## Search found 53 matches

Fri Aug 26, 2011 10:38 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/9
Replies: 10
Views: 8499

### Re: Dhaka Secondary 2009/9

LOL.......
Fri Aug 26, 2011 5:14 pm
Forum: Junior Level
Topic: comparism
Replies: 16
Views: 5539

### Re: comparism

suppose I am not sure about the solution of this one
which one is greater $100^{300}$ or $300!$ (neurone abaro onuronon problem)
I tried inthe same method but not sure of that one..
Fri Aug 26, 2011 4:57 pm
Forum: Junior Level
Topic: comparism
Replies: 16
Views: 5539

### Re: comparism

Yes Here is the wayt o solve it you see always $(x+a)(x-a)<x^2$ (simple) factorial 211 can be explained like this.., $1.2.3.4.5..........211$ or $(1.211).(2.210).(3.209)...........(105.107).106$ as mentioned above $(105.107)<106^2$ $(104.108)<106^2$ ........................ ............................
Thu Aug 25, 2011 6:15 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/9
Replies: 10
Views: 8499

### Re: Dhaka Secondary 2009/9

I think the ans is 0, since the sick player is absent none will be able to play with him, so no other player will play full 100 matches, they all ahve to play 99, isnt it???
Thu Aug 25, 2011 4:50 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/12
Replies: 3
Views: 4565

### Re: Dhaka Secondary 2009/12

the answer is $\sqrt{5}$ here is how I got it $AF=AE=1$ from F, BC er upor FM and from G, AB er upor duita lombo (:P) eke dekhano jay je, $DG=3$, so $EF=\sqrt{2}$ and $GF= 3.\sqrt{2}$ so $EG= 2.\sqrt{5}$ as $EFG$ right angled, so the midpoint of EG will be the centre of it's circumcircle. :?: :?:
Thu Aug 25, 2011 4:34 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/1
Replies: 7
Views: 6086

### Re: Dhaka Secondary 2009/1

Moon vai er solution eo deklam, but why??? please describe....
Thu Aug 25, 2011 4:25 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/11
Replies: 4
Views: 2964

### Re: Dhaka Secondary 2009/11

darun moja pailam eta solve kore here it is:

$4^{502}= 2^{1004}$ and $4^{\frac{2009}{2}}= 2^{2009}$

so the equation is $(2.2^{1004})^2-2^{2009}= 2^k$
or $2^{2010}-2^{2009}= 2^k$
or$2^{2009}(2-1)= 2^k$
so $k=2009$

Thu Aug 25, 2011 4:01 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/6
Replies: 7
Views: 4050

### Re: Dhaka Secondary 2009/6

Yes the answer is correct, but I could not do such multiplications in the hall, so here is another solution frm me $(1!)^3\equiv 1(mod 10)$ $(2!)^3= 2^3\equiv 8(mod 10)$ $(3!)^3=6^3\equiv 6(mod 10)$( 6 last digit amon number er jekono power last digit 6) $(4!)^3=24^3\equiv 4(mod 10)$( 4 last digit h...
Thu Aug 25, 2011 3:18 pm
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/1
Replies: 7
Views: 6086

### Re: Dhaka Secondary 2009/1

But akhon to baki part ta ami buztasi na, x+y=2(x-y) er por ki korlen??
I mean x=3y but what about the logic behind 13 and 31 s?? :
Thu Aug 25, 2011 4:12 am
Forum: Secondary: Solved
Topic: Dhaka Secondary 2009/1
Replies: 7
Views: 6086