Search found 75 matches

by Fatin Farhan
Fri Sep 05, 2014 5:15 pm
Forum: Number Theory
Topic: LTE flavoured divisibility
Replies: 5
Views: 888

Re: LTE flavoured divisibility

$$7^n \mid 9^n-1 \Rightarrow 7 \mid 9^n-1$$.
But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.
by Fatin Farhan
Fri Sep 05, 2014 4:06 pm
Forum: Secondary Level
Topic: X's and 3
Replies: 8
Views: 1399

Re: X's and 3

Here we have finitely many $$x's$$. But where's the prove
No matter how many x's are there, the answer would be $$\sqrt[3]{3}$$
:?
by Fatin Farhan
Wed Sep 03, 2014 1:47 pm
Forum: Secondary Level
Topic: X's and 3
Replies: 8
Views: 1399

X's and 3

what is the largest real value of $$x$$ such that
$$x^{x^{x^{x^{\cdots x^{x^3}}}}}=3$$
Hint
$$f(x)=x \times ln (x)$$ is negative for $$x<1$$
by Fatin Farhan
Tue Aug 26, 2014 12:20 pm
Forum: Number Theory
Topic: Indonesia, 2013
Replies: 2
Views: 549

Indonesia, 2013

An integer $$n$$ is called $$\text{"elephantine"}$$ if there exists a positive integer $$x$$ such that $$x^{nx} + 1$$ is divisible by $$2^n$$. $$1. \text{ Prove } 2015 \text{ is Elephantine. }$$ $$2. \text{ Find the smallest } x\text{ such that }x^{nx} + 1\text{ is divisible by }2^n$$ for $$n = 2015...
by Fatin Farhan
Tue Aug 26, 2014 11:18 am
Forum: Geometry
Topic: I can see a concurrency(Self-made)
Replies: 1
Views: 588

Re: I can see a concurrency(Self-made)

$$AB,BC,CA$$ are tangent to the three circles $$\Rightarrow BX^2+CY^2+AZ^2=XC^2+YA^2+ZB^2$$ Now using the facts, $$(a) X,Y,Z \text{ are points on the sides } BC,CA,AB \text{ of } \triangle ABC.\text{ Then the perpendiculars to}$$ $$\text{ the sides at these points meet in a common point } P \text{ i...
by Fatin Farhan
Mon Aug 25, 2014 7:01 pm
Forum: Combinatorics
Topic: Prove S(n,m)=S(m,n)
Replies: 3
Views: 772

Re: Prove S(n,m)=S(m,n)

$$\begin{eqnarray*} \\ && S(m,n)= \sum_{i=0}^{n-1} (-1)^i \dbinom{n}{i} (2^{n-i}-1)^m \cdots(1) \end{eqnarray*}$$ In the expression of (1), $$\begin{eqnarray*} \\ &&\binom{n}{0} (-1)^m \binom {m}{m}(2^n)^0- \binom{n}{1} (-1)^m \binom {m}{m}(2^n)^0+\binom{n}{2} (-1)^m \binom {m}{m}(2^n)^0 \\ && +\cdo...
by Fatin Farhan
Sat Aug 23, 2014 7:15 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2014 - Day 1 Problem 1
Replies: 2
Views: 882

Re: IMO 2014 - Day 1 Problem 1

$$a_n < \dfrac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}$$ $$na_n < a_0+a_1+a_2+\cdots+a_n \leq na_{n+1}.$$ For uniqueness: Let $$n={n_1,n_2,....n_k}$$ satisfies $$a_0+a_1+a_2+\cdots+a_n \leq na_{n+1}$$ where $$n_1<n_2<\cdots<n_k$$. Let $$n_1+m_i=n_i$$ $$ $$ Now, $$\sum_{k=0}^n a_k\leq n_1a_{n+1}$$. $...
by Fatin Farhan
Mon Mar 03, 2014 5:16 pm
Forum: Junior Level
Topic: RMO-2010/3
Replies: 5
Views: 1264

Re: RMO-2010/3

Nirjhor wrote:Since, $[4,8]=8$, by PIE, the answer is: \[\left.\left(\left\lfloor \dfrac{9999}{4} \right\rfloor-\left\lfloor \dfrac{999}{4} \right\rfloor\right)\right/2~=~\dfrac{2250}2~ = ~\boxed{1125}\]
If you kindly explain what you did. Because the answer is incorrect :?
by Fatin Farhan
Sun Mar 02, 2014 4:38 pm
Forum: Introductions
Topic: Short Composition : Myself
Replies: 6
Views: 1859

Re: Short Composition : Myself

Kiriti wrote:ভাই, এইটা কি তুমি ফেসবুক পাইছো যে ইচ্ছা মত এইসব স্ট্যাটাস দিবা :shock: । এইটা ফোরাম , এইটা প্রবলেম নিয়া আলোচোনা করার জায়গা । এইসসব আবেগময় পোস্ট দেয়ার জায়গা না । আবেগের কথা বলতে হইলে ফেসবুকে যাও :P ;)
কিরীটী , এইটা introduction category. এইটা ফেসবুক স্ট্যাটাস দেওয়ার মতই জাইগা :? :? :?