Mathematical “Knacks”
composed by K. E. Feldman∗
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Introduction
Many problems of elementary mathematics, which are nicely formulated in everyday language, conceal simple logical principles like the Dirichlet Principle, the Maximum Principle, Invariance, etc. Though the principles are simple, the actual use of them requires
a lot of training. Quite often it is not clear what the right principle is to choose for the
solution of a problem, how to apply it properly and, finally, how to find the best way to
explain the solution. If you can do it correctly, then the reward will not just be great
satisfaction but also the ability to find a good strategy in various complicated real life
situations.
A particular thing that you will have to learn is to be as rigorous as possible even if
the style of the problem is very relaxing and confusing. We hope that this task will help
you to achieve it quickly.
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The Dirichlet Principle
Every one knows that it is impossible to put n + 1 pigeons into n pigeonholes so that in
every pigeonhole there would be only one pigeon. This is the simplest formulation of the
Dirichlet Principle.
Example 2.1 At a party with 28 children James ate 13 sweets and all other children ate
less. Prove that at least three children ate the same number of sweets.
Solution. In this example the ”pigeons” are the children at the party and the ”pigeonholes” are numbers of sweets that each child ate at the party. We put into ”pigeonhole 0”
those children who did not eat sweets at all during the party, into ”pigeonhole 1” those
who ate only one sweet, etc., and, finally, we put James alone into ”pigeonhole 13”.
Now we apply the Dirichlet Principle. Please, take note as this is the most important
step.
∗
Please, send comments, corrections and queries at [email protected].
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We prove the statement by contradiction. Assume that no three children ate the same
number of sweets, i.e. in each ”pigeonhole” 0, 1, 2, 3, ..., 12, we have put less than
three children. Then in all these ”pigeonholes” there are at most 26 children. In addition
to that we have James who was the only person who ate 13 sweets. Therefore, in all
”pigeonholes” we can have no more then 27 children, but there were 28 children at the
party. This is a contradiction.
Test Question 1 Prove the general Dirichlet Principle: it is impossible to put nk+1
pigeons into n pigeonholes so that in each pigeonhole we would have no more than k
pigeons.
The Dirichlet Principle gives the key to many text problems and now we discuss some
types of such problems.
2.1
”Friendship”
Two people are either friends with each other or not. Here we think of friendship as
something ”symmetric”: if John is a friend of Tom, then Tom is a friend of John.
Example 2.2 Among any 7 people there are at least two with the same number of friends.
Solution. Let us take 7 ”pigeonholes”: 0,1,...,6. We put into “pigeonhole” k those people
who have exactly k friends. Two situations are possible:
1) “pigeonhole 0” is empty;
2) “pigeonhole 0” is not empty.
In the first case we would have all 7 people being assigned to pigeonholes 1,2,...,6. As
there are less pigeonholes than people, the Dirichlet Principle implies that there are at
least two people in one ”pigeonhole” and these two have the same number of friends.
In the second case we would have ”pigeonhole” 6 empty as if there is a person assigned
to it, then she/he is a friend to every other of these 7 people and in particular to those
who are assigned to ”pigeonhole” 0. But every person from the ”zero pigeonhole” has
no friends at all and, therefore, ”pigeonhole” 6 is empty. This again gives us only 6
pigeonholes for seven people and as earlier we conclude that there are at least two people
in the same ”pigeonhole”, i.e. with the same number of friends.
2.2
Divisibility
”Pigeonholes” for ”pigeons” appear naturally when we study remainders of division by
integer numbers. In such situations remainders are ”pigeonholes” and integer numbers
are the pigeons.
Example 2.3 Prove that among any 100 integers there are always two whose difference
is divisible by 99.
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Solution. There are 99 different remainders of division by 99:
0,
1,
2,
...,
98.
These are our ”pigeonholes”. We have 100 integer numbers. We put each of them into the
”pigeonhole” whose number is the remainder of the division of the chosen integer by 99.
At least two out of 100 integers will be assigned to the same ”pigeonhole”. The difference
of any two integers from the same ”pigeonhole” is divisible by 99.
Example 2.4 Eleven integers are written in a row. Prove that one of the integers, or the
sum of a consecutive subsequence of them is divisible by 11.
Solution. Denote the integers by a1 , a2 , · · · , a11 . Consider 11 numbers:
a1 ,
a1 + a 2 ,
a1 + a 2 + a 3 ,
···
a1 + a2 + · · · + a11 .
If one of them is divisible by 11 then the problem is solved. In the opposite situation there
are only 10 possibilities for the remainders of these numbers with respect to division by
11. Therefore at least two of these 11 integers have the same remainder modulo 11 and
their difference is the required sum of consecutive numbers.
2.3
The Dirichlet Principle in Geometry
We often come across the Dirichlet Principle in geometrical problems, when the question
concerns length or area.
Example 2.5 51 points are thrown into a square of side 1m. Prove that there are at
least three of them that can be covered by a disk of radius 1/7m.
Solution. Let us divide the square into 25 equal squares of size 0.2m × 0.2m. These are
our ”pigeonholes”. Due to the Dirichlet Principle at least one
√ of them contains 3 of our
51 points. This square can be inscribed in a circle of radius 2/10m < 1/7m.
Example 2.6 Several arcs of a circle are painted black. The sum of lengths of black arcs
is less than the half of the length of the circle. Prove that the circle has a diameter of
which both ends are not painted.
Solution. Let us paint blue all the arcs symmetric to the black ones with respect to the
centre of the circle. Because the sum of lengths of blue arcs is equal to the sum of lengths
of the black arcs, the total length of all painted arcs is less than the length of a circle.
Therefore, there is a point which was not painted. The diameter of the circle with one
end at this point has both ends unpainted.
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Test Question 2 How did we use the Dirichlet Principle in the above solution?
Another application of the Dirichlet Principle is based on the following observation.
Consider a segment AB which is covered by some segments of a total length greater than
|AB|. Then there is a point which belongs to at least two of the covering segments.
Example 2.7 Several circles with the total sum of their radius 0.6cm are placed inside
the square ABCD of side 1cm. Prove that there is a line which is parallel to the side AB
and which has common points with at least two distinct circles.
Solution. Let us project all the circles orthogonally onto the side AD. As the total sum
of lengths of the diameters of the circles is 2 · 0.6cm = 1.2cm, there is a point on AD
covered by projections of two points from different circles. The line which goes through
these two points is parallel to AB.
2.4
Sooner or Later...
If we have an infinite process whose state at every discrete moment of time is an element
of a finite set, then in a finite number of steps the process will take a state which was
already attained. This is also the Dirichlet Principle when the “pigeonholes” are possible
states of the process and the “pigeons” are the states of the process at each moment of
time. Let us have a look how it works.
Example 2.8 The Fibonacci series consists of numbers
1,
1,
2,
3,
5,
8,
13,
21,
34,
55,
89,
··· ,
and every next one is the sum of the previous two. Prove that among 100 000 001 first
elements of the Fibonacci series we have a number that ends with four zeros.
Solution. We denote the first 100 000 002 Fibonacci numbers by a1 , a2 , ..., a100000002 . Let
us consider all the pairs (m, n) of integer numbers with 0 ≤ m, n < 10000. These are
our “pigeonholes” and there are 100002 = 100000000 of them. We put each of the pairs
(ak , ak+1 ), 0 < k ≤ 100000001, into one of these “pigeonholes” according to the remainders
the pair gives with respect to the division by 10000. There are 100000001 “pigeons”=pairs
and 100000000 pigeonholes. Therefore, in at least one of these “pigeonholes” we have two
distinct pairs (ak , ak+1 ) and (am , am+1 ), 0 < k < m ≤ 100000001:
ak ≡ am (mod10000),
ak+1 ≡ am+1 (mod10000).
From here it follows that if k > 1, then
ak−1 ≡ ak+1 − ak ≡ am+1 − am ≡ am−1 (mod10000).
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We can continue this chain of identities further on until we get
1 = a1 ≡ am−k+1 (mod10000).
It is now clear that am−k ≡ 0(mod10000).
Test Question 3 Prove that, indeed, am−k ≡ 0(mod10000).
Test Question 4 Prove that for any n ∈ N there is a Fibonacci number that ends with
n zeros.
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3.1
Invariants
Equivalence Relations
We have already mentioned processes as objects of mathematical problems. Here we
discuss one of the most general methods of treating problems which involve finite or
infinite processes. To define a process itself we need a set of states for the process. For
the sake of simplicity, in what follows we will assume the set of states to be a countable
set, i.e. you can count all elements of the set by natural numbers, possibly using infinitely
many of them. States can be numbers, geometrical figures, positions of points on a plane
or of draughts on a board, various combinations of cards, results of football matches and
many many other things. For us it is only important that the states are related to each
other. They have the same nature and you can get from a certain state to another one
by means of special transformations which form the process itself.
Example 3.1 Take 25 sweets and place them on a table in two piles: the first one with
12 sweets and the second one with 13 sweets. You play with your friend following the rule
that at each step each of you in turn either eats two sweets from one pile or moves one
sweet from the first pile to the second one. The loser is the one who cannot make her/his
next step. Prove that the player who starts first always loses the game.
Solution. You can easily recognise the set of states for this game. It is a set of integer
pairs (m, n) which denote numbers of sweets in the first and the second piles. The process
itself consists of changing one state to another one. It is easy to see that a pair of integers
(m, n) goes into one of the following three:
(m, n − 2),
(m − 1, n + 1),
(m − 2, n).
Of course, in the context of the problem the condition should be satisfied that the result
consists of nonnegative integers (do not forget we are playing with non-abstract sweets).
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Observation. Under the action of this process the remainder of the difference between
number of sweets in the second pile and in the first pile with respect to division by 4
always changes by 2.
At the start this remainder was 1 ≡ 13 − 12(mod4). Therefore, we get a sequence of
remainders 1, 3, 1, 3, 1, 3, 1, 3, ... Moreover, each time the second player is about to play,
the remainder of the difference is 3. This means that there is either at least one sweet in
the first pile (the pair of integers (1, 0) gives us 3 ≡ 0 − 1(mod4)) or at least 3 sweets in
the second pile (why?). In the first case the second player can move the sweet from the
first pile to the second one, and in the second case the second player just eats two of at
least three sweets in the second pile. Therefore, the second player can always continue
the game. The game cannot continue infinitely often as on each step the players decrease
either the total number of sweets by two or the number of sweets in the first pile by 1
(why does this guarantee finiteness of the process?).
An important feature of this example is that we found a quantity (the (mod4) difference of numbers of sweets in the piles) that stays unchanged after applications of two
steps of the process. Let us try to formulate this in more mathematical terms.
Let M be the set of all states of the process. The rules by which the process transforms
one state α ∈ M to another β ∈ M is just a set of functions fj : M → M , j = 1, 2, ...,
and fk (α) = β for some k. After k steps of the process state α is transformed into
fik ◦ · · · ◦ fi1 (α).
Transitivity. The first important feature of such a process is transitivity. It means
that if a finite number of steps of the process transforms state α into state β and another
sequence of steps of the process transforms β into state γ then there is a sequence of
transformations of the process which transforms α into γ (for example, we can take a
composition of the previous two transformations: α → β and β → γ).
Symmetry. The process is called symmetric when it can transform state α to state β
if and only if it can transform state β into α.
The process we came across above was not symmetric, but as you might notice, we
really did not work with the process itself. We were studying only remainders of the
difference with respect to the division by four.
Reflexivity. Finally, let us agree that from any state α the process can get into α for
example by doing nothing (this is so called “mathematical abstraction”, by the way).
This property of the process is called reflexivity. It will be very important for the proof
of Theorem 3.1 below.
Equivalence. We call state β related to α:
α→β
if the process can transform α into β following its rules.
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If the process obeys three conditions (see above):
1) transitivity;
2) symmetry;
3) reflexivity,
then the relation ”→” induced by this process on the set of all possible states is called
an equivalence relation and related states are called equivalent: α ∼ β.
Theorem 3.1 The equivalence relation ∼ decomposes the set of all states M into the
union of pairwise disjoint equivalence classes:
[
[
\
M = M α1
M α2
··· ,
M αi
Mαj = ∅, i 6= j,
where equivalence class Mαi consists of all states β ∈ M which are equivalent to αi ,
i = 1, 2, ...
β ∼ αi .
Test Question 5 Prove that α ∈ Mα . What is the property of the process you are using
for it?
Test Question 6 Prove that if α 6∼ β, then Mα
process you are using this time?
T
Mβ = ∅. What is the property of the
Test Question 7 Finish the proof of Theorem 3.1. Where do you use the symmetry
property of the process?
Before we move further we need to say a few words about mathematics, which is still
an alive subject with more than 2000 years of history. You do need a good sense of humour
when you think about mathematics. Abstract things, like we just had at the end of this
section, do not always help in solutions and sometimes even very “rigorous” theorems
turn out to be completely wrong. The only way to overcome this unpleasantness is to
keep very cool and use a high level of common sense. Have a look at the following joke:
Joky Example of P. Dirac. Three fishermen were catching fish. Later in the night
they fell asleep. In the morning the first fisherman woke up, counted the fish, threw one
fish into the river, took 1/3 of the remaining fish and walked away. Then the second
fisherman did the very same thing. He counted the remaining fish, threw into the river
one fish, took 1/3 of the remaining fish and walked away. Afterwards the third fisherman
did exactly the same. How many fish did they catch if each of them had taken the same
number of fish? (We assume that when each of them were taking 1/3 of the fish, the
number of fish was divisible by 3).
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3.2
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Invariant functions
Invariant. A function which is defined on the set of all possible states of a process and
which takes the same value on any state from the same equivalence class is called an
invariant.
This notion turns out to be very useful in studying properties equivalence classes.
Example 3.2 A disk is divided into n sectors. We put n draughts in an arbitrary manner
into the sectors. At each step we are allowed to take any two draughts (possibly from
different sectors) and move each to another sector: one of them into the next sector
clockwise and the other one into the next sector anti-clockwise. Is it possible to get from
the situation when every sector contains a single figure into the situation when all the
figures are gathered in one sector?
Solution. First we consider the case when n = 2m. Let us paint sectors of the disk
white and black so that no two neighbouring sectors are painted in the same colour.
After each move the number of draughts in white sectors either increases by 2, stays the
same or decreases by 2. We introduce a function (invariant) f on the set of all possible
configurations of draughts, which takes value 0 when there is an even number of draughts
in white sectors and value 1 otherwise.
Test Question 8 Check that f is invariant.
At the end we are required to get into the position when all n = 2m draughts are
gathered at the same sector. For this configuration f (α) = 0. If m = 2k + 1, then for
the original configuration we would have 2k + 1 draughts in white sectors and the value
of the invariant f is 1. This proves that for the case n = 4k + 2 there is no way to move
draughts by the given rules from the starting position to the required final position.
When m = 2k, the starting value of the invariant is 0. One might hope that there is a
way to get from the original position to the final one. We shall show that this is not the
case by choosing a more subtle invariant.
Let us assign numbers from 1 to n to each sector clockwise. We denote the number of
draughts in sector k in configuration α of draughts by ak (α). Consider the function
q(α) = 1 · a1 (α) + 2 · a2 (α) + · · · + (n − 1) · an−1 (α) + n · an (α).
(1)
Let us try to check if this function is an invariant for our process. If we move a draught
from sector i > 1 to sector i − 1 and a draught from sector j < n to sector j + 1, then for
the new configuration β
q(β) = · · · + (i − 1)ai−1 (β) + i · ai (β) + · · · + j · aj (β) + (j + 1) · aj+1 (β) + · · · .
(2)
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The expression (1) differs from (2) only in the (i − 1)st, ith, jth and (j + 1)st terms.
Therefore, it is easy to calculate the difference between q(α) and q(β).
q(β) − q(α) = (i − 1) · (ai−1 (β) − ai−1 (α)) + i · (ai (β) − ai (α)) +
+j · (aj (β) − aj (α)) + (j + 1) · (aj+1 (β) − aj+1 (α)) =
= (i − 1) · 1 + i · (−1) + j · (−1) + (j + 1) · 1 = 0.
If only one of the conditions i > 1 or j < n is not satisfied then q(α) − q(β) = ±n (check
this!). If i = 1 and j = n then, obviously, q(β) = q(α). Therefore, the function q(α) is
not an invariant for our process, but the function
r(α) ≡ q(α)(modn).
is an invariant, and this invariant does not even depend on the parity of n.
Let us calculate the value of this invariant for the original configuration α when in
each sector we have only one draught:
r(α) ≡ 1 + 2 + · · · + n ≡
n(n + 1)
(modn).
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If n = 2m, then r(α) ≡ m(2m + 1) ≡ m(modn). If n = 2m + 1, then r(α) ≡ (m +
1)(2m + 1) ≡ 0(modn).
For the final configuration β, when all draughts are gathered in sector k, we deduce
that
r(β) ≡ kn ≡ 0(modn).
Therefore, again for even n the values are different and we cannot achieve the required
configuration from the starting one. The case of odd values of n still remains open, as
the constructed invariant does not distinguish the original configuration α and the final
configuration β.
A universal invariant is an invariant which takes different values on non equivalent
states.
The following theorem helps to check whether a given invariant on the set of equivalence classes is a universal invariant.
Theorem 3.2 Assume that
a) there are l states δ1 , ... δl , such that any state α ∈ M is equivalent to one of them;
b) an invariant f takes l different values.
Then f is a universal invariant and states δj , j = 1, 2, ..., are pairwise not equivalent:
δi 6∼ δj for i 6= j.
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Proof. The condition a) implies that there are no more than l equivalence classes defined
by our process. The condition b) implies that there are at least l different equivalence
classes. Therefore, there are precisely l equivalence classes. Again from condition b) we
see that f is a universal invariant as it takes exactly l different values. Finally from a)
it follows that δ1 , ... δl belong to different equivalence classes and, therefore, they are
pairwise distinct.
Let us complete the solution of Example 3.2. We shall show now that the invariant r
constructed above is a universal invariant. It will imply that for any odd number n we
can transform the original configuration of draughts into the required one.
Test Question 9 Explain how it follows from universality.
Let δi be a configuration of draughts such that n − 1 draughts lie in sector n and the
remaining one lies in sector i. By δn we denote the configuration when all n draughts stay
in sector n. It is clear that any configuration is equivalent to one of δ1 , ..., δn . Indeed,
let us move all draughts into sector n. We take the first draught and move it into sector
n simultaneously moving the second draught in the opposite direction. After the first
draught has reached the nth sector we shall start moving the second draught into the
nth sector moving simultaneously the third draught in the opposite direction. Doing this
subsequently with the 4th, 5th,... (n − 1)st draughts we obtain a configuration when all
but one draught stay in sector n. The remaining nth draught stays in some sector i which
defines the corresponding δi . This means that the original configuration α is equivalent
to δi .
Let us calculate the value of r(δi ), i = 1, 2, ..., n. If i 6= n then
a1 (δi ) = · · · = ai−1 (δi ) = ai+1 (δi ) = · · · = an−1 (δi ) = 0,
ai (δi ) = 1,
an (δi ) = (n − 1)
Therefore,
q(δi ) = i · ai (δi ) + n · an (δi ) = i + n(n − 1),
r(δi ) ≡ i(modn).
In addition r(δn ) ≡ nan (δi ) ≡ 0(modn). We obtain that invariant r takes at least n
different values. By Theorem 3.2 r is a universal invariant and the configurations δ 1 ,...,δn
are pairwise non equivalent.
3.3
Semi-invariants
The condition of “symmetry” plays a very important role in the definition of equivalence
relation which in its turn leads to the notion of an invariant. There are many situations
when you do not have “symmetry” in your process: once you leave a configuration (say,
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eat a sweet) you cannot get back to this situation (it does not help to eat more sweets,
you just decrease the total number of them...) Even if the process does behave properly
and gives rise to an equivalence relation, the number of equivalence classes can be very
large so that the actual invariant is difficult to construct. In such situations it might be
helpful to use semi-invariants instead of invariants.
A semi-invariant of a process is a function defined on the set of states of the process
which strictly decreases (or strictly increases) after any action of the process.
Example 3.3 A rectangular table m × n is filled in with numbers. At any time we are
allowed to change the signs of all numbers that stay in the same row or column. Prove
that by repeating this operation we can get a configuration such that all the sums of
numbers in each row and in each column are non negative.
Solution. The process here does satisfy all the properties: reflexivity, symmetry and
transitivity. But the actual number of states is very large and, moreover, the equivalence
classes are hardly possible to understand.
Let us have a look at the sum of all numbers in the table. If we change signs in any
line: a row or a column, the sum changes too. So this is not an invariant. Assume that
we have a line (row or column) the sum of whose entries is negative. If we change the
signs of these numbers, we will definitely increase the total sum of the numbers in the
table. Therefore, the sum of all the numbers in the table is a semi-invariant with respect
to changes of signs of numbers in lines, the sum of whose entries is negative. Let us
follow the process changing each time the signs of numbers in a line with negative total
sum. The sum of all the numbers in the whole table will increase. It cannot increase
infinitely often, as there is only a finite number of possibilities for the state of the table if
the absolute values of the numbers in the table are fixed. Therefore, in a finite time we
will reach the situation when every line (a row or a column) has a non negative sum of
entries.
Semi-invariants can have a varied nature. Sometimes we may not even have a process
explicitly given in the statement of the problem. In this case we introduce the process by
ourselves and a semi-invariant is the tool which helps one to get the final solution. Let
us look at another example.
Example 3.4 Consider n points on a plane. On the same plane we are given n lines in a
general position, namely, no two of them are parallel to each other. Prove that it is always
possible to drop non intersecting perpendiculars from the given points on the given lines
so that there is a perpendicular on each of the lines and there is a perpendicular from
every point.
Solution. Let us drop one perpendicular on each line from our n points in an arbitrary
manner with exactly one perpendicular from each point. If they do not intersect, then
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the problem is solved. Otherwise there are two perpendiculars, say AA1 and BB1 from
points A and B on lines α and β respectively which intersect at a point P . We then
change perpendiculars AA1 and BB1 to perpendiculars AA2 from A on β and to BB2
from B on α. Let us prove that the total length of all the perpendiculars decreases with
this process. Indeed,
AA2 < AP + P B1 ;
BB2 < BP + P A1 .
Summing these inequalities we obtain
AA2 + BB2 < AA1 + BB1 .
The total sum of perpendiculars can take only a finite number of values for a given combination of points and lines. The process cannot continue infinitely often and, therefore,
at some point we achieve a configuration when all the perpendiculars do not intersect.
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Colourings
Sometimes in a text problem dealing with situations or processes a simple observation
gives a shorter way to solution than the methods we discussed earlier. The nature of
these observations is very often the same as in the case of invariants. We can divide
objects into groups analogously to the equivalence classes. Then by counting elements in
the groups we make a conclusion which solves the problem.
Let us look at a few examples. Probably, the simplest example of this kind is the
following.
Example 4.1 Two squares 1 × 1 are cut off a chessboard 8 × 8. One is the upper right
square and the other one is the lower left square. Prove that the remaining 62 squares
cannot be covered by 31 dominoes 1 × 2.
If we look at the classical colours of the chessboard we can easily come to the solution.
In this situation the colour of a square is an equivalence relation which divides the set of
all squares into two subsets (what are these two subsets?). The covering by the dominoes
is a map from one class of equivalence to another which has to be a one to one map.
Because there are 32 squares of one colour and 30 squares of the other colour we obtain
a contradiction.
The proper colouring of a chessboard can allow us to solve some other complicated
problems.
Example 4.2 Prove that it is impossible to cover a chessboard 8×8 by fifteen Γ-tetramino
(each one consists of four sectors 1 × 1 in the form of the letter Γ) and a single square
tetramino 2 × 2.
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Solution. The original colouring of the chessboard gives nothing in this case. Each
tetramino covers exactly two black and two white squares 1 × 1. Let us change the
colouring. We will colour white all the rows with odd numbers and black all the rows
with even numbers. The single square tetramino 2 × 2 in any position covers two white
and two black squares 1 × 1. On the other hand, any Γ-tetramino covers an odd number
of white and an odd number of black squares 1 × 1. 15 Γ-tetraminos will then cover an
odd number of white squares 1 × 1. Adding two white squares 1 × 1 covered by the square
tetramino 2 × 2 we again obtain an odd number. Therefore, the total number (even) of
32 white squares 1 × 1 cannot be covered.
Sometimes the problem cannot be reduced just to a clever colouring but requires a
more sophisticated use of different methods. In the example below we use parity as an
analog of the colouring.
Example 4.3 A net of regular hexagons of side 1 is drawn on a plane. A spider creeps
along the edges of the net from node A to node B using the shortest route of length 100.
Prove that one-half of the way the spider is creeping parallel to the same direction.
Solution. Let us number the edges which form the route of the spider. We chose one
of the three possible directions in the net and call it horizontal. Let us prove that the
numbers of all horizontal edges of the route have the same parity. Let a and b be two
consecutive horizontal edges on the spider’s route in the sense that there are only edges
of the other directions in between a and b.
Let us draw a vertical line, i.e. the line which is perpendicular to the horizontal
direction. The edges a and b cannot intersect the same vertical line, as in this situation
there is a shorter way from the “first” endpoint of a into the “second” endpoint of b (why?).
This implies that there is only an odd number of edges between a and b. Therefore, all
horizontal edges have numbers of the same parity. This also holds for the edges of the
other two directions.
Because there are only three directions then either all even edges or odd edges are
parallel to the same direction.
Test Question 10 Prove this.
Therefore, 50 edges on the spider’s route are parallel to the same direction.
You might also note that the way to prove that two horizontal directions cannot cross
the same vertical line is a by-product of the idea of a semi-invariant from the previous
section.
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The Maximum Principle
One of the most important mathematical principles is the Maximum Principle which says
that a function with some nice properties can attain its maximum (minimum) only on the
boundary of the region of definition. Here we will not touch on this principle seriously.
We will only take the idea that for solving problems it is often useful to have a look at
what is happening in the most extreme situation.
Example 5.1 777 gangsters are standing in a field so that the distances between any two
of them are pair-wise distinct. At the same moment each of them fires at the gangster
nearest to him. Prove that there is a gangster who was not shot.
Solution. Let us choose two gangsters in such a way that the distance between them is the
shortest among all the distances between the gangsters. Obviously, these two gangsters
fire at each other. If any of the remaining 775 gangsters fires at one of the chosen two,
then the remaining 774 cannot shoot all these 775 gangsters. In this situation the problem
is solved. The difficult situation is when the remaining 775 gangsters do not shoot at any
of the two gangsters chosen at the first step. So they fire at some gangsters from these
remaining 775 gangsters. We can now repeat same argument with 775 gangsters instead
of 777. Continuing in this way we will come to the situation when we have only three
gangsters. Obviously, two of them fire at each other and the remaining one will not be
shot by any other gangster.
In this example we used parts of the Maximum Principle when we considered the
minimum of the distances between two gangsters.
To finish let us “play” football.
Example 5.2 Twenty teams take part in the football Premiership. What is the minimal
number of games which has to be played so that it would be possible to have among any
three teams at least two teams which have already played each other.
Solution. Assume that we have reached the moment when the condition above is satisfied. Choose team A that has played the minimal number of games (this is the Maximum
Principle hidden here). Assume that team A has played k games. Each of the k teams
which played A, has also played at least k games. The remaining (19 − k) teams have
to have already played each of these (19 − k) teams in their joint games as otherwise the
condition of the problem is not fulfilled (two of these (19 − k) teams would not play each
other and they also did not play with team A).
Therefore, twice the number of all games (it can be obtained by summing up the
games played by each team) is greater than or equal to
k + k 2 + (19 − k)(18 − k) = 2k 2 − 36k + 18 · 19 = 2(k − 9)2 + 180 ≥ 180.
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Therefore the number of games is at least 90.
Test Question 11 Give an example of a tournament timetable when the first 90 games
deliver a situation which satisfies the condition of the problem.
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Problem Sheet.
Problem 1 Answer the test questions in the text.
Problem 2 80 million people live in Britain. Every one has at most a million hairs.
Prove that there are at least eighty people with the same number of hairs.
Problem 3 Twenty teams play in the football Premiership. Every team has to play every
other once. Prove that at every moment there are at least two teams which have played
the same number of matches.
Problem 4 Prove that among any 52 integers we can always find two whose sum or
difference is divisible by 100. Is this statement true for 51 integers?
Problem 5 101 arbitrary points are marked inside the square of side 1, so that no three
of them belong to the same line. Prove that there are always three of them which form a
triangle of area no greater than 1/100.
Problem 6 2001 flies are taken inside a cube of side 1. Prove that we can catch at least
three of them with a sphere of radius 1/11.
Problem 7 Prove that there is a power of 29 whose decimal expression ends with 00001.
Problem 8 Prove that there is a natural number n which is a multiple of 1441 and whose
decimal expression consists of only ones and zeros.
Problem 9 There are 123 people in a village. The sum of their ages is 3813. Must it be
possible to find 100 among them whose total age is not less than 3100?
Problem 10 Seven lines are chosen on a plane so that no two of them are parallel. Prove
that there are two of them with an angle between them less than 26◦ . Can we change 26◦
to 25◦ so that the statement would still be true?
Problem 11 Prove that from every tree we can cut off 8/15 of its leaves while keeping
at least 7/15 of the shade that the tree was giving originally (we assume that the number
of leaves is divisible by 15).
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Problem 12 21 children are given 200 nuts. Prove that however they choose to share
them, there are always two of them who get the same number of nuts.
Problem 13 Each of 7 intervals has length more than 10cm and less than 1m. Prove
that there are three of them which may be used to form a triangle.
Problem 14 Prove that among any 101 distinct positive integers which are less than or
equal to 200, there are two such that one divides the other.
Problem 15 Numbers from 0 to 9 are written on 20 cards so that each number appears
on two cards. Is it possible to place the cards in a row so that cards with zeroes are next to
each other, between cards with ones there is exactly one card, ... and between cards with
9 there are exactly nine cards?
Problem 16 A 2 × 2 table is filled in with four integer numbers. It is allowed to make
the following operations with the numbers in the table.
a) in any column we can add 2 to any number and subtract 2 from the other.
b) in any row we can add 3 to any number and subtract 3 from the other.
Describe which tables are equivalent and which are not.
Problem 17 A Spanish king decides to change the portrait room in the round tower of
his castle. At every moment he wants his servants to change only two portraits of previous
kings which hang next to each other. Also he does not want to change portraits of two
kings, one of whom reined immediately after the other, at the same time. Because the
room is round the king does not consider those configurations of portraits, which can be
obtained by rotations from each other, to be different.
Prove that the king can get any configuration of portraits by following only these two
rules.
Problem 18 All natural numbers from 1 to 2n are written in any order in a row. To
every number we add the number of the place on which it stands. Prove that among the
2n sums obtained in this way there are at least two with the same remainder mod 2n.
Problem 19 Prove that any 2n points on a plane are the ends of n non intersecting
intervals.
Problem 20 In the Parliament any member has no more than three enemies. Prove that
we can divide the Parliament into two houses so that every member will have no more
than one enemy in the same house with him.
Problem 21 Let a brick consists of two cubes of size 1 × 1. We are given a cube 3 × 3
in which it is allowed to remove one smaller cube 1 × 1. Which smaller cube should we
remove so that the remaining figure could be filled in completely with 13 bricks?
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Problem 22 On one side of an infinite corridor there is an infinite number of rooms
and in each of them there is a piano. A finite number of pianists live in some of these
rooms. Possibly, in some rooms there is more than one pianist. Every day two of the
pianists, who live in the rooms next to each other, say in rooms k and (k+1), decide that
they disturb each other and they move out to rooms (k-1) and (k+2) respectively. Prove
that in a finite time they will end their moves.
Problem 23 In a library N volumes of Encyclopedia Britannica were placed out of order
in a line. Step-by-step, a robot-librarian takes a single volume with number k, which
occupies a wrong place, and places it on the k-th place in the line. Prove that in a finite
time the robot will put all volumes in order.
Problem 24 13 gnomes live in a village. Each of them has three balloons of three different
colours: one yellow, one red and one blue. They start exchanging balloons and every
gnome wants to have balloons of the same colour. Is it possible that at some point each
of them will have exactly three balloons of the same colour?
Problem 25 A castle has a shape of a regular triangle of side 100m. It is divided into
100 halls of the form of a regular triangle of side 10m. There are doors in the middle of
the internal walls of each hall so that one can get from each hall to the next one. Prove
that if someone wants to pass through the castle, and to visit each hall no more than once,
then it would be impossible to get into more than 91 halls.
Problem 26 Two chessboards of the same size 8 × 8 have the same centre and one of
them is rotated with respect to the other one by an angle of 45◦ . What is the total area
of the intersection of black squares of these chessboards if we assume that the area of a
single square is equal to 1?
Problem 27 What is the minimal number of x-shaped figures, consisting of three squares
1 × 1, may be put inside a square 8 × 8, such that we cannot add another x-shaped figure
in this square without overlapping any of the figures?
Problem 28 Twelve numbers are written along the circle in such a way that each of them
is equal to the absolute value of the difference between the two consecutive numbers which
are clock-wise next to it. The sum of the twelve numbers is one. Find these numbers and
prove that there are no other possibilities.
Problem 29 A tourist goes from town A to the town B most distant from it, then from
B to the town C most distant from B etc. Prove that if A and C are different towns, then
the tourist will never come back to A.
Problem 30 Consider 50 segments on a line. Prove that one of the two following statements is true:
a) there are eight segments which have a common point;
b) there are eight segments which are pairwise disjoint (have no common points).
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References
[1] “Kvant”, http://kvant.mccme.ru/.
[2] N. B. Vasil’ev, A. A. Egorov, Problems of All-Union Mathematical Olympiads, Nauka,
1988.
The materials are prepared and can be used for non commercial purposes only.