Olympiad Drawer Principles Practice 50

2) 9 * 2 + 4 + 1 [9 * 2 + 4 is a variety of colors added together nine pairs and four singles, take one more, there are 10 pairs].

5: 3*3+2+1=12 [take one more when black and white, red, blue and yellow are full, and it must be one of the three colors of red, blue and yellow].

6: (error correction: take any 11 positive integers, at least two (numbers) their difference is divisible by 10).

10 positive integers, their last digits must be 0-9 respectively, and then take a number, its last digit must coincide with a number in 0-9, so the difference is divisible by 10.

7: 2*3=6, a total of 9 columns, 6<9, so in any case there must be two columns of coloring in the same way.

8: Because the average of odd and odd numbers is a positive integer, and the average of even and even numbers is also a positive integer, there are three numbers, and there must be "same sex".

9: 1+2+3+4+5+6+7+8+9+10+11+12+13=93>90, so there must be 2 numbers equal.

In 10:1-20, prime numbers and one are coprime with any number, there are 9 prime numbers, 1 is one, and if you count all and numbers, there is also a number that is prime or 1, so there must be two coprime numbers.

11: Either they don't know each other, or three people don't know each other, and the other three know each other, or they all know each other.

, at least 2 of the 24 positive integers have the same single digit, and their difference is divisible by 10.

7. Each column has 3 grids, and the 3 grids are colored with two colors at most 2 3 = 8 different methods, and now there are a total of 9 columns, so at least 2 columns are colored.

8. In any three positive integers, two of them must be odd or even, and their average numbers are still positive integers.

Each child needs at least 1 candy in different quantities.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 = 91, so 90 candies are divided among 13 children, each with at least 1 candy, no matter how it is divided, there are always two people who share the same amount.

10. Among the 19 positive integers less than 20, there are 8 prime numbers (2, 3, 5, 7, 11, 13, 17, 19), 10 composite numbers, and 1 number is 1, so among the 11 positive integers that are different from each other and are less than 20, at least one number is a prime number or a number 1, and a prime number or number 1 and any positive integer are co-prime numbers. So if there are 11 positive integers that are different from each other, and they are all less than 20, then two of them must be coprime.

11. Use 6 dots (abcdef) to represent 6 people, and the two people who know each other are connected by a solid line, and those who do not know each other are connected by a dotted line. At least 3 of the 5 lines connected with A are solid lines or dashed lines, and there may be 3 solid lines (set AB, AC, AD), now investigate the connection between the three points of BCD, if all are connected by dotted lines, then the three of them do not know each other; If there is a solid line, you might as well set it to BC, then the connection between the three points of ABC is all solid lines, then the three of them know each other. So if there are any six people, or three of them know each other, or three of them don't know each other, they must be one of them.

I hope it can be helpful to you.

It is said that there are no more than 200,000 human hairs, and if there are 36.45 million people in Shaanxi Province, according to these data, do you know how many people in Shaanxi Province have at least the same number of hair roots?

Answer & Analysis:

If there are no more than 200,000 human hairs, they can be regarded as 200,000 "drawers", 36.45 million people can be regarded as 36.45 million "elements", and 36.45 million "elements" can be put into 200,000 "drawers" to obtain.

3645÷20=182……5 According to the generalization law of the drawer principle, it can be seen that k+1=183

A: There are at least 183 people in Shaanxi Province who have the same number of hair roots.

Drawer Principle 1:

If (n+1) objects are placed in n drawers, then there must be at least 2 objects in one drawer.

For example, if you put 4 objects in 3 drawers, you will break 4 into three integers. And, then there are the following four situations:

Looking at the above four ways of placing objects, we will find a common feature: there is always a drawer with 2 or more objects, which means that there must be at least 2 objects in a drawer.

Drawer Principle 2:

If n objects are placed in m drawers, where n > m, then there must be a drawer that has at least:

k=[n m]+1 object: when n is not divisible by m.

k=n m objects: when n is divisible by m.

Understanding the knowledge points:[x] indicates the maximum integer that does not exceed x.

Example[; [Key Questions:Constructing objects and drawers. That is, we find the quantity that represents the object and the drawer, and then calculate it according to the drawer principle.

7 2 = 3 (Ben)....Wang Lun Blind 1 (Ben Tong Touch).

3+1=4 (Ben).

A: There is always a drawer with at least 4 empty books

So the answer is:

Olympiad Drawer Question: If you want to make sure that you find two piles of apples and pears in a large basket of apples and pears, then how many piles should you at least divide these apples and pears into piles?

Answer 5 heaps. Reverse thinking.

On the other hand, this problem can be understood as follows: a pair of two numbers, at least a few pairs of logs, so that there must be two pairs of these logarithms, and the sum of their numbers in the same position is even.

Don't take it lightly to say that this understanding is wrong. Because the two numbers in each logarithm are random, some people will say that the apples and pears in the basket are fixed, what if you can't tell these numbers? In fact, the number of apples and pears is fixed and random at the same time.

If there is such a division, there must be pears and apples that satisfy it, right? The word "big basket" originally meant random numbers.

Now consider each pair that has been divided, all of which satisfy the 4 types (odd, odd) (odd, even) (even, odd) (even, even). To add that if there are no apples or pears in the pile, 0 is also an even number.

If you look at these 4 types first, you will find a rule, as long as the same type appears twice, then their sum must be even. There are only 4 types, so as long as there are more than 4 piles, there must be two logarithmic numbers of the same type, and they add up to the requirements.

Since you want to find the least distribution, think about it again, and divide it into 2 piles, 3 piles, and 4 piles that are full and do not meet the requirements.

Let's start the discussion:

Divided into 2 piles: There are too many cases of dissatisfaction, (odd, odd) + (odd, even) are not satisfied, such as (1, 1) + (3, 4). When there are 4 apples, 5 pears there will be a distribution that does not meet the requirements.

Divided into 3 piles: there are also many situations that are not satisfied, such as: (odd, odd) + (odd, even) +

Even, odd) is not satisfied. I won't give practical examples, it's easy to find.

Divided into 4 piles: In the 4 logs divided into numbers, either one type is counted in one pair, or one type appears twice. As analyzed earlier, the condition must be met twice for one type. And a type that appears once does not meet the condition. The example is given at the beginning.

So, the answer is 5 piles.

551 19 times 29, 29 1 28 not divisible by 3

19 1 18 3 times 6

There are 18 people in total.

18 4 people.

According to the drawer principle.

Use one method to get 5 people.

A: ——

551=19*29=(3*6+1)29

So the teacher took 18 students and planted 29 trees each.

551=19*29

Assuming there are 3x classmates, then the total number of teachers and students is 3x+119=3*6+1, and 29 is divided by three and does not leave 1So the total number of teachers and students is 19, and there are 18 students.

If students are divided into 4 groups, then one group has at least: [(18-1) 4]+1=5 people.

551=19*29

The students happened to be divided into three groups, plus the teacher must be 19 people, and the number of students must be 18 people. If the total number of students is 29, then 29-1=28, the students cannot be divided into three groups exactly.

If you want to divide these students into 4 groups, then at least one group has 5 students.