P20 Distribute wine equally:

Zhang San and Li Si each have a bottle of wine, and they want to divide the wine equally. Li Si first poured the wine from his own bottle into Zhang San's bottle to double Zhang San's wine, then poured Zhang San's wine into his own bottle to double the wine in his own bottle. I poured it twice, but it was still uneven. There is wine160g in Zhang San bottle and wine120g in Li Si bottle. How much wine are there in each bottle?

Analysis: Using backward deduction method.

The second time was "Zhang San poured wine for Li Si and tripled the wine in Li Si's bottle". At this time, there is 120g in Li Si bottle. It can be known that before Zhang poured wine for Li Si, Li Si had wine: 120 ÷ 3 = 40 (grams).

The wine that Zhang San poured for Li Si this time is: 120-40 = 80 (grams).

After Zhang San poured 80 grams of wine for Li Si, Zhang San still had 160 grams of wine, which means that before Li Si poured wine for Zhang San, Zhang San had wine: 160+80 = 240 (grams).

Li Si poured wine for Zhang San for the first time, which doubled the amount of alcohol in Zhang San's bottle. After drinking, Zhang San has 240 grams of wine. Half of the 240 grams of wine is Zhang San's own, and the other half is poured by Li Si.

So Zhang Sanyuan has wine: 240 ÷ (1+1) =120 (g).

Li Si's original wine is 40+ 120 = 160 (g) or160+120-160 (g).

P 16 bottom hole bugs:

A bug accidentally fell into the well. It's climbing every day. Unfortunately, it can climb 3 meters every day and slide down 2 meters at night. But the bug insisted on climbing. This well is 20 meters from the bottom to the wellhead. Insects began to climb up from the bottom of the well in the early morning. How many days will it take to climb out of the wellhead?

Analysis: The last day: Climb three meters out of the wellhead, without losing 20-3= 17 meters.

17m /(3-2) m = 17 days

17+ 1= 18 days

How many zeros are there?

How many zeros are there at the end of the product of 1× 2× 3× …× 100? The answer is 24.

Analysis: Multiply consecutive integers from 1 to 10/0;

1×2×3×4×5×6×7×8×9× 10。

How many zeros are there at the end of a continuous product?

The answer is two zeros. Where, the factor 10 gives 1 zeros, and the factors 2 and 5 are multiplied to give 1 zeros, and two zeros count as * * *.

Just two zeros? Will there be more?

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

Original formula =3628800. You see, there are exactly two zeros at the end of the product, not more than 65438+ zeros.

So, what if we expand the scale and lengthen the team? For example, from 1 times 20:

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

Now the answer has become four zeros. Where we get 1 zeros from the factor 10, 1 zeros from 20, 1 zeros from 5 and 2, and 1 zeros from 15 and 4.

Just four zeros? Will there be more?

Don't worry, that's all. In order to get 0 at the end of the product, prime factor 5 and prime factor 2 must be multiplied in pairs. Among the prime factors of products, 2 is more and 5 is less. There is a prime factor of 5 and a 0 at the end of the product. From 1 multiplied by 20, only 5, 10, 15 and 20 each have a prime factor of 5, and there may be only four zeros at the end of the product, and there will be no more.

Expand the ratio a little, from 1 to 30:

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

Obviously, there are at least six zeros.

You see, from 1 to 30, where 5, 10, 15, 20, 25 and 30 are all multiples of 5. From each of them, we can get 1 zero; They * * * have six numbers and can get six zeros.

Just six zeros? Will there be more?

Whether it can be more or not depends on the number of prime factors 5. 25 is the square of 5, including two prime factors 5, and 1 5. From 1 multiplied by 30, although only 6 of the 30 factors are multiples of 5, it contains 7 prime factors 5. So the product has seven zeros at the end.

Anyone who multiplies by 30 will do it, no matter how big it is.

For example, multiply this time, from 1 to 50, and the answer is 12 zeros.

This "product" question essentially tests the knowledge points of "prime numbers and composite numbers"

Several mathematical theorems involved in this topic include

A prime number is a natural number with only 1 and its own two divisors, such as 2, 3 and 5.

Second, the composite number refers to a natural number with other divisors besides 1 and itself, such as 4, 6 and 8.

Third, 1 is neither a prime number nor a composite number.

4. Integer A can be divisible by Integer B, where A is called a multiple of B and B is called a factor of A..

How many zeros are there in the mantissa of the product of 1×2×3×4×5×…×3000?

Suppose M= 1×2×3×4×5×…×3000.

Because 2× 5 = 10, the final zero can only be obtained by multiplying the prime factors 2 and 5.

Therefore, it is only necessary to calculate how many prime factors 2 and 5 have after M is decomposed into continuous products of prime factors, and which number is small, how many consecutive zeros will be at the end of M..

This solution first calculates the number of m with prime factor of 5.

At 1, 2, 3, …, 2998, 2999, 3000, 3000/5=600.

That is, there are 600 multiples of 5. They are: 5 10, 15, …, 3000.

Of these 600 numbers, 3000/25= 120, that is, 120, which can be divisible by 25, they are 25, 50, 75, …, 3000.

Of this number 120, 3000/ 125=24, that is, 24 can be divisible by 125, and they are 125,250,375, …, 3000.

Of these 24 numbers, 3000/625=4, and there are still four divisible by 625, namely 625, 1250,1875,2500.

So the number of prime factor 5 in m is equal to 600+ 120+24+4=748.

The number of prime factors 2 in m is obviously more than that of prime factors 5.

So 1×2×3×4×5×…×3000 has 748 zeros at the end.