## Search found 75 matches

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 $$n$$.
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 $$\sqrt[3]{3}$$
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
Sat Aug 30, 2014 5:51 pm
Forum: Physics
Replies: 0
Views: 926

Tue Aug 26, 2014 12:20 pm
Forum: Number Theory
Topic: Indonesia, 2013
Replies: 2
Views: 549

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 $$n = 2015... Tue Aug 26, 2014 11:18 am Forum: Geometry Topic: I can see a concurrency(Self-made) Replies: 1 Views: 588 ### Re: I can see a concurrency(Self-made)$$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...
Mon Aug 25, 2014 7:01 pm
Forum: Combinatorics
Topic: Prove S(n,m)=S(m,n)
Replies: 3
Views: 772

$$\begin{eqnarray*} \\ && S(m,n)= \sum_{i=0}^{n-1} (-1)^i \dbinom{n}{i} (2^{n-i}-1)^m \cdots(1) \end{eqnarray*}$$ In the expression of (1), $$\begin{eqnarray*} \\ &&\binom{n}{0} (-1)^m \binom {m}{m}(2^n)^0- \binom{n}{1} (-1)^m \binom {m}{m}(2^n)^0+\binom{n}{2} (-1)^m \binom {m}{m}(2^n)^0 \\ && +\cdo... Sat Aug 23, 2014 7:15 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2014 - Day 1 Problem 1 Replies: 2 Views: 882 ### Re: IMO 2014 - Day 1 Problem 1$$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}. $... Mon Mar 03, 2014 5:16 pm Forum: Junior Level Topic: RMO-2010/3 Replies: 5 Views: 1264 ### Re: RMO-2010/3 Nirjhor wrote:Since,$[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
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 । এইটা ফোরাম , এইটা প্রবলেম নিয়া আলোচোনা করার জায়গা । এইসসব আবেগময় পোস্ট দেয়ার জায়গা না । আবেগের কথা বলতে হইলে ফেসবুকে যাও
কিরীটী , এইটা introduction category. এইটা ফেসবুক স্ট্যাটাস দেওয়ার মতই জাইগা