Post the problem.Ishmam! wrote:BANGLADESH GONIT OLYMPIAD(JUNIOR) ER CTG 2013 ER 9 NO PARTISINA . KEO KI HELP KORBEN?
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- Wed Jan 15, 2014 7:42 pm
- Forum: Junior Level
- Topic: REGIONAL OLYMPIAD PROBLEM
- Replies: 4
- Views: 4070
Re: REGIONAL OLYMPIAD PROBLEM
- Wed Jan 15, 2014 7:38 pm
- Forum: Divisional Math Olympiad
- Topic: Barisal Secondary 2013 / 8
- Replies: 12
- Views: 8669
Re: Barisal Secondary 2013 / 8
What is Jensen's inequality?What is convex and concave function?
- Mon Jan 13, 2014 4:45 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Higher Secondary 2011
- Replies: 2
- Views: 2332
Higher Secondary 2011
Let $f$ be an injective function and $f: \mathbb R^{+} \mapsto \mathbb R^{+}$. $f(1)=2$ and $f(x+\frac{1}{f(y)})=\frac{f(x)f(y)}{f(x)+f(y)}$ for all positive real $x$ and $y$. Find $f(2012)$.
- Mon Jan 13, 2014 4:33 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 8, Higher Secondary 6
- Replies: 8
- Views: 12353
Re: BdMO National 2013: Secondary 8, Higher Secondary 6
We induct on n. The result is trivial for $n=1,2$. We assume that it is true for $n=m$. Now we prove that it is true for $n=m+1$. Among $m+1$ cities, if there exists a city which is connected with 0 or 1 city, then there are $m+1$ or m roads among the rest m cities.But by our induction hypothesis, t...
- Thu Jan 09, 2014 7:24 pm
- Forum: Number Theory
- Topic: function (USAMO 2012, Balkan 2012, IMO proposal 2010)
- Replies: 3
- Views: 3968
Re: function (USAMO 2012, Balkan 2012, IMO proposal 2010)
@asif e elahi You have left out the solutions: $f(n)\equiv 1 \forall n\in \mathbb{N}, f(n)\equiv 2 \forall n\in \mathbb{N}$. And I can't understand how you deduced $f(p-2)=p-2$. :oops: Could you please make it clear? :oops: By Wilson's theorem $p\mid (p-2)!-1\mid f(p-2)!-f(1)$. If $f(p-2)>p-1$ then...
- Thu Dec 26, 2013 6:37 pm
- Forum: Number Theory
- Topic: function (USAMO 2012, Balkan 2012, IMO proposal 2010)
- Replies: 3
- Views: 3968
Re: function (USAMO 2012, Balkan 2012, IMO proposal 2010)
$f(1)=f(1!)=f(1)!$.So $f(1)=1$ or $2$. By Wilson's theorem,for a prime $p$ $(p-1)!\equiv -1\equiv p-1(mod p)$ or $(p-2)!\equiv 1(mod p)$. So $p\mid (p-2)!-1\mid f((p-2)!)-f(1)=f(p-2)!-(1 or 2)$.So $gcd(p,f(p-2)!)=1$. This implies $f(p-2)\leq p-1$.Again $(p-2)!-1\mid f(p-2)!-f(1)$. So $f(p-2)!=(p-2)!...
- Sat Dec 21, 2013 9:26 pm
- Forum: Algebra
- Topic: AN INTERESTING PROBLEM BY SAKAL DA
- Replies: 5
- Views: 4392
Re: AN INTERESTING PROBLEM BY SAKAL DA
How to solve thus recurrent relations?zadid xcalibured wrote:Let $b_n=a_n-2012$ notice that $a_n$ depends totally on $a_{n-1}$.$a_{n-1}-2012=a_n(a_n-2012)$
So $b_{n+1}=b_n(b_n+2012)$
Then the general formula can be established using induction.I see no better way right now.
- Fri Dec 20, 2013 8:41 pm
- Forum: Higher Secondary Level
- Topic: infinitely many primes 1 mod p
- Replies: 2
- Views: 3774
Re: infinitely many primes 1 mod p
Here is a sequence of steps through which you can find infinitely many primes congruent to $1\bmod p$ for any prime $p$. (Of course, without having to use Dirichlet's theorem!) A. Let $a>1$ be an integer. B. Let $N=a^p-1$. C. What is the order of $a\bmod N$? D. This order must divide $\phi(N)$, whe...
- Fri Dec 20, 2013 1:53 pm
- Forum: Number Theory
- Topic: Family Of Functions
- Replies: 5
- Views: 4684
Re: Family Of Functions
Why f(0)=0 ?*Mahi* wrote:
If $f(x)$ is a solution to the given equation, then so is $f(x)+c$, so WLOG let $f(0)=0$.
- Sun Dec 15, 2013 8:56 pm
- Forum: Number Theory
- Topic: IMO Shortlist 2006 N7
- Replies: 1
- Views: 1987
IMO Shortlist 2006 N7
For all positive integers $n$, show that there exists a positive integer $m$ such that $n$ divides $2^{m} + m$.