## Search found 50 matches

- Fri Feb 09, 2018 9:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2008/1
- Replies:
**2** - Views:
**463**

### Re: BdMO National Higher Secondary 2008/1

Tasnood,$0$ is not a positive integer. Now,the sum of the first $n$ even numbers is $n(n+1)$ and for the case of odd numbers,that is $n^2$.So,by the question, the answer is $2008(2008+1)-2008^2$=$2008$.

- Thu Jan 25, 2018 8:44 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 4
- Replies:
**5** - Views:
**469**

### Re: BDMO 2017 National round Secondary 4

ahmedittihad meant $AD$ is perpendicular to $CD$.Let $C'$ be the reflection of $C$ in $M$.So,$CC'$=$13$.$C',A,D$ are collinear.Again,[$ABCD$]=[$AMCD$]+[$BMC$]=[$AMCD$]+[$AC'M$]=[$CC'D$].Suppose $BC$=$a$,$CD$=$b$,$AD$=$c$.Since $\bigtriangleup CC'D$ is a right triangle,it suffices to find $b(a+c)$.Fr...

- Sat Dec 16, 2017 7:37 pm
- Forum: Number Theory
- Topic: Prime numbers
- Replies:
**4** - Views:
**212**

### Re: Prime numbers

$n$=$2,5,7$.

- Wed Dec 13, 2017 8:45 pm
- Forum: Number Theory
- Topic: Yet divisibility...
- Replies:
**6** - Views:
**523**

### Re: Yet divisibility...

This is an IMO $1994$ problem.Look up the problem in $AoPS$.

- Mon Dec 11, 2017 1:55 pm
- Forum: Higher Secondary Level
- Topic: A strange NUMBER THEORY Problem
- Replies:
**4** - Views:
**306**

### Re: A strange NUMBER THEORY Problem

The $3rd$ number begets an answer of $-1$ which isn't a natural number.

- Fri Jul 07, 2017 1:08 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**4294**

### Re: Beginner's Marathon

Problem $26$

Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.

Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.

- Fri Jul 07, 2017 10:45 am
- Forum: Number Theory
- Topic: More primes and more primes!
- Replies:
**1** - Views:
**320**

### Re: More primes and more primes!

Apparently,this is a TST problem from Hong Kong.We rewrite our equation:$2^m*p^2$=$y^7-1$=$(y-1)(y^6+y^5....+y+1)$.Clearly $y$ being odd implies $y^6+y^5...+y+1$ is odd.Hence $2^m$ must divide $y-1$ since $p$ is a positive integer.Now the equation $\frac{y-1}{2^m}$*$(y^6+y^5...+y+1)$=$p^2$ leads us ...

- Wed Jul 05, 2017 7:55 pm
- Forum: Number Theory
- Topic: More primes and more primes!
- Replies:
**1** - Views:
**320**

### More primes and more primes!

Find every triplet of positive integers ($m,p,y$) such that the equation:$2^m*p^2+1$=$y^7$ is satisfied when $p$ and $y$ are primes.

- Tue Jul 04, 2017 7:46 pm
- Forum: Number Theory
- Topic: Infinite solutions
- Replies:
**1** - Views:
**265**

### Infinite solutions

Find with proof every pair of ($a$,$b$) in general form such that $a+b^2$|$a^3+b^3$.

- Mon Jun 19, 2017 3:02 pm
- Forum: Geometry
- Topic: IMO 2009 SL(G-1)
- Replies:
**1** - Views:
**284**

### IMO 2009 SL(G-1)

Let $\bigtriangleup$ $ABC$ be an isosceles triangle with $AB$=$AC$.The bisectors of $\angle BAC$ and $\angle ABC$ intersect $BC$ and $AC$ at $D$ and $E$ respectively.Suppose $\angle BEK$ to be $45$ degree,where $K$ is the incenter of $\bigtriangleup$ $ACD$.Determine all possible values of $\angle BA...