4_Combinatorica

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Chapter IVChapter IVChapter IVChapter IV

COMBINATORICS COMBINATORICS COMBINATORICS COMBINATORICS

Problem nr. 135Problem nr. 135Problem nr. 135Problem nr. 135Consider a 1 × n rectangle and some tiles of size 1 × 1 of four different colours. The rectangle is

tiled in such a way that no two neighboring square tiles have the same colour.

a) Find the number of distinct symmetrical tilings. b) Find the number of tilings such that any consecutive square tiles have distinct colours.

Dan Brânzei, 2002

Problem nr. 136Problem nr. 136Problem nr. 136Problem nr. 136

A square of side 1 is decomposed into 9 equal squares of sides3

1and the one in the center is

painted black. The remaining eight squares are analogously divided into nine squares each, andsquares in the centres are painted in black.

Prove that after 1000 steps the total area of the black region exceeds 0.999.

Cristinel Mortici, Costel Chiteş, 2002

Problem nr. 137Problem nr. 137Problem nr. 137Problem nr. 137A given equilateral triangle of side 10 is divided into 100 equilateral triangles of side 1 by drawing

parallel lines to the sides of the original triangle.Find the number of equilateral triangles, having vertices in the intersection points of parallel lines

whose sides he on the parallel lines.

Dinu Şerbănescu, 2002

Problem nr. 138Problem nr. 138Problem nr. 138Problem nr. 138Show that one can color all the points of a plane using only two colors such that no line segment

has all points of the same color.

Valentin Vorniciu, 2003

Problem nr. 139Problem nr. 139Problem nr. 139Problem nr. 139Consider a cube and let M , N be two of its vertices. Assign the number 1 to these vertices and 0 to

the other six vertices. We are allowed to select a vertex and to increase with a unit the numbersassigned to the 3 adjacent vertices – call this a movement .

Prove that there is a sequence of movements after which all the numbers assigned to the vertices of the cube became equal if and only if MN is not a diagonal of a face of the cube.

Marius Ghergu, Dinu Şerbănescu, 2004



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An array 8 × 8 consists of 64 unit squares. Inside each square are written the numbers 1 or –1 sothat in any 2 × 2 subarray the sum of the four numbers equals 2 or –2. Prove that there exist two rowsof the array which are equal.

Marius Ghergu, 2004

Problem nr. 141Problem nr. 141Problem nr. 141Problem nr. 141In a chess tournament each of the players have played with all the others two games, one time with

the white pieces and then with the black pieces. In each game the winners sets one point and both players receive 0.5 points if the game ends with draw. At the end of the tournament, all the players endwith the same number of points.

a) Prove that there are two players with the same number of draws. b) Prove that there are two players with the same number of losses when playing the white.

Marius Ghergu, 2004

Problem nr. 142Problem nr. 142Problem nr. 142Problem nr. 142A regular polygon with 1000 sides has the vertices colored in red, yellow or blue. A move consists

in choosing to adjacent vertices colored differently and coloring them in the third color. Prove thatthere is a sequence of moves after which all the vertices of the polygon will have the same color.

Marius Ghergu, 2004

Problem nr. 143Problem nr. 143Problem nr. 143Problem nr. 143A country has six cities with airports and two rival flight companies. Any two cities are connected by

flights so that on each route between two cities one may travel with exactly one of the two flight com-

panies. Prove that you can visit 4 cities in a cycle flying with the same air company (that is, there exist

four cities A, B, C , D and a company which operates on the routes A ↔ B, B ↔ C , C ↔ D and D ↔ A).Dan Schwarz, 2005

Problem nr. 144Problem nr. 144Problem nr. 144Problem nr. 144A phone company starts a new type of customer service. A new client can choose k phone numbers

in this network which are call-free – regardless if is called or if calling. A group of n students decide totake advantage of this promotion.

• Show that if n ≥ 2k + 2 then there will exist 2 students which will be charged when speaking.

• Show that if n = 2k + 1 then there exists a way of arranging the free calls so that in this groupeverybody speaks free to anyone else.

Valentin Vornicu, 2005

Problem nr. 145Problem nr. 145Problem nr. 145Problem nr. 145The positive integers from 1 to n

2are placed arbitrarily on squares of an n × n chessboard. Two

squares are called adjacent if they have a common side. Show that two opposite corner squares can be

joined by a path of 2n – 1 adjacent squares so that the sum of the numbers places on them is at least

12

23

+−+

nn

n.

Radu Gologan, 2005



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An 7 × 7 array is divided in 49 unit squares. Find all integers n ∈ N* for which n checkers can be placed on the unit squares so that each row and each line have an even number of checkers.

(0 is an even number, so there may exist empty rows or columns. A square may be occupied by atmost 1 checker.)

Dinu Şerbănescu, 2006

Problem nr. 147Problem nr. 147Problem nr. 147Problem nr. 147A rectangular cardboard is divided successively into smaller pieces by a straight cut; at each step,

only one single piece is divided in two. Find the smallest number of cuts required in order to obtain – among others – 251 polygons with 11 sides.

Marian Andronache, 2007

Problem nr. 148Problem nr. 148Problem nr. 148Problem nr. 148Consider a n × n array divided into unit squares which are randomly colored in black or white.

Three of the four comer squares are colored in white and the fourth is colored in black. Prove that

there exists a 2 × 2 square which contains an odd number of white squares.

Lidia Ilie, 2007

Problem nr. 149Problem nr. 149Problem nr. 149Problem nr. 149Consider the numbers from 1 to 16. A solitaire game is played in the following manner: the

numbers are paired and each pair is replaced by the greatest prime divisor of the sum of the numbersin that pair – for example, (1, 2); (3, 4); (5, 6); …; (15, 16) produces the sequence 3, 7, 11, 5, 19, 23, 3,31. The game continues similarly until one single number is left. Find the greatest possible value of

the number which ends the game.

Adrian Stoica, 2007

Problem nr. 150Problem nr. 150Problem nr. 150Problem nr. 150Eight persons attend a party, and each participant has at most three others to whom he/she cannot

speak. Show that the persons can be grouped in 4 pairs so that each pair can converse.

Mihai Bălună, 2007

Problem nr. 151Problem nr. 151Problem nr. 151Problem nr. 151A set of points is called free if there is no equilateral triangle whose vertices are among the points

in the set. Show that any set of n points in the plane contains a free subset of at least n points.

Călin Popescu, 2007

Problem nr. 152Problem nr. 152Problem nr. 152Problem nr. 152A 8 × 8 square board is divided into 64 unit squares. A “skew-diagonal” of the board is a set of 8

unit squares with the property that each row or column of the board contains only one unit square of the set. Checkers are placed in some of the unit squares so that each “skew-diagonal” has exactly 2

squares occupied by checkers. Prove that there exist two rows or two columns which contain all thecheckers.

Dinu Serbănescu, 2007



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To obtain a square P of side length 2 cm divided into 4 unit squares it is sufficient to draw 3squares: P and another 2 unit squares with a common vertex, as shown below:

Find the minimum number of squares sufficient to obtain a square of side length n cm divided into

n2unit squares (n ≥ 3 is an integer).

* * *, 2009

Problem nr. 154Problem nr. 154Problem nr. 154Problem nr. 154The plane is divided into a net of equilateral triangles of side length 1, with disjoint interiors. A

checker is placed initially inside a triangle. The checker can be moved into another triangle sharing acommon vertex (with the triangle hosting the checker) and having the opposite sides (with respect tothis vertex) parallel. A path consists in a finite sequence of moves. Prove that there is no path betweentwo triangles sharing a common side.

Vasile Pop, 2009

Problem nr. 155Problem nr. 155Problem nr. 155Problem nr. 155Show that there exist (at least) a rearrangement a0, a1, a2, …, a63 of the numbers 0, 1, 2, …, 63,

such that ai – a j ≠ a j – ak , for any i < j < k ∈ {0, 1, 2, …, 63}.

* * *, 2009



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Chapter IV

COMBINATORICS

Solution to problem nr. 135Solution to problem nr. 135Solution to problem nr. 135Solution to problem nr. 135a) If n = 2k , there are no such symmetrical tilings (otherwise the k and k + 1 tiles must have the

same colour).

If n = 2k + 1, the problem is to count the possible tilings for k + 1 squares. There are 4 ⋅

3...33 ⋅⋅⋅ =

= 4 ⋅ 3k

such tilings. b) There are 4 ⋅ 3 ⋅

2...22 ⋅⋅⋅ = 4 ⋅ 3 ⋅ 2

n –2tilings.

Solution to problem nr. 136Solution to problem nr. 136Solution to problem nr. 136Solution to problem nr. 136

The first step give rise to one black square of area9

1

3

12

=

. After the second step we obtain eight

more squares of side9

1, the black region increasing thus by

29

8. In the same manner, the third step

increases the black area by 82

= 64 black squares, each of area27

1, that is at this stage the black area

becomes

3

2

2 9

8

9

8

9

1++ .

We conclude that after 1000 steps, the area of the black region is

=

++

++=++++9992

1000

999

3

2

2 9

8...

9

8

9

81

9

1

9

8...

9

8

9

8

9

1

1000

1000

9

81

9

81

9

81

9

1

−=

−

−⋅= .

It is left to prove that the last number is greater than 0.001. This easy follows by using a binomialexpansion evaluation, that is

1000642

9991000

8

1

2

1000

8

11

8

92

10001000

>⋅⋅

=⋅

>

+=

,

the proof being complete.



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Let us consider the general case, that is to consider the number an of equilateral triangles formed bydivision in n segments. We shall find a recurrence relation.

Consider an equilateral triangle with sides partitioned into n + 1 equal segments and draw the n parallels to each side of the given triangle. We will count all triangles with at least one vertex on BC ;

the remaining ones are triangles counted in an. Consider first the triangles that have two vertices on BC . When choosing two division points on

BC , one counts exactly one triangle, namely that one obtained by drawing parallels from M i, N i to AB,

AC respectively. Hence we add2

)1)(2( ++ nnnew triangles.

Considering triangles with only one vertex on BC , observe that for any point of division, inside theside BC , we count one triangle of side 1, one triangle of side 2, and so on. Hence we add n + (n – 2) +

+ (n – 4) + … triangles with one side on BC . It follows that we have

+−+−++++

+=+ )4()2(2

)1)(2(1 nnn

nnaa nn …

Changing n with n + 1, we get

=−+−+++++

+= ++ )3()1()1(2

)2)(3(12 nnn

nnaa nn …

Adding up, we obtain

5

)53)(2(

2

)2)(1(

2

)2)(3(

2

)1)(2(2

+++=

+++

+++

+++=+

nna

nnnnnnaa nnn .

It follows that

315315...237145

2

)583(10068810 =+==+=+=

+⋅+= aaaaa .

Therefore, the number of triangles is 315.

Solution to problem nr. 138Solution to problem nr. 138Solution to problem nr. 138Solution to problem nr. 138Choose an arbitrary point A in the plane. Points located in the plane at a rational distance from A

are colored in red, while the others are colored in blue. Consider an arbitrary segment PQ. We mayassume that AP < AQ; if not, take instead of P another point of the line segment ( PQ).

Recall that between two real numbers one can find a rational number q and an irrational number r .The circles centered at A and having the radii q and r intersect the segment PQ at the points M and N

respectively. It is obvious that M is colored in red and N in blue, so the claim is proved.

Solution to problem nr. 139Solution to problem nr. 139Solution to problem nr. 139Solution to problem nr. 139Color the 8 vertices of the cube in black or white so that the 4 vertices of the 2 regular tetrahedrons

have the same color; notice that the 3 vertices adjacent to a vertex have its opposite color. Therefore,each movement increases the sum of the numbers assigned to the vertices sharing the same color by 3.Consider the cases:

1) MN is a diagonal of a face of the cube. Then M and N have the same color, say black. Assume by contradiction that there is a sequence of movements after which the same number n is assigned toall vertices. Let k 1 and k 2 be the number of white, respectively black vertices that were selected to perform the movements. Then



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4n = 3k 1 + 2 = 3k 2,

a contradiction.2) MN is a diagonal of the cube. Selecting the vertices M , then N , and performing these 2

movements, to all the vertices the number 1 will be assigned, as needed.3) MN is a side of the cube. The same outcome as in the previous case will occur after 2 movements

when selecting the diagonally opposite vertices of M and N.

This provides us with a full solution.

Solution to problem nr. 140Solution to problem nr. 140Solution to problem nr. 140Solution to problem nr. 140The main idea is to observe that two consecutive rows have exactly 4 equal elements, namely those

lying on the columns 1, 3, 5, 7 or 2, 4, 6, 8. Moreover, on the other 4 columns the elements aredifferent. Wlog, assume that rows 1 and 2 are equal with respect to the columns 1, 3, 5, 7 and different

on the column 2, 4, 6, 8; we call these rows odd equal . If rows 2 and 3 are also odd equal , then rows 1

and 3 are equal, as needed. If not, then rows 2 and 3 are even equal . Now consider the rows 3 and 4;we are done if the rows are even equal , so assume that they are odd equal . Finally, if rows 4 and 5 areodd equal , then rows 3 and 5 are equal, and if rows 4 and 5 are even equal , then rows 1 and 5 areequal. This concludes the proof.

Solution to problem nr. 141Solution to problem nr. 141Solution to problem nr. 141Solution to problem nr. 141Let n be the number of players in the tournament. The total numbers of matches is n(n – 1), hence

each player end up with n – 1 points.a) Assume by contradiction that each player has a different number of draws. As a draw gives 0.5

points, it follows that each player has an odd number of draws. Since the possible cases are: 0, 2, 4, …,2(n – 1), we infer that each of these numbers is assigned to each of the players. Consider A the player

with 0 draws and B the player with 2(n – 1) draws. Each player has played 2(n – 1) matches, hence B obtained a draw in each match played. The match A – B thus ended with a draw, a contradiction, since

A has no draws. b) Suppose the contrary. Then each of the n players has 0, 1, …, n – 1 losses when playing the

white. Let X and Y be the players with 0 and n – 1 losses, respectively. The player Y has no pointswhen playing the white and n – 1 points, so he won all the matches with the black pieces. This impliesthat the match X – Y is won by Y , so is lost by X , a contradiction, since X has 0 losses with the white

pieces.

Solution to probleSolution to probleSolution to probleSolution to problem nr. 142m nr. 142m nr. 142m nr. 142Let A1, A2, …, A1000 be the vertices of the polygon. We start with two lemmas.

Lemma 1. Three of four consecutive vertices have the same color. Then after a sequence of moves

all vertices will have the color of the fourth vertex. Proof . Let the colors be 0, 1 and 2. We have two cases:

a) 1110 → 1122 → 1002 → 2202 → 2112 → 0000.

b) 1011 → 1221 → 0000. Lemma 2. Any 4 consecutive vertices will turn after several moves in the same color. Proof . Form two pairs of consecutive vertices and change them in the same color – if they do not

already have it. Then follow the sequence 1122 → 1002 → 2202 → 2112 → 0000.By the second lemma, after several moves the vertices A1, A2, A3, A4 will have the same color, say



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red. Likewise, A5, A6, A7, A8 will have the same color. Consider now the vertices A4, A5, A6, A7; the first

is red and the other three have the same color. By the first lemma they all will turn red – of course, wedo nothing if they were already red. We move on with this procedure until A1, A2, …, A997 turn red

(note that 997 = 4 + 3 ⋅ 332, so this requires 332 steps). Now consider the vertices A998, A999, A1000, A1; by the second lemma they all will share the same color. If this is red, we are done. If not, say that theyare blue, and taking the vertices A997, A998, A999, A1000 we obtain – using the first lemma – all vertices to

be red, except for A1, which is blue. Now A1, A2, A3, A4 turn blue, then A5, A6, A7, A8 and so on. This

time, after 333 steps, all the 1000 vertices (1000 = 1 + 3 ⋅ 333) will be colored in blue.

Comment . Substituting colors with digits, notice that all moves: 01 → 22, 02 → 11 and 12 → 00

preserve the sum (mod 3). This means that the final color is unique and, of course, is given by the sumof the digits assigned to the vertices of the initial configuration.


The number of routes between 2 cities is

2

6= 15. By the pigeonhole principle one can find that a

company – say M – operates at least 8 routes.

There are

4

6 = 15 subsets of 4 cities and each pair of cities occurs in

2

4= 6 such subsets. Using

again the pigeonhole principle follows that there exists a subset of 4 cities among which at least 4

routes are operated by the company M (3 ⋅ 15 < 6 ⋅ 8).If those routes form a cycle, we are done. If else, then one can easy observe that among these 4

cities X , Y , Z , T , the routes X ↔ Y , Y ↔ Z , Z ↔ X and X ↔ T are operated by the company M (set thenotation accordingly).

The other 2 cities, say P and Q, are connected by at least 4 routes operated by M . Even if P ↔ Q isone of them, from P or Q – say P – at least 2 routes to X , Y , Z , T are operated by M . If both

destinations are from X , Y , Z , a cycle is obtained, so assume that one of the routes is P ↔ T . In this

case we are done if the second route is P ↔ Y or P ↔ Z , hence we are left with the case P ↔ X . From those above, the existence of a third route from P to X , Y , Z or T will provide a cycle. If not,

from Q exist 2 routes to X , Y , Z, T . The same line of reasoning shows that we have to consider that on

the route Q ↔ T is operated by M .

Any of the routes Q ↔ X , Q ↔ Y , Q ↔ Z will close a cycle. To conclude, if the route P ↔ Q isoperated by M , then any route from Q to X , Y , Z or T will provide us with the desired cycle.


1) A person can speak free of charge with at most 2k persons – k that he chooses and other k – atmost – that choose his numbers among their free calls. Since n ≥ 2k + 2, each person will be chargedwhen speaking to (at least) another one.

2) Assume that all 2k + 1 persons are arranged in a circle. Each person will choose to speak free of

charge when calling any of the k persons located – consecutively – at his right. Then any person willspeak free of charge with the k persons located at his left, as all of them will choose him among their “favorite numbers”.



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The sum of all integers is 1 + 2 + … + n2=

2

)1( 22 +nn, so there is a row r having the sum of the

numbers assigned to its squares at least equal to2

)1( 22 +nn. Consider the 2n – 2 numbers written on

the first and last column – except for the two numbers which belong to the row r – and observe thattheir sum is at least 1 + 2 + … + (2n – 2) = (n – 1)(2n – 1).

Now we can select two “complementary parts” of these columns – in order to complete the row r to

a path – so that the sum of the numbers placed on these n – 1 squares is at least2

)12)(1( −− nn. Since

2

)1( 2 +nn+

2

1

22

)12)(1( 23

+−+=−−

nnnnn

, to conclude we only have to notice that

2

3n+ 1 =

=2

1

2

3

+n

– if n odd – and that

2

3n+ n2 – n +1 is the smallest integer greater than

2

1

2

23

+−+ nnn

–

when n is even.

Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 146666One can place 4, 6, …, 40, 42 checkers in the given conditions.

We start by noticing that n is the sum of 7 even numbers, hence n is also even. On a row one can

place at most 6 checkers, hence n ≤ 6 ⋅ 7 = 42.

The key step is to use 2k × 2k squares filled completely with checkers and 2k + 1 × 2k + 1 squares

having checkers on each unit square except for one diagonal. Notice that these types of squaressatisfies the conditions, and moreover, we can glue together several such squares within the problemconditions.

Below we describe the disposal of n checkers for any even n between 4 and 42.

For 4, 8, 12, 16, 20, 24, 28, 32 or 36 checkers 1, 2, 3, 4, 5, 6, 7, 8 or 9 2 × 2 squares; notice that all

fit inside the 7 × 7 array!

For 6 checkers consider a 3 × 3 square – except for one diagonal; then adding 2 × 2 squares we getthe disposal of 10, 14, 18, 22, …, 38 checkers.

For 40 checkers we use a 5 × 5 and five 2 × 2 squares.

Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 147777Let n be the required number. We claim that n = 2007.

With 7 cuts, from the given rectangular piece one can obtain an 11-sided polygon and sometriangles. From a triangle, with 8 cuts one can get an 11-sided polygon and some extra pieces,

sufficiently enough to continue the same procedure. Hence, using 7 + 8 ⋅ 250 = 2007 cuts one canobtain the 251 requested 11-sided polygons.

Denote by k the number of pieces left at the end which are not 11-sided polygons and notice that

each has at least 3 sides. Now, observe that with each cut the number of pieces increases by 1 and totalthe number of vertices increases with at most 4 – actually, with 2, 3 or 4, according to the number of existent vertices through which the cutting line passes.



84848484

Then n = k + 250 and 4n + 4 ≥ v ≥ 11 ⋅ 251 + 3k , where v is the total number of vertices of all

polygons at the end. Hence 4n + 4 ≥ 11 ⋅ 251 + 3(n – 250) = 2011 + 3n, so n ≥ 2007, as claimed.

Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 148888Assign the number 0 to each white square and the number 1 to each black square. The claim is

achieved if we prove the existence of a 2 × 2 square with an odd sum of the 4 numbers inside.

Assume the contrary, so each sum of the 4 numbers inside a 2 × 2 square is even. Summing over allsquares we get an even number S . Notice that each square not sharing a common side with the givenarray occurs 4 times in 5, the squares with only a common side occurs twice, while the 4 squares in thecorners only once. But in the four corners there are three 0’s and one 1, so the sum S is even, acontradiction.

Remark . The given array may have a rectangular form, and the above solution requires noalteration. However, this remark can easily lead to alternative solutions using induction. Here is a

sketch: choose a row with 0 and 1 at endpoints and call it the first row. Suppose that the number below0 is also 0; arguing by contradiction, we notice that all “doubletons” formed vertically from the firsttwo rows have equal numbers inside, so the second row – which starts with 0 – ends with 1. Deletingthe first row of the given rectangular array, the claim is reached by induction. The same line of reasoning is applied to the case when below 0 the number is 1.

Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 14Solution to problem nr. 149999Let a ♥ b be the greatest prime divisor of a + b. At first, notice that from the initial 16 numbers we obtain 8 primes. The largest prime that can be

obtained is 31 = 15 ♥ 16; if this number occurs, the second largest can be 23 = 11 ♥ 12. Otherwise, 29may occurs twice, from 16 ♥ 13 and 15 ♥ 14, followed by 19 – or lower.

From the stage when we are left with 8 primes, and after pairing them we get 4 primes. If a prime is

obtained from two odd primes a and b, then a ♥ b ≤ 2

ba +.

If else, at least one is 2 and let p be the other. The number p + 2 is prime only if p ∈ {3, 11, 17,

29}. Therefore, if p and q are prime with p ≤ q, then p ♥ g ≤ q + 2.We will prove that the largest number which can end the game is 19. One example to obtain it is

exhibit below:

(1, 8); (2, 7); (9, 16); (10, 15); (3, 14); (4, 13); (5, 12); (6, 11) → 3, 3, 5, 5, 17, 17, 17, 17

(3, 3); (5, 5); (17, 17); (17, 17) → 3, 5, 17, 17

(3, 5); (17, 17) → 2, 17

(2, 17) → 19.

Now, we have to show that the game cannot end with a number strictly greater than 19. Since fromthe second stage the number cannot increase with more than 2, and since 31 ♥ 2 = 11, we derive that

the game will end with a prime p ≤ 31. Suppose by contradiction that p ∈ {23, 29, 31}.If p = 29, as 29 is not sum of two primes, then p is obtained from two of 29. In the previous stage

four 29’s are needed, then in the second stage eight 29’s are required, in contradiction with an initialobservation. Moreover, we have obtained a stronger result: 29 cannot end the game and cannot occur even in the last pair, since after 2 steps at most one 29 may occur.

Suppose that p = 31. Two cases are possible: 31 = 2 ♥ 29 or 31 = 31 ♥ 31. The latter result forbids



85858585

the first case, while the second case requires that the last four numbers are 31, 31, 31, 31 or 31, 31, 29,

2. But among the 8 primes obtained after the first step we have at most two 29’s or one 31, not enoughto produce three 31’s or two 31’s and one 29.

Assume that p = 23. Again two cases are possible: 23 = 29 ♥ 17 or 23 = 23 ♥ 23. The first case isimpossible as shown above, while the second case is allowed if the last four primes are 23, 23, 23, 23or 29, 17, 23, 23. If all primes are 23, the previous step has eight numbers with the average of 23,which is a contradiction with

8 ⋅ 23 < 1 + 2 + 3 + … + 16 = 8 ⋅ 17.The second case lead similarly to contradiction, since 29 requires two 29’s and the pair of 23’s are

given by four numbers with the sum 4 ⋅ 23:

2 ⋅ 29 + 4 ⋅ 23 = 150 < 1 + 2 + … + 16 = 136.The solution is now complete.

Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 155550000Consider an arbitrary grouping in pairs. A pair in which the persons cannot speak will be called

“bad”. If there are bad pairs, we prove that some changes can be made to decrease to number of bad pairs. Applying this at most 4 times exhibit a grouping with no bad pairs, and we are done.

Label the pairs A, B, C , D and the persons in pair X by X 1 and X 2. Two persons that cannot converseare called “enemies”, otherwise “friends”. Assume that A is a bad pair. Beside A1, the person A2 has atmost two other enemies. Two cases arise:

a) If the other enemies of A2 belong to the same pair – call it B, then A1 has at least a friend amongC 1, C 2, D1, D2.

Choose C 1 as a friend of A1 and swap A1 with C 2. The new pairs A and C are good, and the claim issatisfied.

b) If not, in at least one of the pairs B, C , D there are only friends of A1. Wlog, say that this pair is B. One of the persons in this pair must be a friend of A2; call this person B1. Now swap A1 with B1 andthe new pairs A and B are good, as desired.

Remark . Consider the graph with vertices in the eight persons and edges corresponding to each pair

of friends. The degree of each vertex is at least 4, so, according to Dirac’s theorem there exists ahamiltonian cycle. Taking 4 edges with ho common endpoint from this cycle, we get 4 good pairs, asneeded.

Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 155551111Given a set X of n points in the plane, consider a maximal free subset Y made of m elements, hence

such that any point in X \ Y completes an equilateral triangle with (at least) a pair of points from Y . (Any X contains free subsets, since any subset with 1 or 2 elements is obviously free.) But for any pair

of points from Y there exist only two points in the plane which complete an equilateral triangle, so

n – m ≤

22

m, that is n ≤ m

2, or m ≥ n .

One checks the validity of this result for small values (1, 2, 3) of n, too (while the coplanarityrestriction is obvious).



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Label the rows from 1 to 8 and the columns from 1 to 8. The unit square which lies on the row iand the column j will be referred as (i, j).

On the skew-diagonal {(i, i) | i = 1, 2, …, 8} there are exactly 2 squares in which checkers were

placed; wlog, assume that the squares are (1, 1), (2, 2). Looking at the 6 × 6 sub-array Q determined bythe rows 3-8 and the columns 3-8, we see that any “skew-diagonal” of Q, togheter with (1, 1), (2, 2), isa skew-diagonal of the initial array. In view of the given conditions, no checkers are placed in thesquares of Q. Now take any skew-diagonal of Q with the squares (2, 1), (1, 2); this is a skew-diagonal

of the initial array, and the two checkers are placed inside (2, 1), (1, 2).Up to the point, we know that checkers are placed in the squares on the rows 1-2 or on the columns

1-2. Suppose by way of contradiction that there exist a square located on the first two rows – say

(i, m), i = 1, 2, m ≥ 3 – and a square on the first two columns – say ( s, j), j = 1, 2, s ≥ 3 – that holdcheckers. Then squares (i, m), ( s, j), (3 – i, 3 – j) belongs to a skew-diagonal, contradiction.


For n ≥ 3, it suffices to draw 2n – 2 squares, as below:

• if n is odd, from each vertex V of P draw squares of sides2

1−nsquares n – 1, n – 2, …,

2

1−+n,

with V as vertex in each square.

• if n is even, from each vertex V of P draw2

n– 1 squares of sides n – 1, n – 2,

2

n+ 1, with V as

vertex in each square and add 2 more squares of side2

nwith vertices in 2 opposite vertices of P .

To show that 2n – 2 square are needed, divide all four sides in unit segments with 4 n – 4 points – other than the vertices of the square – and consider for each point the unit segments perpendicular tothe border of the square. This 4n – 4 segments can be covered by no less than 2n – 2 squares, since asquare cannot cover 3 such segments, so we are done.

Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 1Solution to problem nr. 155554444Two types of triangles arise after such a partition: “<” and “=”. Suppose that the initial triangle

ABC is of type < and notice that all three adjacent triangles – the ones sharing a common side – are of type =. Call A-stripe the portion of plane containing the triangle ABC and bounded by the line BC andthe parallel from A to BC . (It is easy to observe that the plane is divided in stripes parallel to thisA-stripe.) Two =-triangles adjacent to ABC lie in this A-stripe. We claim that no path from ABC toone of these triangles exists.

Indeed, observe that a move changes the type of the triangle hosting the checker, implying that if a path must exists, then it has an odd number of moves. On the other hand, a move changes the positionof the checker from a stripe to an adjacent stripe located upwards or downwards. To reach in the endthe same A-stripe, such a path must consists in an even number of moves, a contradiction.

To conclude the proof, consider the B-stripe and the two =-triangles adjacent to ABC . Repeatingthe argument, we are done.

Alternative Solution. Use four colors – one for ABC and the three adjacent triangles – to indicatethe triangles that can be visited starting from any of these triangles. Justify.



87878787


Suppose A = a0, a1, …, 12 −na is a rearrangement of the numbers 0, 1, 2, …, 2n – 1 which satisfies

the required condition. Then (2 A), (2 A + 1) is a rearrangement of the numbers 0, 1, 2, …, 2n+1 – 1,

where (2 A) = 2a0, 2a1, …, 212 −na and (2 A + 1) = 2a0 + 1, 2a1 +1, …, 2

12 −na + 1. It is obvious that the

rearrangement (2 A), (2 A + 1) satisfies the claim. Indeed, no triples bi, b j, bk , with bi – b j = b j – bk ,

i < j < k may occur in (2 A) nor in (2 A + 1), since either 2

,2

,2

k ji bbbor

2

1,

2

1,

2

1 −−− k ji bbbbelongs to A,

contrary to the fact that A is “free” of triples in arithmetic progressions.Starting with the arrangement 2, 0, 3, 1 for the numbers 0, 1, 2, 3 and applying the above

procedure, in 5 steps one has the required rearrangement.

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