Search found 238 matches
- Fri Feb 04, 2011 12:33 pm
- Forum: Introductions
- Topic: Username Changed(hasib.mo--Hasib)
- Replies: 1
- Views: 2658
Username Changed(hasib.mo--Hasib)
Hi, i'm not new in the forum. I am a regular user since u all saw me as "hasib.mo". But, i was feeling uneasy with this username! So, i requested the moderator bench to change my username to "Hasib". Now feeling comfortable :) again, saying hello to u all with this new username! So, dont confused i ...
- Fri Feb 04, 2011 12:39 am
- Forum: Divisional Math Olympiad
- Topic: Barishal higher sec 10
- Replies: 9
- Views: 5271
Re: Barishal higher sec 10
wheres from 'I' comes in FEI and EFI?
- Fri Feb 04, 2011 12:31 am
- Forum: Divisional Math Olympiad
- Topic: Gopalgang sec 2011
- Replies: 2
- Views: 2406
Re: Gopalgang sec 2011
Input x=79, y =1.
You will have f(79)=79*25
You will have f(79)=79*25
- Thu Feb 03, 2011 10:33 pm
- Forum: News / Announcements
- Topic: Stall for voting Sundarban in BdMO.
- Replies: 5
- Views: 4728
Re: Stall for voting Sundarban in BdMO.
lolz!!!! i requested moon vai infront of u! Forgot?Tahmid Hasan wrote:hey r u hasib from khulna?
ur username is different from b4!!!!!!!
what happened?
- Thu Feb 03, 2011 9:40 pm
- Forum: News / Announcements
- Topic: Stall for voting Sundarban in BdMO.
- Replies: 5
- Views: 4728
Re: Stall for voting Sundarban in BdMO.
i amnt sure it also!
- Thu Feb 03, 2011 1:19 pm
- Forum: News / Announcements
- Topic: Stall for voting Sundarban in BdMO.
- Replies: 5
- Views: 4728
Stall for voting Sundarban in BdMO.
Firstly i am sorry for introducing the matter in news forum. Yes. It's needed to make a voting stall for Sundarban in BdMO. Sometimes, for some simple cases, people couldn't vote for Sundarban. Such as, i use mobile to use internet. I tried to vote for Sundarban, but the process is so though to vote...
- Wed Feb 02, 2011 6:50 pm
- Forum: Secondary: Solved
- Topic: Rangpur Secondary 2011/1
- Replies: 4
- Views: 9448
Re: Rangpur Secondary 2011/1
$(x-40)+(x-39)....+x+....(x+40)=9^5$
so, $81x=9^5$
so, $x=9^3$
median is $9^3$
so, $81x=9^5$
so, $x=9^3$
median is $9^3$
- Wed Feb 02, 2011 6:32 pm
- Forum: Combinatorics
- Topic: Moon vai iz gonna to hack my acount :x :x
- Replies: 11
- Views: 8970
Re: Moon vai iz gonna to hack my acount :x :x
yap! I thought zubaer vai did $\frac{1}{10^{10}}\times 25$
but, i was wrong! But, lets me know, why $\text{2nd times success probability}=\text{1st times unsuccess probability}\times \text{ 2nd time's success probability}$?
but, i was wrong! But, lets me know, why $\text{2nd times success probability}=\text{1st times unsuccess probability}\times \text{ 2nd time's success probability}$?
- Wed Feb 02, 2011 12:36 am
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2009/1
- Replies: 7
- Views: 11614
Re: Dhaka Secondary 2009/1
$[y][x]$ is a prime. So $x$ is odd number. So, $x-2$ is also odd number.
So how its possible that $[\frac{x-2}{2}]$ be a digit.
I assume that $[x][y]$ represent $\overline{xy}$. Is my assume correct? Or, i will try again.
So how its possible that $[\frac{x-2}{2}]$ be a digit.
I assume that $[x][y]$ represent $\overline{xy}$. Is my assume correct? Or, i will try again.
- Tue Feb 01, 2011 11:34 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/5
- Replies: 2
- Views: 8283
Re: Dhaka Higher Secondary 2010/5
$11x^36-21x^10+26x^2=48$
or, $x^2(11x^34-21x^8+26)=48$
so, $x^2|48$
hence, $48=2^4 \times 3$
so, we can only have a square
number(4) which divides 48.(N.B yet dont know if it is a solution)
But, $x>5$.
So, there is no solution!
or, $x^2(11x^34-21x^8+26)=48$
so, $x^2|48$
hence, $48=2^4 \times 3$
so, we can only have a square
number(4) which divides 48.(N.B yet dont know if it is a solution)
But, $x>5$.
So, there is no solution!