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Balkan Mathematical Olympiad 2017 – Problems

May 10, 2017 Leave a comment

Problem 1. Find all ordered pairs of positive integers { (x, y)} such that:

\displaystyle x^3+y^3=x^2+42xy+y^2.

Problem 2. Consider an acute-angled triangle {ABC} with {AB<AC} and let {\omega} be its circumscribed circle. Let {t_B} and {t_C} be the tangents to the circle {\omega} at points {B} and {C}, respectively, and let {L} be their intersection. The straight line passing through the point {B} and parallel to {AC} intersects {t_C} in point {D}. The straight line passing through the point {C} and parallel to {AB} intersects {t_B} in point {E}. The circumcircle of the triangle {BDC} intersects {AC} in {T}, where {T} is located between {A} and {C}. The circumcircle of the triangle {BEC} intersects the line {AB} (or its extension) in {S}, where {B} is located between {S} and {A}.

Prove that {ST}, {AL}, and {BC} are concurrent.

Problem 3. Let {\mathbb{N}} denote the set of positive integers. Find all functions {f:\mathbb{N}\longrightarrow\mathbb{N}} such that

\displaystyle n+f(m)\mid f(n)+nf(m)

for all {m,n\in \mathbb{N}}

Problem 4. On a circular table sit {\displaystyle {n> 2}} students. First, each student has just one candy. At each step, each student chooses one of the following actions:

  • (A) Gives a candy to the student sitting on his left or to the student sitting on his right.
  • (B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps. Find the number of legitimate distributions.

(Two distributions are different if there is a student who has a different number of candy in each of these distributions.)

Source: AoPS

Missing digit – short puzzle

April 15, 2017 Leave a comment

The number 2^{29} has 9 digits, all different; which digit is missing?

Mathematical Mind-Benders, Peter Winkler

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IMO 2015 Problem 1

July 10, 2015 Leave a comment

Problem 1. We say that a finite set {\mathcal{S}} of points in the plane is balanced if, for any two different points {A} and {B} in {\mathcal{S}}, there is a point {C} in {\mathcal{S}} such that {AC=BC}. We say that {\mathcal{S}} is center-free if for any three different points {A}, {B} and {C} in {\mathcal{S}}, there is no points {P} in {\mathcal{S}} such that {PA=PB=PC}.

(a) Show that for all integers {n\ge 3}, there exists a balanced set having {n} points.

(b) Determine all integers {n\ge 3} for which there exists a balanced center-free set having {n} points.

Problem 2. Find all triples of positive integers {(a, b, c)} such that {ab-c, bc-a, ca-b} are all powers of 2.

Problem 3. Let {ABC} be an acute triangle with {AB > AC}. Let {\Gamma } be its cirumcircle., {H} its orthocenter, and {F} the foot of the altitude from {A}. Let {M} be the midpoint of {BC}. Let {Q } be the point on { \Gamma } such that {\angle HQA } and let {K } be the point on {\Gamma } such that {\angle HKQ }. Assume that the points {A,B,C,K }and {Q } are all different and lie on {\Gamma} in this order.

Prove that the circumcircles of triangles {KQH } and {FKM } are tangent to each other.

Source: AoPS

Categories: Olympiad Tags: , ,

Monotonic bijection from naturals to pairs of natural numbers

October 23, 2014 Leave a comment

This is a cute problem I found this evening.

Suppose {\phi : \Bbb{N}^* \rightarrow \Bbb{N}^*\times \Bbb{N}^*} is a bijection such that if {\phi(k) = (i,j),\ \phi(k')=(i',j')} and {k \leq k'}, then {ij \leq i'j'}.

Prove that if {k = \phi(i,j)} then {k \geq ij}.

Proof: The trick is to divide the pairs of positive integers into families with the same product.

\displaystyle \begin{matrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) & \cdots \\ & (2,1) & (3,1) & (2,2) & (5,1) & (2,3) & \cdots \\ & & &( 4,1) & & (3,2) & \cdots \\ & & & & & (6,1) & \cdots \end{matrix}

Note that the {M}-th column contains as many elements as the number of divisors of {M}. Now we just just use a simple observation. Let {\phi(k)=(i,j)} be on the {M}-th column (i.e. {ij = M}). If {n \geq 1} then {\phi(k+n)=(i',j')} cannot be on one of the first {M-1} columns. Indeed, the monotonicity property implies {M = ij \leq i'j'}. The fact that {\phi} is a bijection assures us that {\phi(1),...,\phi(k)} cover the first {M-1} columns. Moreover, one element from the {M}-th column is surely covered, namely {(i,j) = \phi(k)}. This means that

\displaystyle k \geq d(1)+...+d(M-1)+1 \geq M = ij,

where we have denoted by {d(n)} the number of positive divisors of {n}.

IMC 2014 Day 1 Problem 4

July 31, 2014 2 comments

Let {n > 6} be a perfect number and {n = p_1^{e_1}...p_k^{e_k}} be its prime factorization with {1<p_1<...<p_k}. Prove that {e_1} is an even number.

A number {n} is perfect if {s(n)=2n}, where {s(n)} is the sum of the divisors of {n}.

IMC 2014 Day 1 Problem 4

Group actions and applications to number theory

April 6, 2014 Leave a comment

Theorem. Let {G} be a finite group and {p} be a prime number. Then {|G|^{p-1} \equiv |\{x \in G : x^p=e\}|}. (we denote by {|X|} the cardinal of {X}).

Proof: Denote {X = \{(g_1,...,g_p) \in \Bbb{Z}^p : g_1...g_p=e\}}. Then {|X|=|G|^{n-1}} since if we cnoose the first {p-1} elements arbitrarily in {G} then the last element can be chosen in a unique way such that the equation is satisfied. Note that {X} is stable under cyclic permutations of the elements of a vector. Therefore the action of {\Bbb{Z}_p} on {X} by

\displaystyle j\cdot (g_1,...,g_p)=(g_{j+1},...,g_{j+p})

is well defined.

We will need the following lemma:

Lemma. If {G} is a {p} group (with {p} prime) and {G} acts on {X} then

\displaystyle |X| \equiv |\{x \in X : G\cdot x = \{x\}\}| \pmod p.

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IMO 1981 Day 1

December 18, 2013 Leave a comment

Problem 1. Let {P} be a point inside a given triangle {ABC} and denote {D,E,F} the feet of the perpendiculars from {P} to the lines {BC,CA,AB} respectively. Find {P} such that the quantity

\displaystyle \frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}

is minimal.

Problem 2. Let {1 \leq r\leq n} and consider all subsets of {r} elements of the set {\{1,2,..,n\}}. Each of these subsets has a smallest member. Let {F(n,r)} denote the arithmetic mean of these smallest numbers. Prove that

\displaystyle F(n,r)=\frac{n+1}{r+1}.

Problem 3. Determine the maximum value of {m^3+n^3} where {m,n \in \{1,2,..,1981\}} with {(n^2-mn-m^2)^2=1}.

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