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HAVERFORD COLLEGE PROBLEM SOLVING GROUP
October 2007
Solutions (part 1): The Pigeonhole Principle
Exercises
Zeitz 3.3.15 Seven points are placed inside a regular hexagon with side length 1. Show
that at least two points are at most one unit apart.
Partition the hexagon into six equilateral triangles meeting at its center. Each triangle has
a diameter (maximum point-to-point distance) of one unit. At least one of those triangles
contains at least two points.
Zeitz 3.3.16 101 points are placed inside a 1 × 1 square. Show that some three of them
form a triangle with area no more than 0.01.
(1)
Partial solution: Partition the square into a ten by ten grid comprising 100 square
cells of side 0.1. At least one cell contains two points. Our partial solution assumed
there was a point in a neighboring cell. When this is the case, the three points have a
base not exceeding 0.2 and a height not exceeding 0.1, and so the area does not exceed
0.01. Unfortunately, we did not show that such a configuration must exist (and indeed it
does not have to happen).
(2)
The question is about triangles. How can we assure the existence of a triangle
within a pigeonhole? Answer: make just 50 pigeonholes. Therefore, partition the square
into 50 vertical strips of width 0.02. At least one strip receives three points. They form a
triangle (possibly degenerate) whose base cannot exceed length 1 and whose height
cannot exceed 0.02, so its area must be less than 0.01.
Zeitz 3.3.17 Show that among any n + 2 integers, either there are two whose difference
is a multiple of 2n, or there are two whose sum is divisible by 2n.
We first looked at this modulo n, but that didn’t work, so then we tried creating
pigeonholes consisting of sets of residues modulo 2n rather than individual residues. The
sets have representatives 0, ±1, ±2, …, ±(n–1), and n modulo 2n. There are n+1
pigeonholes, so at least one receives two integers. Therefore they are negatives of each
other (modulo 2n) or they are equal (modulo 2n). In the first case their sum is zero and in
the second case their difference is zero (modulo 2n), which is what we needed to show.
Comment. Working “modulo n” for a whole number n means computing with the
remainders upon dividing by n: their “residues.” Equivalently, write all numbers base n
and just look at the one’s digit. This algebraic structure—a commutative ring—is called
Z/nZ. All the usual rules of arithmetic still work: the identity properties of zero and one,
the commutative and associative laws, and the distributive laws of multiplication over
addition. Moreover, although you cannot generally divide one integer by another (and
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HAVERFORD COLLEGE PROBLEM SOLVING GROUP
September 2007
end up with an integer result), often you can with modular arithmetic. For instance, 1/3
modulo 10 equals 7 because 3 × 7 = 1. (Watch out though: 1/2 does not exist modulo 10,
because there is no integer k for which 2 × k ends in a one in base ten!)
Zeitz 3.3.19 People are seated around a circular table at a restaurant. The food is placed
on a circular platform in the center of the table and this circular platform can rotate. Each
person ordered a different entrée, and it turns out that no one has the correct entrée in
front of him. Show that it is possible to rotate the platform so that at least two people will
have the correct entrée.
(1)
Let there be n people. Label the possible rotations (in the positive direction, say)
by the number of people passed by each entrée: 1, 2, …, n–1. There are, however, n
possible matches of entrées to people, so at least one of these n-1 rotations has at least
two matches.
(2)
Averaging over all possible rotations, the expected number of matches per
rotation must be exactly one, because the expectation for any individual is 1/n. In the
current configuration there are no matches, implying the average over the remaining
rotations must be greater than one (indeed, it must be n/(n-1)). By the PPP, there is a
rotation with more than one match.
…
The Fibonacci numbers are the sequence 1, 1, 2 = 1+1, 3 = 2+1, 5 = 3+2, 8 = 5+3, …,
etc.: each new number is formed by adding the last two.
Engel 4.8 Prove that the sequence of last digits of the Fibonacci numbers, 1, 1, 2, 3, 5, 8,
3, 1, 4, …, is periodic. What is the maximum possible length of the period?
Each number is determined by its two predecessors. Therefore, we created 10*10 = 100
pigeonholes from all possible ordered pairs of last digits. After generating 102 Fibonacci
numbers, we would have 101 ordered pairs (Fn, Fn+1) in the sequence, guaranteeing the
identity of two pairs, say (Fn, Fn+1) and (F m, F m+1). The period would therefore not
exceed |m – n|. Since n ≥ 1 and m+1 ≤ 102, we obtain |m – n| ≤ 100. However, it is clear
that (0,0) cannot be included among these pairs, because then the sequence when run
backwards would always have zeros for its last digits, which is impossible. This allowed
us to improve our estimate of the maximum period to 99.
Comments. This exercise raises many interesting questions. Is the period indeed 99? If
not, then what can be said? Do all possible digits 0, 1, …, 9 show up? What about
periods modulo other integers besides 10? What about when we start the Fibonacci
sequence with different values, such as 1, 3, 4, 7, 11, … (the Lucas numbers)? What
about generalized Fibonacci sequences, such as 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, … (the next
number is the sum of the three previous ones)?
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September 2007
Our method shows the period modulo b cannot exceed b2 – 1. Our tests with b = 2, 3
showed it can equal b2 – 1 in those cases.
Here are some partial answers. First, let’s continue the experiment to b = 4. Here is the
Fibonacci sequence modulo 4: 0, 1, 1, 2, 3, 1, 0, 1, 1, … . It has period six, not 4 2 – 1 =
16. The period does not have to equal b2 – 1.
This sequence modulo 4 generates the ordered pairs (0,1), (1,1), (1,2), (2,3), (3,1), and
(1,0). One ordered pair not in this set is (3,3). Let us, then, create a Fibonacci sequence
beginning with (3,3): 3, 3, 2, 1, 3, 0, 3, 3, … . This, too, has period six. It generates the
cycle of ordered pairs (3,3), (3,2), (2,1), (1,3), (3,0), (0,3). So far we have accounted for
13 ordered pairs (including (0,0)). One of the three missing ordered pairs is (0,2), so here
we go: 0, 2, 2, 0, 2, 2, … . This sequence has period three, not six. Its ordered pairs are
(0,2), (2,2), (2,0).
Briefly, we have found that the Fibonacci transform F (a,b) = (b,a+b) acts on
Z/4Z × Z/4Z and has four periodic orbits of periods 1, 6, 6, and 3. Let’s see why.
There is a relationship among the four orbits, which becomes noticeable when we write
them in parallel:
(0,1), (1,1), (1,2), (2,3), (3,1), (1,0)
(0,2), (2,2), (2,0); (0,2), (2,2), (2,0)
(3,3), (3,2), (2,1), (1,3), (3,0), (0,3)
(0,0); (0,0); (0,0); (0,0); (0,0); (0,0)
In an obvious way, the second is twice the first, the third is thrice the first, and the fourth
is four times (or zero times) the first: orbits are acted on by multiplication.
When we double all numbers modulo four, we are really working modulo two. In other
words, the period-three orbit really “comes from” the orbit (0,1), (1,1), (1,0) created by
the Fibonacci numbers modulo two. This has to do with the compositeness of four. Can
you generalize this result? What does it imply about the maximum periods of composite
numbers?
It looks like it will be most interesting to look at the Fibonacci numbers modulo primes.
After all, by the Chinese Remainder Theorem, if you know the remainders modulo two
relatively prime numbers (such as powers of two distinct primes), then you can determine
the remainders modulo their product. It is an interesting exercise to make this statement
concrete by relating periods modulo m and periods modulo n to periods modulo mn.
Some primes are distinguished by actually appearing in the Fibonacci sequence: 2, 3, 5,
13, 89, etc. When you work modulo such a prime, there is a pretty simplification.
Consider 13, for instance (because 5 is a little too small to see the pattern). Here we go:
0, 1, 1, 2, 3, 5, 8, 0, 8 (=8*1), 8 (=8*1), 3 (=8*2), 11 (=8*3), 1 (=8*5), …
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September 2007
Suppose we reach the base prime p at the nth term. (I start numbering with F 1 = 0 and
F2 = 1). Then, working modulo p, the general pattern is …, Fn-1, Fn, Fn+1, Fn+2, … = …,
Fn-1, 0, Fn+1, Fn+2, … . But we can recompute the values starting with Fn-1 and 0: Fn+1 =
Fn-1 + 0 = Fn-1 and Fn+2 = 0 + Fn-1 = Fn -1. In other words, we have arrived at the initial
sequence (0, 1, 1), multiplied by Fn-1 (which cannot be zero modulo p, because it exceeds
p/2). We conclude that the Fibonacci sequence modulo p contains an initial sequence
0, 1, 1, 2, …, Fn-1 (which contains no subperiods) followed by the same sequence
multiplied by Fn-1, which then must be followed by that sequence multiplied by Fn-1
(which therefore is the initial sequence multiplied by Fn-12), etc. Fermat’s Theorem says
this will continue until p-1 distinct subsequences are produced, because p-1 is the first
power k for which Fn-1k = 1 modulo p. At this point we will have returned to the initial
0, 1, 1, 2, … and the sequence repeats for the first time. Its period therefore equals n-1
times p-1.
Let’s count things. Notice that for p ≥ 13 there are enough gaps in the Fibonacci
sequence to guarantee there are fewer than p numbers in 0, 1, 1, 2, …, Fn-1. Therefore the
period is less than (p-1)*(p-1) < p2 – 1. (We will look at p = 5 in just a moment.)
There are other interesting primes, the first of which are 7 = 2 + 5 and 11 = 3 + 8. These
are the primes formed by adding Fn-1 to Fn+1. Others of this form include 29 = 8 + 21, 47
= 13 + 34, and 199 = 55 + 144. It is amusing to investigate the Fibonacci periods for
these primes. What happens? (For example, the Fibonacci transform modulo 11 has ten
orbits of period 10 starting 0,1; 0,2; …, 0,10; another period-10 orbit starting 5,7; and two
period-5 orbits starting 2,8 and 9,3 (= -1*2, -1*8).)
Finally, let’s draw some conclusions in the original situation modulo 10. We have seen it
is fruitful first to look modulo 2 and 5, because 10 = 2*5 is its prime power
decomposition. The Fibonacci sequence modulo 2 just goes 0, 1, 1, 0, 1, 1, … with
period three. Modulo 5 it’s longer, but we have already analyzed that: the initial
subsequence 0, 1, 1, 2, 3 gets multiplied by 3, then by 32 = 4, then 3 3 = 2, and then we
repeat. Thus, the period equals 5*4 = 20. Because the least common multiple of 3 and
20 is 60, the period of the Fibonacci sequence modulo 10 is 60, not 99.
Exercise: Can you find a simple, low-computation way to find and describe the other
periods of the Fibonacci transform modulo 10?
We’re not yet done! Consider the periods modulo one of the special primes we have
investigated so far. What happens for the other primes 17, 19, 23, 31, etc., that we
haven’t yet investigated? Can we say anything in general? (I don’t know, but I would
start by noting that 21 = 3*7 and 34 = 2*17 and doing something with that. Does every
prime appear as a factor of some Fibonacci number?—you can answer this!)
At this point you should have no trouble investigating a related question: What is the set
of residues of the Fibonacci numbers modulo a whole number n? Do you obtain all the n
equivalence classes or can there be fewer? If so, can you predict what they would be?
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