Search found 16 matches

by Cryptic.shohag
Thu Feb 16, 2012 6:01 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National 2012: Higher Secondary 02
Replies: 11
Views: 4800

Re: BdMO National 2012: Higher Secondary 02

Number of ways when he doesn't land is 1. Number of ways when he lands once is 11C0. Number of ways when he lands twice is 11C1. Number of ways when he lands thrice is 11C2. Thus for 4,5,.....,11 landings the number of ways would be 11C3, 11C4,......, 11C10 respectively. So, in total the number of w...
by Cryptic.shohag
Fri Feb 18, 2011 3:23 am
Forum: Social Lounge
Topic: Mathematical Jokes
Replies: 15
Views: 6116

Re: Mathematical Jokes

Coz if the 2nd differential is negative u'll get the highest peek value.... :)
by Cryptic.shohag
Thu Feb 17, 2011 3:34 pm
Forum: Algebra
Topic: functional equation australia
Replies: 9
Views: 2633

Re: functional equation australia

Solution for both of the functions should be $f(x)=0$
by Cryptic.shohag
Thu Feb 17, 2011 2:41 pm
Forum: National Math Camp
Topic: Problems of the day (Day 1)
Replies: 9
Views: 2986

Re: Problems of the day (Day 1)

Opera 4 diyei dkha jay. Tomar ki image off kora naki?
Jodi ta na hoy tahole web address lekhar jaygatay opera:config likhe click koro. Then last option ta no theke yes kore save koro. R jodi tateo na hoy tahole Allah tomar shohay houn....
by Cryptic.shohag
Thu Feb 10, 2011 2:29 am
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Higher Secondary 2007/8
Replies: 5
Views: 2172

Re: BdMO National Higher Secondary 2007/8

Let O be the center of the circle and AL, BM, CN be the 3 chords such that AL=14, BM=\[\sqrt{a}\] and CN=10. Let OD, OE, OF be perpendiculars on AL, BM, CN respectively. Let OD=P, then OE=P+3 and OF=P+6[considering that they are all on the same side of the diameter]. As OD, OE, OF are perpendiculars...
by Cryptic.shohag
Tue Feb 08, 2011 5:36 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Higher Secondary 2007/4
Replies: 5
Views: 2194

Re: BdMO National Higher Secondary 2007/4

Actually my logic was if a and b r back to back common integers then $(a,b)=1$ and ab can't be a square. Is there any problem in my logic moon vaia? If so, plz give me an example....
by Cryptic.shohag
Tue Feb 08, 2011 12:54 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Higher Secondary 2007/14
Replies: 5
Views: 2463

Re: BdMO National Higher Secondary 2007/14

Oops!! I missed one solution. In last solution I considered only $a,b>0$....
As $a>b$ and $p^a-p^b=24$, a's value cannot be 0. So, if we consider $b=0$, then we get,
$p^a-1=24$
$p^a=25$
$p^a=5^2$....
So, another value of p is 5....
by Cryptic.shohag
Tue Feb 08, 2011 12:29 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Higher Secondary 2007/14
Replies: 5
Views: 2463

Re: BdMO National Higher Secondary 2007/14

let's name the eqs $(i)$ and $(ii)$ and do $(i)-(ii)$ than divide three cases 1.$a>b$ 2.$a<b$ 3.$a=b$ Actually it's not necessary as it's clear that $a>b$. $m+12=p^a$....(i) $m-12=p^b$....(ii) So, $p^a-p^b=24$. As $p^a$ and $p^b$ both are multiples of p, we can write $p^a-p^b=kp$ where k is an inte...
by Cryptic.shohag
Tue Feb 08, 2011 12:12 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Higher Secondary 2007/4
Replies: 5
Views: 2194

Re: BdMO National Higher Secondary 2007/4

Masum wrote:There is no common factor between $2^2,3^2$ but their product is $6^2$
$2^2,3^2$ are not back to back integers but n,(n+1) are. Putting square makes difference here....