Download Pigeonhole principle

CPSC 313
Spring 2016
Pigeonhole principle
If n pigeons go into n + 1 holes, then some hole must contain at least two pigeons.
If n pigeons go into m holes, and n > m, then some hole must contain at least two
pigeons.
1. Prove that, if we select six distinct numbers from {1, 2, . . . 9}, then two of these numbers
must sum to 10.
Solution. Consider the sets {1, 9}, {2, 8}, {3, 7}, {4, 6}, {5}. The sets will play the role
of the holes, and the pigeons will be the numbers we select. No matter how we choose
the six numbers, we must take from one set with two elements both numbers, and these
numbers will sum to 10.
2. There are n persons present in a room. Prove that among them there are two persons
who have the same number of friends in the room. (If A is friend with B then also B is
friend with A).
Solution. A person (pigeon) goes into the hole labelled i if she has i friends. The holes
are numbered 0, 1, . . . n − 1. However, the hole with number 0 and the hole with number
n − 1 cannot be both occupied (if a person X has n − 1 friends, then each person in the
room has at least one friend – namely X). Thus we have n pigeons and n − 1 holes, so
there is at least one hole with more than one pigeon.
3. Prove that, if we select n + 1 integers from {1, 2, ..., 2n}, then at least two of the selected
numbers have the gcd 1. Here n is a positive integer.
Solution. Let the holes be {1, 2}, {3, 4}, . . . {2n − 1, 2n}. In other words, the holes are
the pairs of consecutive integers. There are n such holes. The pigeons will be the n + 1
numbers we select. Therefore, when choosing n + 1 integers, we must have at least one
pair of consecutive integers, which have the gcd 1.
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