How many zeros are there in a row at the end of 2005? Can there be a detailed calculation method?

A few zeros are to see how many 10s are multiplied

How did 10 come about? 5*2＝10

So look at 2005! You can decompose a few 5s and a few 2s.

Let's take a look at 5 first.

First of all, from 5 to 2005, there are 401 multiples of 5s, so 401 5s can be broken down first

Looking at the multiples of 25 to 2000, there are 80 multiples of 25, which should be able to decompose 80*2 160 5s, but the previous step has already been calculated once, so it has to be counted again, which is actually 80 5s

Then look at the 125 to 2000, there are 16 multiples of 125, which should be able to decompose 16*3 48 5s, but the last two steps have been calculated twice, so they have to be counted again, which is actually 16 5s

Looking at the difference between 625 and 1875, there are 3 multiples of 625, and there are actually 3 5s (for the same reason as above).

Add up the number of 5 that comes up to 401 + 80 + 16 + 3 500

Let's take a look at 2.

From 2 to 2004, there were 1002 of them, so at least 1002 2s can be broken down.

At this point, although the number of 2 has not been calculated, the problem has been solved.

Here's an example: I bought 10 apples and 10 pears and planned to eat 2 apples and 1 pear a day for up to 5 days (because the apples were gone, my plan couldn't go on).

Just like in this problem, 500 5's have formed 500 0's at the end of 500 with 500 2's, and although there are many more 2s, the 0's at the end can no longer be formed.

Summary: There are 500 zeros at the end

That's it, thank you!

A 0 at the end, multiplied by a 2 and a 5 after the prime factor is decomposed.

It is easy to see that there are many 2s, so we only need to count the number of 5s after the prime factor is broken down. The following is divided and rounded:

2005 5=401, there are 401 numbers that are multiples of 5, they contribute 401 5, 2005 25=80 There are 80 numbers that are multiples of 25, they can contribute a total of 2 5s for each number, and they have already contributed one on it, and they can also contribute a 5 for each;

2005 125=16 There are 16 numbers that are multiples of 125, and each of them can contribute a total of 3 5s, and they have already contributed two on it, and they can also contribute a 5 to each;

2005 625=3 There are 3 numbers that are multiples of 625, and each of them can contribute a total of 4 5s, and they have already contributed three on it, and they can also contribute a 5 to each.

Therefore, there is a total of 401 + 80 + 16 + 3 = 500 5's, i.e. there are 500 zeros at the end.

Ask for 1999!How many consecutive zeros are at the end of .

There are 496 zeros at the end of the factorial of 1999, when 0 < n < 5, f(n!) 0;When n >=5, f(n!) k + f(k!where k = n 5 (rounded) example: f(5!)=1 + f(1!

1 f(10!)=2 + f(2!)=2 f(20!

4 + f(4!)=4 f(100!)=20 + f(20!

20 + 4 + f(4!)=24 f(1000!)=200 + f(200!

200 + 40 + f(40!)=240 + 8 + f(8!)=248 + 1 + f(1)=249 f(1999)=399+f(399!

399 + 79 + f(79!)=399 + 79 + 15 + f(15!)=399 + 79 + 15 + 3 + f(3!

496 So there are 496 zeros at the end of the factorial for 1999.

Of these 24 multipliers, there are five at the end of the three numbers, and multiplying the two numbers with an even number will also cause a 0 at the end of the product.

So, 2012! ÷1988！At the end of the calculation, there are a total of 7 consecutive zeros.

500 pcs. First of all, the first point should be clear ... Why look for the factor 5?

Because the 2 factors of the 2005 factorial must be more than the 5 factors, the number of 10 factors (i.e. 2*5) we are looking for depends on the number of 5 factors.

The next thing to consider is which numbers have a factor of 5. Yes, 5, 10, 15... The source.

2005, a total of 2005 5=401. But some of these numbers don't have only one 5-factor, such as 25, 50, 125, etc., with two or more 5-factors. Numbers below 2005 can have up to 4 5 factors (because 5 is 5 to the 5th power in 2005).

The number containing 2 pentatonic factors is high. i.e. 25, 50, 75 ... A total of 2005 25 = 80 hall feet. The 5 factors of these numbers just now are only counted as one, so we have to add them again, that is, 401 + 80 = 481.

A number that contains 3 five-factor. i.e. 125,250 ,。。 In total, 2005 125 = 16. The 5 factors of the previous numbers are only remembered twice, and with the addition of these, 481 + 16 = 497.

In the same way, there are finally 4 five-factor numbers, 625, 1250, 1875, 3 (2005 625). 497 + 3 = 500 pcs.

From simple to, look at 10 first!

2 and 5 are a celebration to sell 0

10 is a 0, so there are no 10 consecutive natural fast widths, and there are two zeros

So 2000! There are 200*2=400 zeros

2002*2005 another 0

So Yuchang teased a total of 401 0s

Each time a multiple of 5 is encountered, there is an additional 0

So, 2005 divided by 5

Calculate 2005!There are a few 5s in it, and there are a few zeros when you multiply it.

And then it's all less than one.

So the answer is 401 + 80 + 16 + 3 = 500

There are 500 of them.

Because, a 0 was generated by a 5 2, and in 2005! There are 401 multiples of 5, 80 multiples of 25, 16 multiples of 125, and 3 multiples of 625.

So, put 2005! Factoring, where 5 has 401 + 80 + 16 + 3 = 500, and the number of 2 must be more than the number of 5, so 2005! has 500 consecutive zeros at the end of the .

Depends on the number of times of 5. It can also be simply buried and converted to the number of multiplied hoods divided by 5 + the number of multipliers divided by 25 + .The number of multipliers divisible by 5 n.

Specifically. [a]

Represents the largest integer not greater than a.

n=[2002/5]+[2002/25]+[2002/125]+[2002/625]

Hello, because 25 = 5 * 5 has 2 5 factors, 125 = 5 * 5 * 5 has 3 5 factors, 625 = 5 * 5 * 5 * 5 * 5 has 4 5 factors, 5 * 5 * 5 * 5 = 3025>2005 so do not consider.

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Calculate 2005!There are a few early closures, and there are a few zeros when you multiply them.

And then it's all less than fission.

So the answer is 401 + 80 + 16 + 3 = 500