Volume 1, Number 1
January -February, 1995
Pigeonhole Principle
Kin-Yin Li
What in the world is the pigeonhole principle? Well. this famous principle states that if n+ 1 objects (pigeons) are taken from n boxes (pigeonholes), then at least two of the objects will be from the same box. This is clear enough that it does not require much explanation. A problem solver who takes advantage of this principle can tackle certain combinatorial problems in a manner that is more elegant and systematic than case-by-case. To show how to apply this principle, we give a few examples below.
Note that the two examples look alike, however the boxes fornled are quite different. By now, the readers must have observed that forming the right boxes is the key to success. Often a certain amount of experience as well as clever thinking are required to solve such problems. The additional examples below will help beginners becomefarniliar with this useful principle.
Olympiad Corner
The 35th International Mathematical Olympiad was held in Hong Kong last summer. The following are the six problems given to the contestants. How many can you solve? (The country names inside the parentheses are the problem proposers.) -Editors Problem 1. (France)Let m and n be positive integers. Let aI, a2, ..., am be distinct elements of {I, 2, ..., n} such that whenever at + aJ ~ n for some i,j, 1 ~ i ~ j ~ m, there exists k, 1 ~ k ~ m, with at + aJ = at. Prove that
al+~+'..+am
n+l
~-m
2
Example 3. Show that aInQng any nine distinct real numbers, there are two, say a and b, such that
Example 1. Suppose
51 numbers are
chosen from 1, 2, 3, ...,99, 100. Show
that there are two which do not have any
common
prime divisor.
0 < (a-b)! (1 +ab) <.J2 -1.
Solution. The middle expression (a-b)/(I+ab) reminds us of the fonnula for tan(x-y). So we proceed as follow. Divide the interval (-1&12,1&12] into 8 intervals (-1&/2,-31&/8], (-31&/8,-1&14], ..., (1&14,31&18], (31&/8,1&12]. Let the numbers be ap a2, ..., agandletxj=arctanaj, i=I,2,...,9. By the pigeonhole principle, two of the x/s, say Xj and Xk with xi> Xk' must be in one of the 8 subintervals. Then we have 0 < Xj -Xk < 1&/8, so 0 < tan(Xj-xJ = (aj-aJ/(1 +aJaJ < tan(1&/8) = Ii -1.
Solution. Let us consider the 50 pairs of consecutive numbers (1,2), (3,4), ..., (99,100). Since 51 numbers are chosen, the pigeonhole principle tells us that there will be a pair (k, k+l) among them. Now if a prime number p divides k+ 1 and k, thenp will divide (k+ 1) -k = 1, which is a contradiction. So, k and k+ 1 have no common prime divisor.
Problem 2. (Armenia/Australia)
ABC is an isosceles triangle with AB = AC. Suppose that
(i) M is the midpoint of BC and 0 is the point on the line AM such that OB is perpendicular to AB;
(ii) Q is an arbitrary pOint on the segment BC different from Band C;
(iii) E lies on the line AB and F lies on the line AC such that E, Q and F are distinct and collinear.
Prove that OQ is perpendicular to EF if and only if QE = QF.
(continued on page 4)
Example 2. Suppose 51 numbers are chosen from 1, 2, 3, ..., 99, 100. Show that there are two such that one divides the other.
Example 4. Suppose a triangle can be placed inside a square of unit area in such a way that the center of the square is not inside the triangle. Show that one side of the triangle has length less than 1. (This example came from the XLI Mathematical Olympiad in Poland.)
Editors: Li, Kin- Yin, Math Dept, HKUST Ko,Tsz-Mei, EEE Dept, HKUST Ng, Keng Po Roger, lTC, HKP
Artist: Yeung, Sau- Ying Camille, Fine Arts Dept, CU
Acknowledgment: Thanks to Martha A. Dahlen, Technical Writer, HKUST, for her comments.
Solution. Consider the 50 odd numbers 1, 3, 5, ..., 99. For each one, form a box containing the number and all powers of 2 times the number. So the first box contains 1,2,4,8, 16, ...and the next box contains 3,6,12,24,48, ...and so on. Then among the 51 numbers chosen, the pigeonhole principle tells us that there are two that are contained in the same box. They must be of the form 2mk and 2nk with the same odd number k. So one will divide the other.
Solution. Through the center C of the
square, draw a line L) parallel to the
closest side of the triangle and a second
line ~ perpendicular
to L) at C. The lines
L) and ~ divide the square into four
congruent quadrilaterals.
Since C is not
(continued on page 2)
The editors welcome contributions from all students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in TeX, MS Word and WordPerfect, are encouraged. The deadline for receiving material for the next issue is January 31, 1995. Send all correspondence to:
Dr. Li, Kin-Yin Department of Mathematics
Hong Kong University of Science and Technology Clear Water Bay, Kowloon
Hong Kong Fax: 2358-1643 Email: [email protected]
Mathematical Excalibur Vol. 1, No.1, lan-Feb, 95
Page 2
Pigeonhole Principle
(continued
from page 1)
The Game of "Life"
Tsz-Mei Ko
inside the triangle, the triangle can lie in at most two (adjacent) quadrilaterals. By the pigeonhole principle, two of the vertices of the tiiangle must belong to the same quadrilateral. Now the furthest distance between two points in the quadrilateral is the distance between two of its opposite vertices, which is at most 1. So the side of the triangle with two vertices lying in the same quadrilateral must have length less
than 1.
The game of "ljfe" was flfst introduced by John Conway, a mathematician and a game hobbyist currently working at Princeton University. The game is played on an infinite chessboard, where each cell has eight neighboring cells. Initially, an arrangement of stones is placed on the board (the live cells) as the flfSt generation. Each new generation is determined by two simple generic rules:
respectively (Death Rule). The empty cells marked b will become live cells in the next generation (Birth Rule). The second generation is shown in Figure 2.
What will happen in the third, fourth, and nth generation? Is there an initial
generation that will grow infinitely?
Below we provide some exercises
for
the active readers.
The Death Rule: Consider a live cell (occupied by a stone). If it has 0 or 1 live neighbors (among the eight neighboring cells), then it dies from isolation. If it has 4 or more live neighbors, then it dies from overcrowding. If it has 2 or 3 live neighbors, then it survives to the next generation.
1. Eleven numbers
are chosen
from 1, 2,
3, ..., 99, 100. Show that there are two
nonempty disjoint subsets
of these eleven
numbers ~hose elements
have the same
sum.
2. Suppose nine points with integer coordinates in the three dimensional space are chosen. Show that one of the segments with endpoints selected from the nine points must contain a third point with integer coordinates.
The Birth Rule: Consider a dead (unoccupied) cell. If it has exactly 3 live neighbors, then it becomes a live cell (with a stone placed on it) in the next generation.
Here is an example. The six circles in Figure 1 indicate the live cells in the first generation. Those marked i and c will die due to isolation and overcrowding
3. Show that among any six people,
either
there are three who know each other or
there are three, no pair of which knows
each
other.
!GIQII,
Figure
2
4. In every 16-digit number, show that there is a string of one or more consecutive digits such that the product of these digits is a perfect square. [Hint: The exponents
of a factorization of a perfect square into prime numbers are even.] (This problem is from the 1991 Japan Mathematical
Olympiad.)
Proof Without Words
kth Power of a Natural Number n as the S~ of n Consecutive Odd Numbers (k = 2,3, ...)
(Answers
can be found on page 3.)
,
~
k-l
~14 n .,
n -n
n
Mathematical Excalibur Vol. 1, No.1, lan-Feb, 95
Page 3
Problem Corner
We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceeded by the solver's name, address and school affiliation. Please send submissions to Dr. Kin r: Li, Department
of Mathematics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is January 31st, 1995.
Problem 1. The sum of two positive integers is 2310. Show that their product is not divisible by 2310.
color c. Otherwise, all edges of T are colored opposite to c. In both cases, there is a triangle with all edges the same color.
4. Let d1, d2, ..., dl6 be the digits ofa 16-digit number. H one of the 16-digits of the sixteen digits is either 0 or 1 or 4 or 9, then the problem is solved. So, we may assume each of the digits is 2, 3, 5, 6=2x3, 7 or 8=23. Let Xo = 1 and Xi be the product of d J' d2,...,difori=I,2,...,16. Now each Xi = ~'x3qlx5"x7" fori=O, 1,2, ..., 16. Each of the Pi' qi' rj, $t is either even or odd. So there are 24 = 16 possible parity patterns. By the pigeonhole principle, the Pi' qt, rj, $j for two of the seventeen x/s, say Xj and Xk with j < k, must have the same parity pattern. Then dj+1 x ...X dk = X~Xj is a perfect square.
Problem 2. Given N objects and B(~2) boxes, find an inequality involving N and B such that if the inequality is satisfied, then at least two of the boxes have the same
number of objects.
Mathematical Application: Pattern Design
Roger Ng
Problem 3. Show that for every positive
integer n, there are polynomials P(x) of
degree n and Q(x) of degree n-l such that
(P(x)f- 1 = (r-l)(Q(x)f.
Mathematics is by far the most poweiful tool that human race has created. We invite articles which can share with us different areas of applications in mathematics. We wish that this column will inspire students to study mathematics. -Editors Problem 4. If the diagonals of a
quadrilateral in the plane are perpendicular. show that the midpoints of its sides and the feet of the perpendiculars dropped from the midpoints to the opposite sides lie on a circle.
Problem 5. (1979 British Mathematical Olympiad) Let ai' a2, ..., an be n distinct positive odd integers. Suppose all the differences la/-ail are distinct, I ~ i <j ~ n. Prove that a1 + ~ + ...+ an ~ n(n2+2)/3.
Answers
to
Exercises
"Pigeonhole Principle"
in
1. The set of eleven numbers have 211_2 = 2046 nonempty subsets with less than eleven elements, and the maximal sum of the elements in any of these subsets is 91 + 92 + ...+ 99 + 100 = 955. So, by the pigeonhole principle, there are two rionempty subsets with the same sum. If they have common elements, then remove them from both subsets and we will get two nonempty disjoint subsets with the same sum.
CO
---~
Same SlopeT~=:~~
1C
2. For the nine points, each of the threecoordinates is either even or odd. So, there
are 23=8 parity patterns for the coordinates.
Figure
1
By the pigeonhole principle, two of thenine points must have the same parity coordinate patterns. Then their midpoint must have integer coordinates.
3. Let the six people correspond to the six vertices of a regular hexagon. If two people know each other, then color the
segment with the associated vertices red, otherwise blue. SolVing the problem is equivalent to showing that a red triangle or a blue triangle exists.
Take any vertex. By the pigeonhole principle, of the five segments issuing from this vertex, three have the same color c. Consider the three vertices at the other ends of these segments and the triangle T with these vertices. If T has an edge colored c, then there is a triangle with
Mathematical Excalibur Vol. 1, No.1, lan-Feb, 95
Page 4Olympiad Corner
(continued
from page 1)
From Fermat Primes to Constructible Regular Polygons
Tsz-Mei Ko
Pierre de Fermat (1601-1665), an amateur mathematician, once guessed that all numbers in the form 22" + 1 are prime numbers. If we try the fIrst five n's (n = 0,
1,2,3,4), they are in fact all primes:
decided to devote his life to mathematics. After his death, a bronze statue in memory of him standing on a regular 17-gon pedestal was erected in Brauschweig-the hometown of Gauss.
Problem 3. (Romania)
For any positive integer k, let f( k) be the number of elements in the set {k+l, k+2, ..., 2k} whose base 2 representation has precisely three 1 s.
(a) Prove that, for each positive integer m, there exists at least one positive integer k such thatf(k)=m.
(b) Determine all positive integers m for which there exists exactly one k with f(k)=m.
22" + 1
Which
regular
polygons are
constructible? From Gauss's result, we know that the regular triangle, pentagon,
17-gon, 257-gon and 65537-gon are
constructible. (How?) We also know that regular polygons with 7, 11, 13, 19, ... sides are not constructible since they are primes but not Fermat primes. In addition, we know how to bisect an angle and thus regular polygons with 4,8,16,32, ...or 6, 12, 24,48, ...sides are also constructible. What about the others? Is a regular 15-gon constructible? The answer turns out to be
yes since 1/15 = 2/5 -1/3 and thus we can divide a circle into 15 equal parts. What about a regular 9-gon? It can be proved that a regular 9-gon is not constructible. Can you find a general theorem on which regular polygons are constructible?
0
1
2 3 4 3 5 17 257 65537Problem 4. (Australia)
Detennine all ordered pairs (m,n) of
positive integers such that
n3 +1
mn -1is an integer.
It was later discovered by Leonhard Euler (1707-1783) in 1732 that the next Fermat number (n = 5) can be factored as
s
22 + 1 = 641 x 6700417 and thus not a prime. The story would have ended here if without an ingenious discovery by Carl Friedrich Gauss
(1777-1855).
In 1794, at the age of seventeen, Gauss found that a regular 'p-gon" (a polygon
with p sides), where p is a prime, is constructible (i.e., using only ruler and compass) if and only if p is a "Fermat prime" (a prime number in the foim22" + 1). He proved this by considering the solutions of certain algebraic equations. (The interested reader may refer to the book, "What Is Mathematics?" written by Courant and Robbins, Oxford University Press.) The young Gauss was so overwhelmed by his discovery that he then Problem 5. (United Kingdom)
Let S be the set of real numbers strictly greater than -1. Find all functions/." S -+ S satisfying the two conditions:
(i) f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all x and y in S;
(ii) f(x)/x is strictly increasing on each of the intervals -1 < x < 0 and 0 < x.
Are there any other constructible p-gons (where p is a prime) besides the five mentioned? This question is equivalent to asking whether there are any other Fermat primes. To date, no other Fermat number has been shown to be prime, and it is still
not known whether there are more than five Fermat primes. Perhaps you can discover a new Fermat prime and make a note in the history of mathematics. Problem 6. (Finland)
Show that there exists a set A of positive integers with the following property: For any infinite set S of primes there exist two positive integers mEA and n $" A each of which is a product of k distinct elements of S for some k ? 2.
([)
0
Right: A photo of the six members of the Hong Kong Team and one of the editors (far right) taken at the Shatin Town Hall after the closing ceremony of the 35th International Mathematical Olympiad.
From left to right are: SueD Yun-Leung, Chu Hoi-Pan, Tsui Ka-Hing, Wong Him- Ting, Ho Wing-Yip. Poon Wai-Hoi Bobby, and Li Kin- Yin.
