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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, and there is only a 0 at the end of the 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, they contain 7 prime factors of 5.

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 500:

1×2×3×4×…×99×..500。How many zeros are there at the end of the product now?

The answer is 124.

The number with a factor of 5 is: 500 5=100

The number with a factor of 25 is: 500 25=20

The number with a factor of 125 is: 500 125 = 4

So there are total: 100 + 20 + 4 = 124.

Other than that. Question 1x for landlord....There are 24 x100s.

Obviously, in 1 2 3 4 5 6 ......In 99 100, there are more prime factors 2 than 5, so the number of zeros at the end is determined by the number of prime factors 5.

There are 100 5 = 20 multipliers with prime factor 5, but note that in 1 to 100, there are multipliers with two prime factor 5s, and such multipliers have 100 25 = 4, which are 25, 50, 75 and 100.

There are 20+4=24 in total.

10 100 has 11 zeros

There are 5, multiplied by even numbers, and there are 10 zeros

So there are 21 zeros

The upstairs method is correct, but there are loopholes.

10 20 100 has 11 zeros

In the number of 5, note that 25 75 and the multiplication of even numbers have 2 zeros, so there should be 12 zeros here

So there should be 25 zeros in total

1 question and.

Question 2 and the number of decimal places are different.

3 questions 100

4 questions and. 5 questions and 6 questions, the first three and six sub-questions cannot be simplified, and the second question: subtract two numbers according to a number, you can subtract the sum of these two numbers.

Question 4: 25x204=24x(200+4)=25x200+25x4=5000+100=5100 according to the multiplicative distribution law.

Question 5: According to the law of addition and associativity.

These questions are not actually difficult questions.

In fact, this question is not difficult, as long as you do it carefully step by step, first do it separately from the bottom, as follows.

Start with the lowest fraction = 11 3

Hehe, the waiter's 2 pieces are now included in 27 pieces! The three of them gave a total of 30 yuan, gave 25 yuan to the boss, and the waiter stole 2 yuan, and they still had 3 yuan left!

3 times 9 = 27 This is all the money that three people actually go out of The formula is =30-25-2=3

1. Let the boy have x

, 90-x for girls, the total score is: 90x73 = 6570 points, 70x + (90-x) x 80 = 90x73

This gives us x=63

So there are 63 boys and 17 girls.

2. Set the number of rainy days as x, because Gongling Hui picked 112, an average of 14 per day, so a total of 112 14 = 8 days.

The number of sunny days is: 8-x

days, (8-x) x20 + 12x

Derived. x=6 days of green, rainy days for 6 days of foot bend.

3. If there are x kinds of five-character quatrain poems, then the seven-character quatrain poems are x-13 kinds, (x-13)x(7x4)-(4x5)x=20

x=48, so there are 48 and 35 of the two kinds, respectively.

The column is a jujube douque grinding group. There are x boys and y girls. x+y=90,(70x+80y) 90=73, and the two equations can be combined to get the number of boys and girls. The same is true for the following two questions.

Wine 7 glasses 2, grapes 1 grain, so 7 2 + 2 2 5 = 34.