How many zeros are there at the end of the product of 101 by 102 by 103 by 104 until 199 by 200 Be more detailed!

There is a point right in the above, but the consideration is not comprehensive
There are two sources for mantissa 0. The first is to encounter a multiple of 10, which is 110,
Another source is a multiple of 5, for example, 115, which is multiplied by 112 and the mantissa is zero
125 is multiplied by three even numbers to get three zeros. Even numbers are enough
From the above analysis, so the answer depends on how many factors are there for all numbers from 110 to 200
There are 105, 110, 115, 120, 125,. 200 numbers with 5 factors, 20 in total
There are 125150175200 with two five factors
There are only 125 with three five factors
So the answer is 20 + 4 + 1 = 25

How many zeros are there at the end of the product of two times three times four times 125

There are three 5's in the factor, there is one; there are two 5's, there are four; there are one 5's, there are 20. So the factor of 5 has 3x1 + 4x2 + 1x20 = 31
There are 31 zeros

How many zeros are there at the end of 1 * 2 * 3 * 4 . 50 product?

A 0 at the end is obtained by multiplying a 2 and a 5 after prime factor decomposition
It is easy to see that there are many 2, so we only need how many 5 after prime factor decomposition
50÷5=10
There are 10 numbers that are multiples of 5, and they contribute 10 5's
50÷25=2
There are two numbers that are multiples of 25. Each of them can contribute two 5's in total. They have contributed one above, and they can contribute another 5
So there are 12 5
There are 12 zeros at the end

Judgment question (there are two zeros at the end of the product of 250 * 8)

Wrong
There should be three zeros
）

55.5 () 99.9 (of which 5 and 9 have 20 each) can be divided by 7, so what is the number in the middle bracket?
Tell me, hurry!

If the absolute value of the difference of the number obtained by subtracting 2 times of the last digit of a number from the original number and removing the last digit can be divided by 7, then the number can be divided by 7 (if it is zero, of course, the same), for example, 14,4x2 = 8,8-1 = 7
21:1x2 = 2, 2-2 = 0, for example, 169, 9x2 = 18, 18-16 = 2, so 169 can't be divisible by 7. If it's 189, it must be divisible by 7

A 29 digit number: 55 55 () 99.99 can be divided by 7 (there are 14 5's and 14 9's)

55555 can be divided by 7 (column vertical, first 5 / 7, divide inexhaustible, add 5 after 5, keep on finding 6 5 can be divided by 7, actually because 111111 can be divided by 7), then 55.500.. 0 (14 5, 15 0) can be divided by 7, similarly 9999 can be divided by 7, 999... 9 (14 9) can be divided by 7
So 55... 5 () 99... 9-999... 9 (14 9) - 55.500.. 0 (14 5, 15 0) = 55 () 99000000000000000000 = 55 () 99 * 1000000000000000 can be divided by 7 and 10 ^ 18 is coprime with 7, so
55 () 99 can be divided by 7
0123456789 only six can be brought in

27 digits 55 5□99…… 9 (of which 5 and 9 have 13 each) can be divided by 7, and the number in is（

If the number of one digit of an integer is truncated, and then two times of the number of one digit is subtracted from the remaining number, if the difference is a multiple of 7, then the original number can be divided by 7. If the difference is too large or it is difficult to see whether it is a multiple of 7 by mental arithmetic, we need to continue the above process of "truncation, multiplication, subtraction and difference checking" until we can make a clear judgment. For example, the process of judging whether 133 is a multiple of 7 is as follows: 13-3 × 2 = 7, So 133 is a multiple of 7; for example, the process of judging whether 6139 is a multiple of 7 is as follows: 613-9 × 2 = 595, 59-5 × 2 = 49, so 6139 is a multiple of 7, and so on

This 41 digit 555.5 () 999.9, of which 20 are 5.9, can be divided by 7, then what is the number in the middle bracket?

7
There are many ways to judge whether a number can be divided by 7. The data of your topic is relatively large, which is suitable for the following method
If the data can be divided by 7 even after cutting 1001, then the number can be divided by 7
If we cut 1001 or multiple of 1001, we can cut 9009 or 90090 or 9000000
Or 5005. The results are all reduced, leaving only the number in brackets, so we can only fill in 7 or 0

The following 41 digit 55.x99.9 (of which 5 and 9 have 20 each) can be divided by 7, so what does x represent? It can be divided by 7
The answer should be 6
The characteristics of numbers divisible by 7: the difference between the last three digits of an integer and the numbers before the last three digits (greatly reduced) can be divisible by 7. The rule it summarizes is wrong
For example, 84533754 matches, but cannot be divided by 7

That's right --
754-533=221
221 / 7 = 31. The remaining 4 cannot be divided

The last three digits of a five digit number are 999. If the number can be divided by 23, what is the minimum number of five digits?

Let two digits be m, then m999 can be divisible by 23, 1000m + 999 = 43 × 23m + 11m + 43 × 23 + 10 = (43 × 23m + 43 × 23) + (11m + 10) can get (43 × 23m + 43 × 23) + (11m + 10) can be divisible by 23; because 43 × 23m + 43 × 23 can be divisible by 23, so 11m + 10 can be divisible by 23; suppose 11m + 10 = 23n, then M = (22n-11) △ 11 + (n + 1) △ 11, obviously n + 1 is divisible by 11, the minimum n is 10, and the minimum m is: (23 × 1) To sum up, the minimum number of five digits is 20999. A: the minimum number of five digits is 20999

How many zeros are there at the end of the product of 101 by 102 by 103 by 104 until 199 by 200 Be more detailed!