Search found 176 matches

by Thanic Nur Samin
Thu Feb 23, 2017 11:10 pm
Forum: Algebra
Topic: When the solution is easier than the question
Replies: 1
Views: 1120

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 $(N + 1)! - 1$.
by Thanic Nur Samin
Thu Feb 23, 2017 6:09 pm
Forum: Combinatorics
Topic: Combi Solution Writing Threadie
Replies: 10
Views: 8096

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 ...
by Thanic Nur Samin
Thu Feb 23, 2017 3:19 pm
Forum: Combinatorics
Topic: Combi Marathon
Replies: 48
Views: 22238

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...
by Thanic Nur Samin
Thu Feb 23, 2017 2:58 pm
Forum: Combinatorics
Topic: Combi Marathon
Replies: 48
Views: 22238

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. ...
by Thanic Nur Samin
Tue Feb 21, 2017 11:33 pm
Forum: Algebra
Topic: ISL 2003 A1
Replies: 1
Views: 1016

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...
by Thanic Nur Samin
Tue Feb 21, 2017 10:58 am
Forum: Combinatorics
Topic: Combi Marathon
Replies: 48
Views: 22238

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.
by Thanic Nur Samin
Tue Feb 21, 2017 10:54 am
Forum: Combinatorics
Topic: Combi Marathon
Replies: 48
Views: 22238

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 ...
by Thanic Nur Samin
Mon Feb 20, 2017 6:45 pm
Forum: Number Theory
Topic: IMO Shortlist 2011 N5
Replies: 4
Views: 1868

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...
by Thanic Nur Samin
Mon Feb 20, 2017 6:30 pm
Forum: Number Theory
Topic: IMO Shortlist 2011 N5
Replies: 4
Views: 1868

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 $f(m)$.