The math of a strange problem? How does this math problem work?

The mathematical principle is that "binary" means "decimal".

We all know that different "decimal systems" are just one way of counting, e.g. 5 (10) = 0000000101 (2) and 500 (10) = 0111110100 (2). The number n in the hundred place in the decimal system represents n*10 2, and the number n in the "hundred" place in binary represents n*2 2.

Speaking of which, I don't know if I've explained it clearly, welcome to ask.

Each of the 10 bags contains pearls, and these 10 numbers can be combined to form 1-1000. The first 9 digits are 2 to the power of 2, and the 10th digit is 2 to the 9th power minus 23. Because 2 n - 1 = 2 (n-1), so the sum of the first n-1 numbers is exactly 1 smaller than the number immediately following it, from the first number 1 to the n-1 number, the maximum number they combine is the sum of all of them, and the sum obtained by adding all of them, starting from 1, is continuous to the n-1 power of 2, and the number that requires 1 greater than this sum is exactly the nth power of the next number 2.

There is only the 10th number, which is 2 to the 9th power minus 23, because when added from 2 to the 9th power of 2, it is 23 more than 1000

It is also known that when x is a rational number, f(x)=x is satisfied. Can you think of it when you are bored, please think about the problem when you are bored! Because it is continuous, it is a line. Yes.

Here's how: this old man did it deliberately from the beginning, because half plus a quarter plus a sixth is equal to eleven twelve, which is not enough to make up a whole. In other words, it can be understood that the old man took the remaining twelfth as a hidden condition, that is, the property that did not belong to any of the three brothers.

But this non-existent sheep did not exist in the first place. So I gave them a sheep in the title so that they could make a whole whole, which is twelve sheep. The allotment based on twelve sheep can meet the old man's condition, and the one-twelfth hidden condition, that is, the unwarranted sheep, will be returned to me.

This is a virtual hypothesis, in order to make the question complete without affecting the conditions of the original question, it is just a way of thinking to assist in doing the problem.

Flow firefly ginseng on.

In fact, this is the case, that is, the old man's 11 sheep cannot be killed, but must be divisible, so you can borrow one of their sheep and become 12, and then divide them. 12*2 1=6 12*1 4=3 12*1 6=2 It adds up to 11, and then return the last one to someone else, isn't that just right? (Pure hand fighting, don't blame if you don't like it).

Borrowing one to repay one, in fact, is 1 2 + 1 4 + 1 6 = 11 12, in fact, it is nothing.

Binary to decimal method.

Starting from the last digit, it is listed as the first placeBit.

The number in the nth place (0 or 1) is multiplied by 2 to the nth power.

The sum of the results obtained is the answer.

For example: 01101011Turn decimal:

Bit 0: 1 times 2 to the power of 0 = 1

1 times 2 to the power of 1 = 2

0 times 2 to the power of 2 0

1 times 2 to the power of 3 8

1 times 2 to the power of 4 16

0 times 2 to the power of 5 0

1 times 2 to the power of 6 64

0 times 2 to the power of 7 0

Then: 1 2 0 8 16 0 64 0 91 binary 01011011 decimal 90

For example, if the binary of 3 is 11, then there is the following:

1*2 1+1*2 0=3 *1 times 2 to the power of 1 + 1 times 2 to the power of zero * and then for example the binary of 10 is 1010, then the conversion to decimal has the following:

Or anyway when you convert binary to decimal system.

n*m x-1)+(n*m x-1) until x equals 0, x represents the total number of binary bits.

n represents what the nth digit of binary is (n is either 0 or 1) m represents the real number 2, this number does not change and will always be 2

n*m^x-1）+（n*m^x-1）……n is 10 bits to the power of x-1 of m, so if there are 10 boxes, you can represent any number from 1 to 999.

According to the title, the remaining greenhouses are 240, which can only be completed within 4 days two days in advance, and the addition of a seeder is 4, so 240 4 4 15, each seeder should sow 15 greenhouses every day.

The actual sowing of half of the greenhouse, the number of days taken t.

It was originally planned that 3 seeders should sow x greenhouses every day, x=480 12=40

t=240/40=6

That is, the remaining half should be completed in 4 days.

Add one of the same seeders, each seeder should sow y greenhouses per day y=240 (4*4)=15

Add one of the same seeders, each of which should seed 15 greenhouses per day.

480 2 (12-6-2) (3+1)=240 4 4=15 (pcs).

15 greenhouses are seeded per day.