Pigeon Nest principle (pigeonhole Principle)

basic theorem of pigeon nest principle (pigeonhole Principle)：
If N + 1 objects is distributed into N boxes and then at least one box contains and more of the objects.
A simple example: the birthdays of at least two people in 13 people are in the same January.
related principles of pigeon nest principle: If n objects is put into N boxes and no box gets more than one object, then each box has a object in it.
Let X and Y is finite sets and let F:x-y is a function from X to Y. There is following conclutions:
if X has more elements than Y, and then F was not one-to-one.
if X and Y have the same number of elements and F are onto (mapped), then F is one-toone.
if X and Y have the same number of elements and F are one-to-one, then f is onto.
Application:
Congruence sequence problem: 2,4,6,3,5,5,6 is a series that defines sum (i) =a1+a2+......+ai, then the corresponding sum (i) should be: 2,6,12,15,20,25,31.sum series mod (7) The result is: 2,6,5,1,6,4,3.  The sum of two numbers is the same, that is, SUM (2) =q1*7+r. SUM (5) =q2*7+r. Then there is sum (5)-sum (2) = (Q2-Q1) *7. Hence: (6+3+5)%7=0.
Application: A chess player has 11 weeks to prepare for a game, it decides to train at least one at a time, in order to guarantee not excessive, no more than 12 games per week. Proving that there is a continuous M-day this contestant will train 21 games.
Answer: Set SI is from the first day to the day of the training times, there are 1<=s1<s2<s3<......<s77<=132. -->22<=s1+21<s2+21<s3+21<......<s77+21<=153. S1,s2......s1+21,s2+21......s77+21 A total of 154 numbers, however all the numeric values must be no more than 153, so there must be two numbers are equal. Sj+21=si---j+1,j+2,......, I have played 21 games in a day.
Application: from Integer to ... 200 to choose 101 number, there must be 2 numbers a, a, can meet such conditions: a%b=0 or b%a=0.
Solution: By the knowledge of the decomposition of the element factor, a number can be written as 2^k*a, where the k>=0,a is odd. The analysis shows that there are odd numbers within the range of 1--200:1,3,5 ... 1991 a total of 100. The 101 numbers selected must have two of the same odd factor A. The result of dividing these two numbers is 2^ (K1-K2), which satisfies the dry conditions of the problem.
Pigeon Nest Principle Enhanced Version [Pigeonhole Principle (Strong Form)]：
Let QL, Q2,......, qn is positive integers. If QL + q2 + ... + qn-n + 1 objects is distributed into N boxes, then either the first box contains at least QL objects , or the second box contains
At least Q2 objects, ..., or the nth box contains at least QN objects.
Inference: Let N and r be positive integers. If N (r-1) + 1 objects is distributed into N boxes and then at least one of the boxes contains r or more of the objects.
Another argument: If the average of n nonnegative integers ml, m2, ..., MN is greater than r-1, which is, (ml + m2 + ... + mn)/n>r- 1, then at least one of the integers is greater than or equal
to R.
Another averaging principle:if the average of n nonnegative integers ml, m2, ..., MN is at least equal to r, then at Le AST one of the integers ml, m2, ..., MN satisfies Mi >= r.
Example: Try to prove that the number of n+1 is selected from {1,2,3......kn}, there are always two numbers with a maximum difference of k-1.
Answer: Divide all the numbers into n groups: {1,2,3......k},{k+1,k+2,k+3 ... 2k},{2k+1,2k+2 ... 3k} ... {(n-1) k+1, (n-1) k+2......kn}. The selected n+1 number must have two belong to the same group, then their difference is the largest k-1.
Example: Louis Possa was a Hungarian mathematician, and at the age of 11 Hungarian mathematician Ernie Deuce gave him a question: "If you have n+1 integers on hand, these integers are less than or equal to 2n, then you are bound to have a pair of numbers that are mutual." Do you know what the reason is? Posa only half a minute to answer the question skillfully.
Answer: Suppose this statement is wrong, then take the maximum number and the minimum number: a (n+1) and A1, there is A1>=2,a (n+1) must be greater than or equal to (n+2) * *.
Example: 1 to 10 of the ten words randomly into a circle, proving that there must be a three adjacent number of the sum greater than or equal to 17.
Answer: 1 to 10 of these 10 numbers randomly into a circle, from which to take three adjacent numbers there are 10 ways. The sum of the 10 species equals: 1+2+3+......+10 =3*11/2*10. So, the average number of neighbors is 3 and 3*11/2*10/10=16.5>16. So, there must be a sum of three adjacent numbers greater than or equal to 17.
Application 9.Show that every sequence AI, A2, ..., an2+1 of N2 + 1 real numbers contains either an increasing subsequence of length n + 1 or a decreasing subsequence of length n + 1. Proof See figure:

Using the remainder classification to construct "Pigeon Nest":
Try to prove that any given 52 integers, they must have 2 numbers, and the difference is a multiple of 100 (that is, divisible by 100)
Solution: Use the 100 remainder of 0,1,2,...,99 to construct the pigeon nest, divide them into 51 groups, and construct 51 pigeon nests: {0},{1,99},{2, 98}, ... {49,51},{50}. By the pigeon Nest principle, these 52 integers are divided by 100 to produce 52 remainder R1,R2,... R52 must have two remaining chatter in the same group.

The popularization of pigeon nest principle---the specific contents of RAMSEY theorem can be consulted "introductory combinatorics fifth Edition"--the pigeonhole principle--a theorem of RAMSEY