Search found 550 matches
- Sat Mar 23, 2013 11:45 pm
- Forum: Number Theory
- Topic: Pairs of integers
- Replies: 2
- Views: 2956
Pairs of integers
Find infinitely many pairs of integers $(a,b)$ such that $1<a<b$ and $ab$ divides $a^{2}+b^{2}-1$. Also find all positive integers $k$ such that there exists $(a,b)$ such that $\frac {a^{2}+b^{2}-1}{ab}=k$.
- Sat Mar 23, 2013 11:33 pm
- Forum: Geometry
- Topic: Medians and Side Lengths
- Replies: 2
- Views: 2338
Re: Medians and Side Lengths
pls give me the solution of this problem: Find the ratio of the area of a given tringle to that of a tringle whose sides have the same lengths as the medians of the original triangle. Let $ABC$ be the triangle and let $AD,BE,CF$ be its medians.Extend $FE$ to $M$ such that $FE=EM$.Join $A,M;C,M;D,M$...
- Sat Mar 23, 2013 11:09 pm
- Forum: Junior Level
- Topic: Solve this PLEASE!
- Replies: 5
- Views: 5138
Re: Solve this PLEASE!
If $a , b , c > 0$, prove that $\displaystyle \frac{9}{a + b + c} \leq 2(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ According to the AM-HM inequality,we know that,for positive real numbers $x_1,x_2,...,x_n$, $\displaystyle\frac{n}{\frac{1}{x_1...
- Thu Feb 28, 2013 11:11 am
- Forum: Algebra
- Topic: Vietnam 2012- Surjective function
- Replies: 4
- Views: 3470
Re: Vietnam 2012- Surjective function
The mistake in Zadid vai's solution was perhaps to write $a_n=ap^n+bq^n$. As far as I know the right form should be $a_n=pa^n+qb^n$ and this yields a valid solution $f(x)=4x$
- Tue Feb 26, 2013 11:35 pm
- Forum: Number Theory
- Topic: Divisors...
- Replies: 1
- Views: 2381
Divisors...
Find all positive integer $m$ such that $d(2m^{3})=2m$ where $d(x)$ means the number of divisors of $x$.
- Sun Feb 24, 2013 11:20 pm
- Forum: Algebra
- Topic: INEQUALITY PROBLEM
- Replies: 2
- Views: 2865
Re: INEQUALITY PROBLEM
Expand the RHS. The inequality will turn into $3abc+3\geq ab+bc+ca+a+b+c$.
Now we can again transform it into this: $0\geq ab-abc+c-1+bc-abc+a-1+ca-abc+b-1$
$\Rightarrow 0\geq (ab-1)(1-c)+(bc-1)(1-a)+(ca-1)(1-b)$
Which is absolutely true from the given condition.
Now we can again transform it into this: $0\geq ab-abc+c-1+bc-abc+a-1+ca-abc+b-1$
$\Rightarrow 0\geq (ab-1)(1-c)+(bc-1)(1-a)+(ca-1)(1-b)$
Which is absolutely true from the given condition.
- Mon Feb 18, 2013 11:28 pm
- Forum: Number Theory
- Topic: Help to prove!
- Replies: 5
- Views: 3643
Re: Help to prove!
অপেক্ষা করার কি দরকার ছিল? এখন দিয়ে দেন না।Masum wrote:এতদিন পরে দেইখা শান্তি লাগলো। এখন পোলাপান তাইলে এদের মাহাত্ম্য বুঝছে!!
খালি কি এইটাই? সাথে আর কি কি আছে? জানতে ইচ্ছা হইলো। আমি কয়েকটা সাজেশন দিতে পারি চাইলে।
- Fri Feb 15, 2013 11:08 pm
- Forum: Social Lounge
- Topic: Happy VALENTINE
- Replies: 4
- Views: 5752
Re: Happy VALENTINE
নাহ...আমার মনে হয় বেশিরভাগ।FahimFerdous wrote:Khali tui? :@
- Wed Feb 13, 2013 11:05 pm
- Forum: Secondary Level
- Topic: A theorem ..!!!..
- Replies: 1
- Views: 2416
Re: A theorem ..!!!..
This is actually the proposition that the number of twin primes are infinite. You may see here: http://en.wikipedia.org/wiki/Twin_primesfamim2011 wrote:Prove that There are infinitely many prime-pairs (p,p + 2).
- Wed Feb 06, 2013 10:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2010 higher secondary Q. 6
- Replies: 7
- Views: 6042
Re: BdMO 2010 higher secondary Q. 6
$a^2\equiv0^2,1^2,2^2,3^2,4^2\equiv0,1,-1,-1,1 (mod5)$ There are $804$ numbers of form $5n+2$ and $5n+3$ less then $2010$. There are $804$ numbers of form $5n+1$ and $5n+4$ less then $2010$. There are $401$ numbers of form $5n$ less then $2010$. So that number of pair $(a,b)$ be $2.804^2+401^2$. An...