## Search found 176 matches

Sat Feb 25, 2017 12:37 am
Forum: Social Lounge
Topic: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
Replies: 10
Views: 7598

### 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: 1534

### 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: 9654

### 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: 28532

### 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: 28532

### 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: 1438

### 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: 28532

### 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.
Tue Feb 21, 2017 10:54 am
Forum: Combinatorics
Topic: Combi Marathon
Replies: 48
Views: 28532

$\text{Solution to problem 5:}$ 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 $0\le k\le ... Mon Feb 20, 2017 6:45 pm Forum: Number Theory Topic: IMO Shortlist 2011 N5 Replies: 4 Views: 2623 ### Re: IMO Shortlist 2011 N5 If$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: 2623

### 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 $f(m)$.