## Search found 550 matches

- Fri May 10, 2013 12:59 am
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2013 Problem 5
- Replies:
**1** - Views:
**2038**

### Re: APMO 2013 Problem 5

(I got the idea to use Harmonic Division in this problem from Nadim Ul Abrar vai after the exam. This problem actually made me learn Harmonic Division) Let $BR\cap \omega=E', CE'\cap AR=S$. Now by the first lemma of Zhao, $AC$ is a symmedian of $\triangle ABD$. Thus $(C,A;D,B)=-1$. Thus $E'(C,A;D,B)...

- Fri May 10, 2013 12:15 am
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2013 Problem 3
- Replies:
**2** - Views:
**2085**

### Re: APMO 2013 Problem 3

(This is a solution I saw later in the math camp.) Let $\displaystyle\sum_{i=1}^{k}a_i=A,\displaystyle\sum_{i=1}^{k}b_i=B$. Also let $X_i=X_1+(i-1)d$. Since $X_1,X_2$ are integers, $d=X_2-X_1$ is also an integer. Then, $\displaystyle\sum_{i=1}^{k}(a_in+b_i+1)>\displaystyle\sum_{i=1}^{k}\lfloor a_in+...

- Thu May 09, 2013 11:47 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2013 Problem 2
- Replies:
**1** - Views:
**1503**

### Re: APMO 2013 Problem 2

It's easy to check that $n$ can't be a perfect square. Then let $n=m^{2}+r$, where $1\leq r\leq 2m$. Then, the expression becomes, $\displaystyle\frac{(m^{2}+r)^{2}+1}{m^{2}+2}=\displaystyle\frac{m^{2}(m^{2}+2)+2(r-1)(m^{2}+2)+r^{2}-4r+4+1}{m^{2}+2}$ $=m^{2}+2(r-1)+\displaystyle\frac{(r-2)^{2}+1}{m^...

- Wed May 01, 2013 10:33 pm
- Forum: Geometry
- Topic: China TST 2013 Day 5 Problem 2
- Replies:
**0** - Views:
**1018**

### China TST 2013 Day 5 Problem 2

The circumcircle of triangle ABC has centre $O$. $P$ is the midpoint of $\widehat{BAC}$ and $QP$ is the diameter. Let $I$ be the incentre of $\triangle ABC$ and let $D$ be the intersection of $PI$ and $BC$. The circumcircle of $\triangle AID$ and the extension of $PA$ meet at $F$. The point $E$ lies...

- Sun Apr 14, 2013 12:16 pm
- Forum: Number Theory
- Topic: Cool Number Theory
- Replies:
**1** - Views:
**1020**

### Re: Cool Number Theory

There are primes as powers,primes as divisors-I can't think of sth else than Fermat' Little Theorem! :twisted: Let us assume,by symmetry,that $p\leq q$. Now $p|(5^p-2^p)$ or $p|(5^q-2^q)$. By Fermat's theorem,if a prime $p$ divides $5^p-2^p$,then $5^p-2^p\equiv 5-2\equiv 3\equiv 0\Rightarrow p=3$. A...

- Sat Apr 13, 2013 11:59 pm
- Forum: Number Theory
- Topic: 2002 Czech-Polish-Slovak
- Replies:
**4** - Views:
**1699**

### Re: 2002 Czech-Polish-Slovak

My solution's quite the same.

- Thu Apr 11, 2013 11:42 pm
- Forum: Number Theory
- Topic: BdMC-Not So Easy II problem 5
- Replies:
**1** - Views:
**992**

### BdMC-Not So Easy II problem 5

Find all pairs of $(a,b)$ such that for all $n\in \mathbb{N}$,

$a\left \lfloor bn \right \rfloor=b\left \lfloor an \right \rfloor$

$a\left \lfloor bn \right \rfloor=b\left \lfloor an \right \rfloor$

- Wed Apr 10, 2013 11:49 pm
- Forum: Number Theory
- Topic: m and n
- Replies:
**4** - Views:
**2098**

### m and n

Let $m,n$ be positive integers. Prove that $(2^m-1)^2|(2^n-1)$ if and only if $m(2^m-1)|n$.

Source:Russia 1997

Source:Russia 1997

- Mon Apr 08, 2013 11:52 pm
- Forum: Number Theory
- Topic: Yammy...GCD
- Replies:
**0** - Views:
**664**

### Yammy...GCD

Find all triples $(m,n,l)$ of positive integers such that $m+n=gcd(m,n)^2,m+l= gcd(m,l)^2,n+l=gcd(n,l)^2$.

Source: $1997$ Russian Mathematical Olympiad

Source: $1997$ Russian Mathematical Olympiad

- Sat Apr 06, 2013 11:53 pm
- Forum: Algebra
- Topic: f(f(n))=3n
- Replies:
**1** - Views:
**1091**

### Re: f(f(n))=3n

(I saw a similar problem in IMO math's functional equation tutorial that asked for the value of $f(2006)$. I have used the same procedure here.) We claim first that, :arrow: $f(2\cdot 3^{n})=3^{n+1}$ :arrow: $f(3^{n})=2\cdot 3^{n}$ for $n=0,1,2,....$ We use induction. For $n=0,f(1)$ can't be $1$ oth...