Analysis: This question seems to be confusing. Where should we start? What's the point? In fact, in "two numbers", one of them is an integer multiple of the other. We need to construct a "drawer" so that any two numbers in each drawer are integer multiples of the other. Only by putting the whole geometric series whose common ratio is a positive integer in the same drawer can we get the basic knowledge of natural number classification: any positive integer can be expressed as the product of an odd number and the power of 2, that is, if m∈N+, K∈N+, N. and this representation is unique, such as1=1x2. ...
Prove that any positive integer can be expressed as an odd number multiplied by the power of 2, and this expression is unique, so we can divide the positive integer 1- 100 into the following 50 drawers (because there are 50 odd numbers in * * in1-kloc-0/00):
( 1){ 1, 1×2, 1×22, 1×23, 1×24, 1×25, 1×26};
(2){3,3×2,3×22,3×23,3×24,3×25};
(3){5,5×2,5×22,5×23,5×24};
(4){7,7×2,7×22,7×23};
(5){9,9×2,9×22,9×23};
(6){ 1 1, 1 1×2, 1 1×22, 1 1×23};
……
(25){49,49×2};
(26){5 1};
……
(50){99}。
In this way, the positive integer of 1- 100 will be put into these 50 drawers without repetition or omission. Take any number 5 1 from 100, that is, take any number 5 1 from 50 drawers. According to pigeonhole principle, at least two of them must belong to the same drawer, that is, one of the drawers from (1) to (25). Obviously, any one of the 25 drawers is the same.
Description:
(1) As can be seen from the above proof, this problem can be extended to the general case: from the natural number 1-2n, any number of n+ 1 must have two numbers, one of which is an integer multiple of the other. Think about it, why? Because * * in 1-2n contains n odd numbers of 1, 3, ..., 2n- 1, and N+ 1 > N, the conclusion is inevitable according to pigeonhole principle. By giving n a specific value, different themes can be constructed. In Example 2, the value of n is 50, and the opposite topic can be compiled, such as: "How many numbers should be taken out of the first 30 natural numbers (regardless of these numbers in any way) to ensure that two numbers can be found, and the larger number is a multiple of the smaller number?"
(2) The conclusions of the following two questions are all negative (n is a positive integer). Think about it, why?
① If any number of n+ 1 is selected from 2, 3, 4, …, 2n+ 1, is it necessary to have two numbers, one of which is an integer multiple of the other?
② If any number of n+1 is selected from1,2, 3, …, 2n+ 1, is there necessarily two numbers, and one of them is an integer multiple of the other?
Can you give a counterexample to prove that the conclusions of the above two questions are negative?
(3) If any number of n+ 1 in the two problems in (2) is increased by 1, they are all changed to any number of n+2, are their conclusions positive or negative? Can you judge the evidence?
Example 3. Take 7 numbers from the first 25 natural numbers at will, and prove that there must be two numbers in the number taken, and the largest of these two numbers is not more than 0.5 times of 65438+ decimal.
Proof: Divide the first 25 natural numbers into the following 6 groups:
1; ①
2,3; ②
4,5,6; ③
7,8,9, 10; ④
1 1, 12, 13, 14, 15, 16; ⑤
17, 18, 19,20,2 1,22,23, ⑥
Since the number 7 is randomly selected from the first 25 natural numbers, at least two of the above groups ② ~ ⑥ are selected from the same group, and the largest number of these two numbers does not exceed the decimal of 1.5 times.
Description:
(1) This problem can be changed as follows: take 7 numbers from the first 25 natural numbers at will, and prove that there are two numbers, including their ratios.
Obviously, we must find a way to divide the first 25 natural numbers into six sets (7- 1 = 6), but there is a restriction when classifying: the ratio of any two numbers in the same set is included, so the numerical difference of elements in the same set cannot be too big. In this way, we can use the above special classification: recursive classification:
Starting from 1, it is obvious that 1 can only be used as 1 set alone {1}. Otherwise, the restriction condition is not met.
The number that can belong to a set with 2 is only 3, so {2,3} is a set.
Through this recursion, several consecutive natural numbers belong to the same set, and the largest number does not exceed the times of the smallest number, and six sets that meet the conditions can be obtained.
(2) If we make "drawers" in turn according to the recursive method in (1), the seventh drawer is
{26,27,28,29,30,3 1,32,33,34,35,36,37,38,39};
The eighth drawer is: {40, 465, 438+0, 42, …, 60};
The ninth drawer is: {6 1, 62, 63, …, 90, 91};
……
Then, we can turn Example 3 into the following series of topics:
(1) Select 6 natural numbers from the previous 16 natural numbers;
(2) Choose 8 natural numbers from the first 39 natural numbers;
(3) Choose 9 natural numbers from the first 60 natural numbers;
(4) Select 10 natural numbers from the previous 9/kloc-0 natural numbers; …
You can get the same conclusion: there are two numbers, and their ratio is in.
The above proposition (4) is the test question of the 49th Mathematics Competition in Kiev, the former Soviet Union. If we change the value of the endpoint of the interval [] (p > q), we can construct a series of new problems.