Search found 411 matches
- Sat Feb 08, 2014 7:10 pm
- Forum: Primary Level
- Topic: Easy sum
- Replies: 3
- Views: 48713
Re: Easy sum
@Asif Please do elaborate how you eliminated other possible answers when posting solutions like these. Here is a explanation to why the only possible answer is $125$. Notice that - $1000 = 2^3*5^3$. Assume that one of the two factors is $a = 2^n*5^m$ where $0\leq n,m\leq 3$. If $n \geq 1$ and $m \ge...
- Sat Feb 08, 2014 6:56 pm
- Forum: Secondary Level
- Topic: Fraction
- Replies: 2
- Views: 2885
Re: Fraction
The solution is correct. Just one correction needed - the fraction is $\frac 1 2$, not $2$.
You'll get my point if you read the statement carefully. Nice job with the solution though.
You'll get my point if you read the statement carefully. Nice job with the solution though.
- Sat Feb 08, 2014 6:48 pm
- Forum: Junior Level
- Topic: Confusing geometry
- Replies: 4
- Views: 4138
Re: Confusing geometry
@Thanic Nur Samin Please post full solution or hints with the answer.
@Raiyan I do not know about others but I am having difficulty viewing the image properly.
If you it is possible, could you upload the image in a better format like .png or .jpeg ?
You can use Geogebra or Paint to get this job done.
@Raiyan I do not know about others but I am having difficulty viewing the image properly.
If you it is possible, could you upload the image in a better format like .png or .jpeg ?
You can use Geogebra or Paint to get this job done.
- Sat Feb 08, 2014 5:53 pm
- Forum: Junior Level
- Topic: Monster problem
- Replies: 2
- Views: 2907
Re: Monster problem
@Thanic nur Samin, Please practise posting full solutions or at least hints for your solution unless asked explicitly for just the answer to the problem. Because most often, people asking for help are actually asking for ideas/keys to solve the problem and not just the answer. Here's a possible way ...
- Mon Feb 03, 2014 11:40 am
- Forum: News / Announcements
- Topic: Divisional Informatics Olympiad 2014 : Registration Open!!
- Replies: 1
- Views: 8740
Re: Divisional Informatics Olympiad 2014 : Registration Open
We have an IOI sub-forum in this forum as well!
You can ask us questions/ for help over there as well!
You can ask us questions/ for help over there as well!
- Mon Feb 03, 2014 11:34 am
- Forum: News / Announcements
- Topic: Divisional Informatics Olympiad 2014 : Registration Open!!
- Replies: 1
- Views: 8740
Divisional Informatics Olympiad 2014 : Registration Open!!
1601474_10152147731033788_1065518367_n.jpg Hi everyone! I'm glad to announce that the registration process for Divisional Informatics Olympiad 2014 has finally started. Anybody eligible for this competition can register by going to the following link: REGISTRATION If you are thinking what this olym...
- Sat Feb 01, 2014 10:21 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 6
- Replies: 12
- Views: 8874
Re: BdMO National 2013: Junior 6
Please use Bangla or English as your language.
'Banglish' is not appreciated (and most probably prohibited) in this forum.
'Banglish' is not appreciated (and most probably prohibited) in this forum.
- Fri Jan 31, 2014 2:31 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 6
- Replies: 12
- Views: 8874
Re: BdMO National 2013: Junior 6
I'm not sure if I can make it sound any easier than this Atiab. :-( I'd leave it to someone else who could come up with an easier solution. Meanwhile, I'd give you a heads up. Please do not use "Banglish" in the forum. (I believe it is prohibited) If you wish to right bangla, please use Avro/some ot...
- Wed Jan 29, 2014 5:31 pm
- Forum: Divisional Math Olympiad
- Topic: Re: Dhaka-2 Higher Secondary 2013 / 8
- Replies: 3
- Views: 2879
Re: Re: Dhaka-2 Higher Secondary 2013 / 8
I just copied the problem from the problem set.
If you think the statement is wrong, why don't you post your solution?
It will help everyone.
If you think the statement is wrong, why don't you post your solution?
It will help everyone.
- Wed Jan 29, 2014 5:26 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Junior 2010/2
- Replies: 1
- Views: 1849
Re: Junior 2010/2
Here's a solution: Let's assume that the rectangle has a base of length $B$ and a height of length $H$ and the square has sides of length $A$. Now, $BH = A^2 \Rightarrow B = \frac {A^2}H$ We know, the square of a real number cannot be negative. So, $(A-H)^2 = A^2+H^2-2AH \geq 0 $ $\Rightarrow A^2 + ...