## Search found 176 matches

- Sat Feb 25, 2017 12:37 am
- Forum: Social Lounge
- Topic: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
- Replies:
**10** - Views:
**7384**

### Re: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!

Altho its a bit late, congrats on the 1000th post.

- Thu Feb 23, 2017 11:10 pm
- Forum: Algebra
- Topic: When the solution is easier than the question
- Replies:
**1** - Views:
**1472**

### When the solution is easier than the question

Consider all possible subsets of $\{1, 2, . . . ,N\}$ which contain no neighbouring elements. Prove that the sum of the squares of the products of all numbers in these subsets is $(N + 1)! - 1$.

- Thu Feb 23, 2017 6:09 pm
- Forum: Combinatorics
- Topic: Combi Solution Writing Threadie
- Replies:
**10** - Views:
**9410**

### Re: Combi Solution Writing Threadie

$\text{Problem 6:}$ Each square of a $2^n - 1 \times 2^n - 1$ square board contains either $+1$ or a $-1$. Such an arrangement is deemed successful if each number is the product of its neighbours. Find the number of successful arrangements. Solution to problem 6 First, lets define some terms. 1. A ...

- Thu Feb 23, 2017 3:19 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**27599**

### Re: Combi Marathon

$\text{Problem 7}$ Elmo is drawing with colored chalk on a sidewalk outside. He first marks a set $S$ of $n>1$ collinear points. Then, for every unordered pair of points $\{X,Y\}$ in $S$, Elmo draws the circle with diameter $XY$ so that each pair of circles which intersect at two distinct points are...

- Thu Feb 23, 2017 2:58 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**27599**

### Re: Combi Marathon

Solution to problem 6 First, lets define some terms. 1. A arrangement with at least one $-1$ will be called a nontrivial arrangement. 2. A arrangement that remains unchanged when its reflected by its middle column will be called vertically symmetric. Similarly, horizontally symmetric is defined. 3. ...

- Tue Feb 21, 2017 11:33 pm
- Forum: Algebra
- Topic: ISL 2003 A1
- Replies:
**1** - Views:
**1379**

### Re: ISL 2003 A1

Plug in $y=-f(x)$ to get that $f(0)=2x+f(\text{something})$, so $f$ is surjective. So, there exists $t$ so that $f(t)=0$. Plug $x=t$. We get that $f(y)=2t+f(f(y)-t)$. Set $f(y)=x$ here, and due to surjectivity $x$ ranges over all reals. So, $x=2t+f(x-t)$, equivalently $x+t=2t+f(x)$. So, general solu...

- Tue Feb 21, 2017 10:58 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**27599**

### Re: Combi Marathon

$\text{Problem 6:}$

Each square of a $2^n - 1 \times 2^n - 1$ square board contains either $+1$ or $−1$. Such an arrangement is deemed successful if each number is the product of its neighbours. Find the number of successful arrangements.

Each square of a $2^n - 1 \times 2^n - 1$ square board contains either $+1$ or $−1$. Such an arrangement is deemed successful if each number is the product of its neighbours. Find the number of successful arrangements.

- Tue Feb 21, 2017 10:54 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**27599**

### Re: Combi Marathon

$\text{Solution to problem 5:}$ We denote the $n^{th}$ triangular number by $T_n$. Now, consider the pairs: $(1,m+1),(2,m+2),(3,m+3)\cdots (m,2m)$. If we take $k$ of the second elements, we see that this pairing can only achieve the values of form $T_m+km$ where $k$ is an integer so that $0\le k\le ...

- Mon Feb 20, 2017 6:45 pm
- Forum: Number Theory
- Topic: IMO Shortlist 2011 N5
- Replies:
**4** - Views:
**2501**

### Re: IMO Shortlist 2011 N5

If $f(m)=f(n)$ then $f(m)|f(n)$. Now, assume that $f(m)<f(n)$. We consider $3$ cases. Case $1$: $f(m-n)<f(m)$. We know that, $f(m-n)|f(m)-f(n)$ and $f(m-(m-n))|f(m)-f(m-n)\Rightarrow f(n)|f(m)-f(m-n)$. So, $f(m-n)\le |f(m)-f(n)|=f(n)-f(m)$. $f(n)\le |f(m)-f(m-n)|=f(m)-f(m-n)$. Adding them, $f(m-n)+f...

- Mon Feb 20, 2017 6:30 pm
- Forum: Number Theory
- Topic: IMO Shortlist 2011 N5
- Replies:
**4** - Views:
**2501**

### IMO Shortlist 2011 N5

Let $f$ be a function from the set of integers to the set of positive integers. Suppose that for

any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for

all integers $m, n$ with $f(m) \le f(n)$ the number $f(n)$ is divisible by $f(m)$.

any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for

all integers $m, n$ with $f(m) \le f(n)$ the number $f(n)$ is divisible by $f(m)$.