## Search found 24 matches

- Sun Jun 28, 2020 6:56 pm
- Forum: Combinatorics
- Topic: EGMO '18 P3
- Replies:
**1** - Views:
**271**

### Re: EGMO '18 P3

Let $f(n)$ denote the maximum number of euros the Jury can make if there are $n$ contestants. We will prove that $f(n)= 2^n-(n+1)$ for every $n\geq1$ which proves part (a). If a contestant $C_i$ makes a move and gets ahead of a contestant $C_j$, we will say $C_i$ crossed $C_j$. Also, let $a_k$ denot...

- Sun Jun 28, 2020 5:28 pm
- Forum: Combinatorics
- Topic: EGMO '18 P3
- Replies:
**1** - Views:
**271**

### EGMO '18 P3

The $n$ contestants of EGMO are named $C_1,C_2,...,C_n$. After the competition, they queue in front of the restaurant according to the following rules. 1. The Jury chooses the initial order of the contestants in the queue. 2. Every minute the Jury chooses an integer $i$ with $1\leq i\leq n$. i) If c...

- Sun Oct 20, 2019 11:04 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO '94 P3
- Replies:
**1** - Views:
**36443**

### Re: IMO '94 P3

Solution: (a) Let $A= a_1,a_2,a_3,.... $ be the sequence of positive integers whose base $2$ representation has precisely three $1s$ and for all $k\in N$, $a_{k+1}>a_k$. Claim 1 : $0\leq f(k+1)-f(k)\leq 1$ Proof: Note that $2k+2 \in A$ iff $k+1 \in A$. Now, $f(k+1)-f(k)<0$ implies $k+1$ is in $A$ an...

- Sun Oct 20, 2019 8:58 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO '94 P3
- Replies:
**1** - Views:
**36443**

### IMO '94 P3

For any positive integer $k$, let $f(k)$ be the number of elements in the set ${k+1,k+2,...,2k}$ whose base $2$ representation has precisely three $1s$. (a) Prove that for each positive integer $m$, there exists at least one positive integer $k$ such that $f(k)=m$. (b) Determine all positive integer...

- Tue Jul 23, 2019 12:31 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P4
- Replies:
**1** - Views:
**6471**

### Re: IMO 2019/P4

Solution: We will show that $(1,1)$ and $(3,2)$ are the only possible pairs. Claim1: $\lfloor\frac{k}{3}\rfloor =\lfloor\frac{n}{2}\rfloor$ Proof: $k!=(2^n-1)(2^n-2)(2^n-4)......(2^n-2^{n-1})$ $ =2^{\frac{n(n-1)}{2}}(2^n-1)(2^{n-1}-1)(2^{n-2}-1)....(2-1)$ ...............(1) As $2$ is a primitive roo...

- Wed May 15, 2019 2:34 pm
- Forum: Number Theory
- Topic: Difference Between Divisors
- Replies:
**1** - Views:
**4561**

### Re: Difference Between Divisors

Solution: We will prove that there are infinitely many positive integers $n$ and $a$ such that $n^2+1=a(a-n)$. LEMMA: There are infinitely many positive integer solutions to the equation $5x^2-4=y^2$. Proof: Assume that $(x',y')$ is a solution of this equation.Then we follow the transformation: $(x,...

- Wed May 15, 2019 2:12 pm
- Forum: Number Theory
- Topic: Difference Between Divisors
- Replies:
**1** - Views:
**4561**

### Difference Between Divisors

Show that there are infinitely many positive integers $n$ such that $n^2+1$ has two positive integer divisors whose difference is $n$.

- Thu Mar 28, 2019 9:14 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2019/9
- Replies:
**2** - Views:
**6648**

### Re: BdMO National Higher Secondary 2019/9

Let $B'$ be a point on the line $AB$ such that $AB'=AC$ and $C'$ be a point on the line $DC$ such that $DC'=BD$. So, it suffices to prove that $BB'=CC'$. $\triangle AB'P \cong \triangle ACP$ $ \Rightarrow \angle APB'=\angle APC$ $ \Rightarrow \angle BPB'=\angle APD$. Similiarly, $\triangle DBP \cong...

- Tue Mar 19, 2019 4:57 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2019/5
- Replies:
**1** - Views:
**6284**

### Re: BdMO National Higher Secondary 2019/5

We will prove it by strong Induction. Call a permutation $a_1,a_2,....,a_n$ "GOOD" if the average of any two numbers doesn't appear in between them and call it "VERY GOOD" if it is "GOOD" and for every integer $x\in {1,2,...,n} $, there exists a $i$ such that $a_i=x$ Lemma: If $a_1,a_2,....,a_n$ is ...

- Sat Jan 19, 2019 9:17 pm
- Forum: Combinatorics
- Topic: ISL 2010 C2
- Replies:
**1** - Views:
**4030**

### Re: ISL 2010 C2

Solution: We claim that the smallest positive integer $M$ is $2^{N-2}+1$. We will prove that the highest positive integer $K$ such that we will not find a $DIVERSE$ set of $K$ flags is $2^{N-2}$. A set of $K$ flags is not $DIVERSE$ if and only if there exist two positive integers $m$ and $n$ such th...