Tuymaada math competition

The International Olympiad '"Tuymaada" is an annual mathematics competition for secondary school students held in Republic of Saka, Russia. The contestants compete individually. The participating teams (national and local teams) can have up to three students. . The contest is held in two days of competitions, in July.
Here are a few interesting math problems from Tuymaada:

1. Positive integers a < b are given. Prove that among every b
consecutive positive integers there are two numbers whose product
is divisible by ab.

2. What minimum number of colours is sufficient to colour all positive real
numbers so that every two numbers whose ratio is 4 or 8 have different colours?

3. Several knights are arranged on an infinite chessboard.
No square is attacked by more than one knight (in particular, a square
occupied by a knight can be attacked by one knight but not by two). Sasha
outlined a 14x16 rectangle. What maximum number of knights can
this rectangle contain?

4. Prove that there exists a positive c such that for every positive
integer N among any N positive integers not exceeding 2N there
are two numbers whose greatest common divisor is greater than cN.

5. Seven different odd primes are given. Is it possible that the
difference of 8th powers of every two of them is divisible by each of the
remained numbers?

6. 100 boxers of different strength participate in the Boxing Championship
of Dirtytrickland. Each of them fights each other once. Several boxers
formed a plot: each of them put a leaden horseshoe in his boxing-glove
during one of his fights. When just one of two boxers has a horseshoe,
he wins; otherwise, the stronger boxer wins. It turned out after
the championship that three boxers won more fights than any of the three
strongest participants.
What is the minimum possible number of plotters?

7. Organizers of a mathematical congress found that if they
accomodate any participant in a single room the rest can be accomodated
in double rooms so that two persons living in every double room know each
other.
Prove that every participant can organize a round table on graph theory
for himself and even number of other people so that each participant of
the round table knows both his neighbours.

8.Several rooks stand in the squares of the table shown in the figure.
The rooks beat all the squares (we suppose that a rook beats the square
it stands in). Prove that one can remove several rooks so that not more
than 11 rooks are left and these rooks still beat all the squares.

Tournament of Towns problems and solutions

The International Mathematics Tournament of Towns is a mathematics problem solving competition in which towns throughout the world can participate on an equal basis.

Students participate in their own towns which involves minimal transport and administrative costs.

The Tournament is conducted each year in two stages - Autumn and Spring (northern hemisphere time). The southern hemisphere academic year coincides with this structure.

Each stage has two papers, an "O" level and an "A" level, which are spaced roughly one week apart. The A level paper is more difficult, but offers more points. Students and their towns may participate in either stages or levels, or in all levels and stages.

The Tournament is open to all high school students, with the highest age of students being about 17 years old.

Students are awarded points for their best three questions in each paper, and their annual score is based on their best score in any of the four papers for the year.

There are two versions of each paper, known as the Senior and Junior papers. Students in Years 10 and 11 (the final two years of high school in the Russian nomenclature) are classified as Senior participants and therefore attempt the Senior paper. So that Year 10 students are not disadvantaged their scores are multiplied by 5/4. Younger students, in Years 9 and below, attempt the Junior paper. To ensure that the scoring is fair to all levels of students, Year 8 students have their scores multiplied by 4/3, Year 7 students have their scores multiplied by 3/2 and Year 6 students and below have their scores multiplied by 2.

A town's score is based on the average of its best N students, where the population of the town is N hundred thousand (see below for definition of town). There is a minimum of N=5. If a population is less than 500,000 then the score is multiplied by an appropriate compensatory factor.

Students who exceed a certain minimum score are awarded a Diploma by the Russian Academy of Sciences. Local organising committees also present their own awards. The Tournament is managed by a central committee in Moscow, which is a subcommittee of the Russian Academy of Sciences.

You can find problems and solutions here.

Tournament of Towns 2007 problems and solutions junior A-level

International mathematics Tournament of Towns problems and solutions:
Fall 2007
Junior A-level problems:
1.Let ABCD be a rhombus.Let K be a point on the line CD, other than C or D, such that AB = BK.Let P be a point of intersection of BD with the perpendicular bisector of BC.Prove that A, K and P are collinear.
2. (a) Each of Peter and Basil thinks of 3 positive integers.For each pair of his numbers, Peter writes down the greatest common divisor of the 2 numbers.For each pair of his numbers, Basil writes down the least common multiple of the 2 numbers.If both Peter and Basil write down the same 3 numbers prove that these 3 numbers are equal to each other.
(b) Can the analogous result be proved if each of Peter and Basil thinks of 4 positive integers instead?
3.Michael is at the centre of a circle of radius 100 meters. Each minute he will announce the direction in which he will be moving.Catharine can leave it as is or change it to the opposite direction. Then Michael moves exactly 1 metre in direction determined by Catherine.Does Michael have a strategy which guarantees that he can get out of the circle, even though Catherine will try to stop him?
4.Two players take turns entering a symbol in an empty cell of a 1 x n chessboard where n is an integer greater that 1. Aaron always enters the symbol x and Betty always enters the symbol O.
Two identical symbols may not occupy adjacent cells.A player without a move loses a game.If Aaron goes first, which player has a winning strategy?
5.Attached to each of a number of objects is a tag which states the correct mass of the object.The tags have fallen of and have been replaced on the objects at random.We wish to determine if by fact all tags are in fact correct.We may use exactly once a horizontal lever which is supported at its middle. The objects can be hung from the lever at any point on either side of support. The lever either stays horizontal or tilts to one side. Is this task always possible?
6.The audience arranges n coins in a row.The sequence of heads and tails is chosen arbitrary. The audience also chooses a number between 1 and n inclusive. Then the assistant turns one of the coins over and magician is brought in to examine the resulting sequence. By an agreement with the assistant beforehand, the magician tries to determine the number chosen by audience.
(a) Prove that if this is possible for some n then it is also possible for 2n.
(b) Determine all n for which this is possible.
7.For each letter in the English alphabet, William assigns an English word which contains that letter.His first document consists only of the word assigned to the letter A.In each subsequent document, he replaces each letter of the preceding document by its assigned word. The fortieth document begins with "`Till whatsoever start that guides my moving".Prove that this sentence reappears later in this document.
Solutions

Tournament of Towns 2007 problems and solutions junior O-level

International mathematics Tournament of Towns problems and solutions:
Fall 2007
Junior O-level problems:
1.Black and white checkers are placed on an 8 x 8 chessboard, with at the most one checker on each cell.What is the maximum level of checkers that can be placed such that each row and each column contains twice as many white checkers as black ones?
2.Initially , the number 1 and an non-integral number x are written on a blackboard.In each step we can choose 2 numbers on a blackboard, not necessarily different, and write their sum or their difference on the blackboard.We can also choose a non-zero number of the blackboard and write its reciprocal on the blackboard.Is it possible to write x^2 on a blackboard in a finite number of moves?
3.D is the midpoint on the side BC of triangle ABC.E and F are points on CA and AB respectively, such that BE is perpendicular to CA and CF is perpendicular to AB.If DEF is an equilateral triangle, does it follow that ABC is also equilateral?
4.Each cell of a 29x29 table contains one of the integers 1,2,3...29 and each of these integers appears 29 times.The sum of all the numbers above the main diagonal is equal to 3 times the sum of all the numbers below this diagonal.Determine the number in central cell of the table.
5.The audience chooses 2 of 5 cards numbered from 1 to 5 respectively.The assistant of a magician chooses 2 of remaining 3 cards, and asks the member of audience to take them to magician, who is in another room.The 2 cards are presented to magician in arbitrary order.By an arrangement with the assistant beforehand, the magician is able to deduce witch 2 cards the audience has chosen only from the 2 cards he receives.Explain how this may be done.

Solutions

Coins

Here are some interesting problems I encountered during the last week:
1.Several identical coins are placed on the table without overlapping. Prove that there is a coin which touches no more than three other coins.
2.Several coins are placed on the table without overlapping.Prove that there is a coin that touches no more than five other coins.
3.Several coins are placed on the table without overlapping.Prove that one can slide one of the coins to the edge of table without moving any other coin.
In next problem one can apply the method of small perturbations which is related to the method of extreme case (see previous post):
4.A straight line intersects a polygon exactly 2001 times.Prove that there is a straight line which is not parallel to any side of polygon and has more than 2001 common points with it.
5.There are 100 chords in a circle such that any two of them intersect.Is it always possible to draw one more chord such that it intersects all of them?

How to solve problems? - Extreme case

Here are some of my tips, and some problems from MathClub:
Sometimes in a solution of a problem it is crucial to consider extreme objects: The largest number, the closest point or vertex, a degenerate circle, any other limit case.
The minimal counterexample method is a combination of extreme case and a proof by contradiction: Assuming that some statement A is not true, there exists minimal (in some sense) counterexample.If we can somehow "decrease" it, then we get a contradiction.
Problems:
1.Prove that any polyhedron has at least 2 faces with the same number of sides.
2.A traveller went from his city A to the city B of his country, furthest away from A.Then from B he went to the city C, which is furthest away from B, and so on.Prove that if C and A are different cities, then the traveller will never return to A (all distances between cities of this country are different).
3.Numbers are placed on a chessboard.It is given that any number is equal to an arithmetical mean of its neighbours by a side.Prove that all numbers are equal.
4.Prove that for any integer n>1, 1 + 1/2 + 1/3 +...+1/n is not an integer.
5.Let us consider a point P inside a convex polygon and drop the perpendiculars from P to each side or its extension.Prove that at least one of the perpendiculars meets a corresponding side.
6. A six digit number is called lucky if 7 divides the sum of its digits.Are there any 2 consecutive lucky numbers?

Triangle problems

This is a very tricky problem:
1.Triangle ABC is given.Divide it by a broken line BDEFG into 5 triangles of equal area.
This one is from MathBattle:
2.Point E is given on the diameter AC of the circle.Draw a chord BD through E in such way that an area of the quadrilateral ABCD is maximal.
3.Find the maximal area of the right-angled triangle.

Proof by contradiction, and a few problems

It is known that m^2 + n^2 + m is divisible by mn for some positive integers m and n.Prove that m is a perfect square.
Proof to this problem is proof by contradiction.If you can`t solve this one, try solving one of these:
1.Prove that there is no largest real number.
2.Prove that there is no largest prime number.
3.5 children have picked up 9 flowers.Prove that at least 2 of them picked up the same number of flowers.
4.Prove that (n-1)(n)(n+1) is divisible by 6.
5.Prove that (n-2)(n-1)(n)(n+1)(n+2) is divisible by 2*3*4*5.
6.Prove that if n divides (n-1)! + 1 then n is a prime.

3! = 1*2*3

How to solve problems?

Here are some very good tips from Math club about problem solving:
If problem is hard, try to solve simpler problem related to it.This often gives you a clue to the original problem.
Try to consider a special case and generalize the idea of it`s solution afterwards.
Break the problem down into the smaller ones.
Generalize the problem.
Reduce the problem to simpler one.

Interesting problems

1.Here`s conversation between liars and truthtellers:
B to A:You are a liar.
C to B:It`s you who is a liar.
D to C:Both of them are liars.
D to C:And you are a liar too.
Can you tell who is who?
2.There are 2 ropes.It is known that each of them burns out completely in exactly 1 hour.
How one can measure 45 minutes using these ropes and a box of matches, if pace of burning is not uniform?
These 2 problems are very good examples of good problems.

Analysis

2n+1 people are on the party, and for every n people there is one who knows them all.Prove that one man (or woman:)) knows all 2n+1 people.
This is very hard problem, and although it seams easy it can be quite tricky.
Good luck trying to solve it.

JBMO geometry problem

This one`s from Junior Balkan Math Olympiad.It`s quite easy to find solution.
Given a (acute angled)trianle ABC and the tangent at A to the circumcircle intersects BC at P.
M is the midpoint of AP. BM intersects the circumcircle again at R and PR intersects the circumcircle again at S prove that AP || CS.

Geometry problems

Here are 3 geometry problems, and they`re not very hard.
1.Determine angle x
2.Determinate angle x
3.Determinate angle x if AB=BC






First 2 are easy, third is a bit harder .

1st Geometry problem

Here`s another one...
If AP/BP=q,
what is the value of AC/BC?
Notice that angle cao = angle oap and angle cbo = angle obp.

Seven millenium problems

In order to celebrate mathematics in the new millennium, The Clay Mathematics Institute of Cambridge, Massachusetts (CMI) has named seven Prize Problems. The Scientific Advisory Board of CMI selected these problems, focusing on important classic questions that have resisted solution over the years. The Board of Directors of CMI designated a $7 million prize fund for the solution to these problems, with $1 million allocated to each.
This is one of the coolest ways of getting rich.
Solve one of these and ...
Birch and Swinnerton-Dyer Conjecture
Hodge Conjecture
P vs NP Problem
Poincaré Conjecture
Riemann Hypothesis
Yang-Mills and Mass Gap

7th one is solved, so it`s possible.
More info on claymath.org.

Goldbach conjecture, twin prime conjecure

Goldbach conjecture is one of the oldest unsolved problems in number theory.It states:
Every integer greater than 2 can be written as sum of 2 primes.
Here are some examples:6=3+3;12=7+5;32=29+3...
Twin prime conjecture states that there are infinitely many twin primes.
Examples:3,5;5,7;11,13...
I`ve been trying to solve these two, but you can`t solve them with primary school education.Although they seem simple, these are one of the hardest problems I tried to solve.

Fermat last theorem

Prove Fermat's Last theorem for n=3 : X^3 + Y^3 = Z^3 where X, Y, Z are rational integers, then X, Y, or Z is 0.
This is interesting one.Here`s something about Fermat from wiki:
is the name of the statement in number theory that:

It is impossible to separate any power higher than the second into two like powers,

or, more precisely:

If an integer n is greater than 2, then the equation aⁿ + bⁿ = cⁿ has no solutions in non-zero integers a, b, and c.

In 1637 Pierre de Fermat wrote, in his copy of Claude-Gaspar Bachet`s translation of the famous Arithmetica of Diophantus, "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." (Original Latin: "Cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.")

Fermat's Last Theorem is strikingly different and much more difficult to prove than the analogous problem for n = 2, for which there are infinitely many integer solutions called Pythagorean triples (and the closely related Pythagorean theorem has many elementary proofs). The fact that the problem's statement is understandable by schoolchildren makes it all the more frustrating, and it has probably generated more incorrect proofs than any other problem in the history of mathematics. No correct proof was found for 357 years, when a proof was finally published by Andrew Wiles in 1995. The term "last theorem" resulted because all the other theorems proposed by Fermat were eventually proved or disproved, either by his own proofs or by other mathematicians, in the two centuries following their proposition. Although a theorem now that it has been proved, the status of Fermat's Last Theorem before then, in spite of the name, was that of a conjecture, a mathematical statement whose status (true or false) has not been conclusively settled.

Fermat's Last Theorem is the most famous solved problem in the history of mathematics , familiar to all mathematicians, and had achieved a recognizable status in popular culture prior to its proof.

USA math olympiad problem

Show that for any fixed integer n >=1, the sequence: 2, 2^2, 2^(2^2), 2^(2^(2^2))... mod n $$
is eventually constant. [That is, the sequence is defined: a_1=2, ai+1 = 2a_i. and you need to show that, for any positive integer n, the sequence a_1 mod n, a_2 mod n, .... is eventually constant.]

More number theory

Number theory is my favorite, and most of these problems are about numbers.Here`s another one:
if 11 divides (a+13*b) and 13 divides (a+11*b)
what is the least value of (a+b)?
This one`s easy, so here`s solution:
a+13b=11r =>a+2b=11p, where p=r-b
a+11b=13s =>a-2b=13q, where q=s-b
2a=11p+13q,
4b=11p-13q, thus p>q.
The least solution is: p=3 and q=1, a=23 and b=5, a+b=28,

Another math problem

If k*(a*b+1)=a^2+b^2, prove that k=n^2.
This one is very hard, so sharpen your pencils.

Learning math

Best and only way to learn math is to practice, practice and practice.Try solving some math problems here and you`ll see the difference.
A few simple math problems:
2 cats eat 2 mice in 2 days.How many cats will eat 100 mice in 100 days?

How much is 2+2/2-2*2/2-2?

How much is 2^2007 / 2^2008?(note: 2^3=2*2*2)

Is 2^2007+1 prime?

Number theory

I have been practiced number theory, so expect some number theory problems.

Power of 2 , another math problem

If 2^23 + 2^22 + ... + 2^14 + 2^13 - (x/5+6) is divisible by 1755, find x.
This one is mine too.
It`s similar to my first math problem, so if you can solve that, you can solve this.

Another math problem

Create two six-sided dice, such that the probability of each sum from 2 to 12 is the same as two standard dice. Each side must have at least one dot. Negative numbers are not allowed. There is another answer besides two standard {1,2,3,4,5,6} dice.

Try solving this!

Math Jokes

Here are some jokes I`ve stumbled upon recently:

Teacher: What is 2k + k?
Student: 3000!

Q: How does one insult a mathematician?
A: You say: "Your brain is smaller than any >0!"

A woman in a bar tries to pick up a mathematician.
"How old, do you think, am I?" she asks coyly.
"Well - 18 by that fire in your eyes, 19 by that glow on your cheeks, 20 by that radiance of your face, and adding that up is something you can probably do for yourself..."

Life is complex: it has both real and imaginary components.

Q: How does a mathematician induce good behavior in her children?
A: `I've told you n times, I've told you n+1 times...'

An investment firm is hiring mathematicians. After the first round of interviews, three hopeful recent graduates - a pure mathematician, an applied mathematician, and a graduate in mathematical finance - are asked what starting salary they are expecting.
The pure mathematician: "Would $30,000 be too much?"
The applied mathematician: "I think $60,000 would be OK."
The math finance person: "What about $300,000?"
The personnel officer is flabberghasted: "Do you know that we have a graduate in pure mathematics who is willing to do the same work for a tenth of what you are demanding!?"
"Well, I thought of $135,000 for me, $135,000 for you - and $30,000 for the pure mathematician who will do the work."

Many more at math.ualberta.ca/~runde/jokes.html.

The Riemann hypothesis

The Riemann hypothesis (also called the Riemann zeta-hypothesis), first formulated by Bernhard Riemann in 1859, is one of the most famous and important unsolved problems in mathematics.It has been an open question for almost 150 years, despite attracting concentrated efforts from many outstanding mathematicians. Unlike some other celebrated problems, it is more attractive to professionals in the field than to amateurs.

The Riemann hypothesis (RH) is a conjecture about the distribution of the zeros of the Rieman zeta function ζ(s). The Riemann zeta-function is defined for all s ≠ 1. It has zeros at the negative even integers (i.e. at s = −2, s = −4, s = −6, ...). These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that:

The real part of any non-trivial zero of the Riemann zeta function is ½.

Thus the non-trivial zeros should lie on the so-called critical line ½ + it with t a real number and i the imaginary unit. The Riemann zeta-function along the critical line is sometimes studied in terms of the Z-function whose real zeros correspond to the zeros of the zeta-function on the critical line.

The Riemann hypothesis is one of the most important open problems of contemporary mathematics, mainly because a large number of deep and important other results have been proven under the condition that it holds. Most mathematicians believe the Riemann hypothesis to be true. A $1,000,000 prize has been offered by the Clay Mathematics Institute for the first correct proof.Wow.

My first mathematics problem

This one is mine:
Prove that (2^2008 + 2042)^3 is divisible by 511.
2^3=2*2*2.
I`ll post solution later.

A few interesting math problems - JBMO 2007

Here are problems from Junior Balkan MO 2007:

1	Let be positive real number such that . Prove that the equation has no real solution.	S
2	Let be a convex quadrilateral with , and . The diagonals and intersect at point . Determine the measure of .	S
3	Given are points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least scalene triangles with vertices of that color.	S
4	Prove that if is a prime number, then is not a perfect square.

These are not very hard, but it took me 2 hours to solve 3. one, cuz I`m not very good at analysis.
Here are solutions to all of them(click on here :)).

Hello

I`ve started this blog, and I`ll post here some interesting math problems and interesting facts about mathematics.