×…×29×30。Now how many zeros are there at the end of the product?

Obviously, there are at least 6 zeros.

You see, from 1 to 30, the sum of 30 is a multiple of 5. From each of them, one 0 can be obtained; They have a total of 6 numbers, and you can get 6 zeros.

Exactly 6 zeros? Will there be more?

Whether you can have more or not depends on the number of prime factor 5. 25 is 5 squared, and it contains two prime factors of 5, and here there is an extra 5. Multiplying from 1 to 30, although only 6 of the 30 factors are multiples of 5, there are 7 prime factors of 5 in the band.

So there are 7 zeros at the end of the product.

Multiply it to 30 and do it, no matter how wide it is.

For example, this time multiply a little more, from 1 to 100:

1×2×3×4×…×99×100。How many zeros are there at the end of the product now?

The answer is 24.

200 5 = 40, and 25 contains two factors 5,200 25 = 8, and 125 contains three factors 5,200 125 1 (rounded) 40 + 8 + 1 = 49, i.e. 1 2 3 4 ....At the end of the product of 200 there are 49 consecutive zeros, so the answer is: 49

Look at 1 200 and break it down to 5.