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CGMO-2012 (China Girls Math Olympiad 2012) Problem-3

Find all pairs (a,b) of integers satisfying: there exists an integer  d \ge 2 such that a^n + b^n+1  is divisible by d for all positive integers n


IMO 2012 problems

This year IMO problems !!!!

Problem 1 :

Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST.

Problem 2 :

Let {n\ge 3} be an integer, and let {a_2,a_3,\ldots ,a_n} be positive real numbers such that  {a_{2}a_{3}\cdots a_{n}=1}  Prove that

\displaystyle \left(a_{2}+1\right)^{2}\left(a_{3}+1\right)^{3}\dots\left(a_{n}+1\right)^{n}>n^{n}.

Problem 3 :

The liar’s guessing game is a game played between two players {A} and {B}. The rules of the game depend on two positive integers {k} and {n} which are known to both players.

At the start of the game {A} chooses integers {x} and {N} with {1 \le x \le N.} Player {A} keeps {x}secret, and truthfully tells {N} to player {B}. Player {B} now tries to obtain information about {x} by asking player {A} questions as follows: each question consists of {B} specifying an arbitrary set {S} of positive integers (possibly one specified in some previous question), and asking {A}whether {x} belongs to {S}. Player {B} may ask as many questions as he wishes. After each question, player {A} must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any {k+1} consecutive answers, at least one answer must be truthful.

After {B} has asked as many questions as he wants, he must specify a set {X} of at most {n}positive integers. If {x} belongs to {X}, then {B} wins; otherwise, he loses. Prove that:

1. If {n \ge 2^k,} then {B} can guarantee a win. 2. For all sufficiently large {k}, there exists an integer {n \ge (1.99)^k} such that {B} cannot guarantee a win.

Problem 4 :

Find all functions {f:\mathbb Z\rightarrow \mathbb Z} such that, for all integers {a,b,c} that satisfy {a+b+c=0}, the following equality holds:

\displaystyle f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).

 Problem 5 :

Let {ABC} be a triangle with {\angle BCA=90^{\circ}}, and let {D} be the foot of the altitude from {C}. Let {X} be a point in the interior of the segment {CD}. Let {K} be the point on the segment {AX}such that {BK=BC}. Similarly, let {L} be the point on the segment {BX} such that {AL=AC}. Let {M} be the point of intersection of {AL} and {BK}.

Show that {MK=ML}.

Problem 6 :

Find all positive integers {n} for which there exist non-negative integers {a_1, a_2, \ldots, a_n} such that

\displaystyle \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.

You may download the PDF version :
IMO 2012

Thank you and good luck for these delicious problems

Infinite number of primes :)

In today’s post , I will present a very well known proof. What makes this proof especially appealing is that it is not too complex. Even so, it is very powerful. This theorem was first presented in Euclid’s Elements (Book IX, Proposition 20).

Theorem: There are an infinite number of primes.


(1) Assume that there is only a finite number of primes.

(2) Then, there exists a prime pn that is the largest prime.

(3) Let p1, p2, …, pn be the list of all primes that exist.

(4) Let x = p1*p2*…*pn + 1.

(5) By the fundamental theorem of arithmetic (see Theorem 3, here), we know there is at least one prime that divides x. Let us call this prime p*.

(6) But none of the primes p1 … pn divide x since x ≡ 1 (mod pi) for any of the primes p1 … pn

(7) Therefore, we have a contradiction. We have a prime p* that is not in the complete list of primes.

(8) So, we reject our assumption in step#1 and conclude that there are an infinite number of primes that exist.


Bumper Problems

These are the set of 5 problems  each of  5 marks

The person who get more than 20 points will get a mathematics book on any topic he want…
So try these beautiful problems to test your mathematics abilities and getting new things…

Ways by which you can answer :
I am adding a form in the end of this post  , you can answer there ….

Note : You have give solution of your answer too..

Problem 1)

The bisectors of the angles A and B of the \bigtriangleup ABC  meet the sides
BC and CA at the points D and E , respectively.
Assuming that AE+BD =AB, determine the angle C

Problem 2)

Given a  \bigtriangleup ABC and  D be point on side  AC such that  AB = DC,
\angle BAC= 60-2X ,   \angle DBC= 5X and  \angle BCA= 3X
Find the value of  X

Problem 3)

If p and q are natural numbers so that

Prove that p is divisible by 1979 .

Problem 4)

Find highest degree n of 1991 for which 1991ⁿ  divides the number :

Problem 5)

Let ƒ(n) denote the sum of the digits of n. Let N = 4444⁴⁴⁴⁴
Find ƒ(ƒ(ƒ(n))))

You can use these symbols to write solutions more conveniently

Mathematical Operators
Exponents  :   ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋ ₌ ₍ ₎ 

Modular Arithmetic

Modular arithmetic is a notation and set of mathematics that were first introduced by Carl Friedrich Gauss. 🙂

The major insight is that equations can fruitfully be analyzed from the perspective of remainders. Standard equations use the ‘=’ sign. Modular arithmetic uses the ‘‘ sign. Two values that are ‘≡’ to each other are said to be congruent relative the modulus. In the case below, the modulus is 3.  🙂

Here’s an example of a modular equation:

7 ≡ 1 (mod 3).

By definition, this means that 3 divides 7 – 1.

Definition 1: a ≡ b (mod c) if and only if c divides a – b.

This definition tells us the following is true:

7 ≡ 1 ≡ 10 ≡ -2 (mod 3).

Now, one of the most interesting things about ‘≡’ is that it follows many of the same relations as ‘=’ .  🙂 🙂

Notice 1: For any value a,b,c,d,n where a ≡ b (mod n) and c ≡ d (mod n):  🙂

(a) a + c ≡ b + d (mod n)
 Proof :  We know that n divides (a + c) – (b + d) since this is equal to: (a -b) + (c – d).

(b) a – c ≡ b – d (mod n)
Proof : We know that n divides (a – c) – (b – d) since this is equal to: (a – b) – (c – d).

(c) ac ≡ bd (mod n)
Proof : We know that n divides ac – bd since this is equal to : c(a – b) + b(c – d).


Notice  2: If a ≡ b (mod n) then:

(a) a + c ≡ b + c (mod n)
Proof : We know (a) since n divides a + c – (b + c) = a – b.

(b) a – c ≡ b – c (mod n)
 Proof : We know (b) since n divides a – c – (b – c) = a – b.

(c) ac ≡ bc (mod n)
Proof: We know (c) since n divides ac – bc = c(a – b)


Corrolary 2.1: a ≡ d (mod n), b ≡ e (mod n), c ≡ f (mod n), then:
a + b + c ≡ d + e + f  mod n 

(1) We know that a + b ≡ d + e from above.
(2) We therefore know that (a + b) + c ≡ (d + e) + f.


Notice  3: a + b + c ≡ 0, a ≡ 0 (mod p), then b + c ≡ 0 (mod p).

(1) a + b + c ≡ 0 (mod p) [Definition of ≡ ]

(2) b ≡ c (mod p) → a + b ≡ a + c (mod p) [See above]

(3) So, 0 ≡ a + b + c ≡ 0 + b + c ≡ b + c (mod p).


I will cover modular arithmetic in depth 🙂 🙂
Today you must we wondering about its application ..
But after whole course You will surely love this topic
Thank you 🙂 🙂

If any doubts , then dont hesitate in asking Your doubts 🙂 🙂
I will surely love to solve your doubts

Problem of the day !

 We have started one more interesting feature of the blog 
I will be posting problem of the day , and the best answer will be highly appreciated 

If you win problem of the day for more than 4 times in a  week then I will send you a mathematics very interesting E-Book 

So here is Question 1 )


How many digits does the number 21000 contain?


You have to give solution of your answer too