Search found 155 matches
- Thu Jul 05, 2012 6:41 pm
- Forum: Physics
- Topic: Congrats, CERN :)
- Replies: 2
- Views: 3067
Congrats, CERN :)
On behalf of the entire BDMO Forum, I heartily congratulate all the hard-working people at CERN for making one of the most important discoveries of this era- Discovering the "Higg's Boson" . :) :) :D They have set another milestone in unraveling the mysteries of nature. Hats off ...!! :mrgreen:
Re: নিরুদক
ভাইয়া, আমার মনে হয় পরীক্ষণ বা Experimentation ছাড়া জানার কন সহজ উপায় নেই।
- Tue Jul 03, 2012 2:27 pm
- Forum: Algebra
- Topic: Another Problem on Flooring
- Replies: 3
- Views: 2443
Re: Another Problem on Flooring
Intuitions:
- Mon Jun 18, 2012 6:10 pm
- Forum: Algebra
- Topic: Another Problem on Flooring
- Replies: 3
- Views: 2443
Re: Another Problem on Flooring
Well, I could not go much far with it. :? I only fiddled with it a little. :roll: WLOG, we can suppose that $n$ is Odd . And, we can write the expression in this manner: \[\left \lfloor (n/2^1)+\frac{1}{2} \right \rfloor+\left \lfloor (n/2^2)+\frac{1}{2} \right \rfloor+\left \lfloor (n/2^3)+\frac{1}...
- Mon Jun 18, 2012 6:01 pm
- Forum: Algebra
- Topic: Another Problem on Flooring
- Replies: 3
- Views: 2443
Another Problem on Flooring
Prove that,\[\left \lfloor \frac{n+2^0}{2^1} \right \rfloor+\left \lfloor \frac{n+2^1}{2^2} \right \rfloor+\left \lfloor \frac{n+2^2}{2^3} \right \rfloor+.............+\left \lfloor \frac{n+2^{n-1}}{2^n} \right \rfloor=n\]
\[\forall n\in \mathbb{N}\]
\[\forall n\in \mathbb{N}\]
- Thu Jun 14, 2012 6:51 pm
- Forum: Algebra
- Topic: Flooring Square-Roots
- Replies: 3
- Views: 2708
Re: Flooring Square-Roots
Finally, a breakthrough!...... :D (after a hideous amount of struggle :P ): First, let, $a^2 \leq n<(a+1)^2$, as before. Now, suppose, $n=a^2+z$, and furthermore, $\sqrt{n}=a+f_1$ and $\sqrt{n+1}=a+f_2$, where, $f_1$ and $f_2$ are the fractional parts of $\sqrt{n}$ and $\sqrt{n+1}$ respectively. Cas...
- Tue Jun 12, 2012 10:00 pm
- Forum: Algebra
- Topic: Flooring Square-Roots
- Replies: 3
- Views: 2708
Re: Flooring Square-Roots
Partial Solution: Let, $a^2\leq{n}<(a+1)^2$ Then, Case 1: $n=a^2$. So, $\sqrt{n}=a$, and, $\sqrt{n+1}=a+f$, where, $0<f<1$. Therefore, \[\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor=\left \lfloor a+a+f \right \rfloor=2a\] On the other hand, ${4n+2}=4a^2+2=(2a)^2+2$. Here, $(2a)^2+2$ is surely no...
- Tue Jun 12, 2012 5:26 pm
- Forum: Algebra
- Topic: Flooring Square-Roots
- Replies: 3
- Views: 2708
Flooring Square-Roots
Prove that, \[\displaystyle \left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor\]\[\forall n \epsilon \mathbb{N}\]
- Sun Jun 10, 2012 10:55 pm
- Forum: Computer Science
- Topic: Hacking and Cracking
- Replies: 7
- Views: 6339
Re: Hacking and Cracking
Sorry, Sorry, Sorry................ :oops: :? :oops: এতদিন ইন্টারনেট ছিল না বলে ফোরামে বসতে পারি নি। আর, সবাইকে ভুল ধরিয়ে দেওয়ার জন্য ধন্যবাদ। আমি প্রথম আলোর Article-টা অনেক আগে পরেছিলাম। এজন্য সঠিক তথ্যটা দিতে পারি নাই। এর জন্য সবার কাছে আমি আবারও দুঃখিত। :( :cry: @আদীবঃ ভাই, ওই কথাটার জন্য মাইন্...
- Fri Jun 01, 2012 12:55 am
- Forum: Combinatorics
- Topic: Comby for robbers
- Replies: 3
- Views: 2997
Re: Comby for robbers
Hey, where's Sanzeed ? Please post the solution of the problem.