## Search found 16 matches

- 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...

- 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....

- 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 $f(x)=0$

- 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....

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....

- 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...

- 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....

- Tue Feb 08, 2011 12:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2007/14
- Replies:
**5** - Views:
**2463**

### Re: BdMO National Higher Secondary 2007/14

So, p=2,3,5....

- 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....

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....

- 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...

- 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

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