Solution: Take 366 days in a year as 366 drawers and 400 people as 400 objects. From the pigeon hole principle 1, we can know that at least two people have the same birthday.
For another example, we randomly find 13 people on the street, and we can conclude that at least two of them belong to the same genus.
"Choose 6 pairs of gloves from any 5 pairs of gloves, of which at least 2 pairs are just a pair of gloves."
"From 1, 2, ..., 10, and there are at least two parity differences. "
2:
Kindergarten bought a lot of plastic toys for white rabbits, pandas and giraffes, and each child chose two at random, so no matter how to choose, two out of any seven children always chose the same toy. Try to explain the truth.
solve
Choose two toys from three kinds, and the matching methods can only be the following six kinds: (rabbit, rabbit), (rabbit, panda), (rabbit, giraffe), (panda, panda), (panda, giraffe), (giraffe, giraffe). Considering that each collocation method is a drawer and seven children are the objects, according to the principle of 1, at least two objects should be placed in the same drawer, that is, at least two people should choose toys in the same collocation method and choose the same toys.
three
It is proved that if eight natural numbers are taken, the difference between the two numbers must be a multiple of 7.
Analysis and solutions
There is such a property in the related problems of divisibility. If two integers A and B have the same remainder divided by the natural number M, then their difference a-b is a multiple of M. According to this property, this question only needs to prove that there are two natural numbers in these eight natural numbers. The remainder of their division by 7 is the same. We can divide all natural numbers into seven categories, that is, seven drawers, according to seven different remainders 0, 1, 2, 3, 4, 5 and 6 obtained by dividing by seven. According to the pigeon hole principle, there must be two numbers in the same drawer, that is, the remainder of their division by 7 is the same, so the difference between these two numbers must be a multiple of 7.
4. For any five natural numbers, it is proved that the sum of three numbers must be divisible by three.
It is proved that the remainder obtained by dividing any number by 3 can only be 0, 1, 2, which can be constructed into three drawers respectively:
[0],[ 1],[2]
(1) If the remainder obtained by dividing these five natural numbers by 3 is distributed in these three drawers respectively, we take 1 from these three drawers, and the sum can be divisible by 3.
(2) If these five remainders are distributed in two drawers, there must be three remainders in one drawer (pigeon hole principle), and the sum of these three remainders is either 0, 3 or 6, then the sum of the corresponding three natural numbers is a multiple of 3.
If these five remainders are distributed in one of the drawers, obviously, the sum of the three natural numbers must be divisible by 3.
5: For any integer 1 1, it is proved that there must be 6 numbers in it, and their sum can be divisible by 6.
Proof: Let this 1 1 integer be: a 1, a2, A3...A 1 1.
6=2×3
(1) Let's consider the case that it is divisible by 3.
According to Example 2, among the 1 1 arbitrary integers, there must be:
3|a 1+a2+a3, let's say a1+a2+a3 = b1;
Similarly, among the remaining 8 arbitrary integers, from Example 2, there must be: 3.
|
A4+a5+a6。 Let A4+A5+A6 = B2;
Similarly, among the other five arbitrary integers, there are 3|a7+a8+a9, and let: a7+a8+a9=b3.
② Consider again that b 1, b2 and b3 are divisible by 2.
According to the pigeon hole principle, at least two of the three integers b 1, b2 and b3 are the same odd or even, and the sum of these two same odd (or even) integers must be even. Let's set 2|b 1+b2.
Then: 6|b 1+b2, which means 6|a 1+a2+a3+a4+a5+a6.
Any integer of ∴ 1 1, where the sum of 6 numbers must be a multiple of 6.