Pigeonhole Principle Problems and solutions
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Pigeon hole principle(PHP)
Questions:
1. There are 12 children in the family.
(a) Prove that at least two of the children were born on the same day of the week;
(b) Prove that at least two family members (including mother and father) are born in the
same month;
2. Pigeonhole Elementary School has 500 students. Show that at least two of them were
born on the same day of the year.
3. There are 800,000 pine trees in a forest. Each pine tree has no more than 600,000
needles. Show that at least two trees have the same number of needles.
4.Show that among any n + 1 distinct numbers one can find 2 numbers so that their difference is
divisible by n.
5.There are five points inside an equilateral triangle of side length 2. Show that at least two of
the points are within 1 unit distance from each other.
6. In a room of 500 people, there exist two who share a birthday.
7. In every set of 100 integers, there exist two whose difference is a multiple of 37.
8. For any five points inside a unit square, there are two points at distance less than 1$
2√2
9. Given any 6 integers from 1 to 10, show that some two of them have an odd sum.
Solutions
a)There are 12 children (pigeons) which we are placing into 7 days of the
week(pigeonholes), so by the PHP, some day of the week has two children.
b. There are 14 family members (pigeons) and only 12 months they can be
born in (holes), so some two family members must be born in the same
month by the PHP.
2. Solution: There are 500 students, and only 366 days they could have been
born on, so by the PHP some two students were born on the same day.
4. Solution: Consider n boxes labelled 0, ..., n − 1. Given the n + 1 numbers,
place each in a box according to its remainder upon division by n. By the
pigeonhole principle, some box must contain at least two numbers. That
means those two numbers have the same remainder upon division by n,
and hence their difference is divisible by n.
Question: There are five points inside an equilateral triangle of side length
2. Show that at least two of the points are within 1 unit distance from each
other.
Solution: Partition the triangle into four smaller equilateral triangles by
connecting the midpoints of the three sides of the big triangle. Each of the
small triangles has side length 1. By the pigeonhole principle at least two of
the points must be in the same small triangle, hence their distance is
bounded by 1.
6. Solution. The pigeons are the 500 people. The pigeonholes are 366 possible birthdays. Map
each person to his or her own birthday. Since there 500 people and 366 birthdays, some two
people must have the same birthday by the Pigeonhole Principle.
7. Solution. The pigeons are the 100 integers. The pigeonholes are the numbers 0 to 36. Map
integer k to k rem 37. Since there are 100 pigeons and only 37 pigeonholes, two pigeons must
go in the same pigeonhole. This means k1 rem 37 = k2 rem 37, which implies that k1 − k2 is a
multiple of 37.
8. Solution. The pigeons are the points. The pigeonholes are the four subsquares of the unit
square, each of side length 1/2 . There are five pigeons and four pigeonholes, and a pigeon maps
to the subsquare it is in (points on a boundary get assigned to the leftmost lowest possible
subsquare) so more than one point must be in the same subsquare. The points in the same
subsquare are at distance less than 1$
, because the most distant points in the quadrant are
2√2
at opposite corners at exactly this distance, but only one of these corners can be inside the
square (not on the boundary), so the distance between the points in the quadrant must actually
be less than 1$
.
2√2
9. Possible pigeonholes: {1,2}, {3,4}, {5,6}, {7,8}, {9,10}. There are 5 pigeonholes and 6 pigeons, hence
there exists a pigeonhole with 2 pigeons.