Search found 186 matches
- Wed Feb 05, 2014 12:20 pm
- Forum: Combinatorics
- Topic: rotating a colored square
- Replies: 5
- Views: 6064
Re: rotating a colored aquare
you can rotate to any side , it doesn't matter for calculation .
- Mon Feb 03, 2014 10:28 pm
- Forum: Number Theory
- Topic: positive cube = $p^2-p-1$
- Replies: 7
- Views: 6158
Re: positive cube = $p^2-p-1$
Let $p^2 - p - 1 = a^3$. LHS is positive. So, $a \geq 1$. Now, $p^2 - p = a^3 + 1 \Leftrightarrow p(p - 1) = (a + 1)(a^2 + a + 1)$.......... It will be $p(p-1)=(a+1)(a^2-a+1)$ . for $a=1 , p=2 ; a=2$ has no result for prime $p$ . Following your way $a^2-a+1>a+1$ [for $a>2$]. Now not necessarily $a+...
- Mon Feb 03, 2014 1:03 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Higher Secondary 4
- Replies: 5
- Views: 5931
Re: BdMO National 2013: Higher Secondary 4
$\frac {11}{6} $ is the answer , I had found it by trial and error method .
- Sun Feb 02, 2014 9:37 pm
- Forum: Number Theory
- Topic: 2 quadratic equations
- Replies: 2
- Views: 2155
Re: 2 quadratic equations
I couldn't solve it yet , good approach ... you see $(x-m)(x-n)=0 \Rightarrow x^2-(m+n)x+mn=0$ - for $m,n$ as solution the coefficient of $x$ is $-(m+n)$ . Hence for $a=2p$ , the solutions should be $-p-q,-p+q$ of the first equation. Though it didn't influence on $b$ 's magnitude . For even number $...
- Mon Jan 27, 2014 9:30 pm
- Forum: Number Theory
- Topic: 2 quadratic equations
- Replies: 2
- Views: 2155
2 quadratic equations
Show that there are infinitely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both the equations $x^2 + ax + b=0 , x^2+2ax+b=0$ have integer roots.
[ Source-INMO-1995 ]
[ Source-INMO-1995 ]
- Tue Jan 14, 2014 2:25 pm
- Forum: Divisional Math Olympiad
- Topic: Barisal Secondary 2013 / 8
- Replies: 12
- Views: 8659
Re: Barisal Secondary 2013 / 8
I solved it before , so giving a hint -
- Thu Jan 02, 2014 8:53 pm
- Forum: Number Theory
- Topic: positive cube = $p^2-p-1$
- Replies: 7
- Views: 6158
positive cube = $p^2-p-1$
Find all prime numbers $p$ such that the number $p^2-p-1$ is a cube of some positive integer .
- Sat Dec 21, 2013 11:13 pm
- Forum: Geometry
- Topic: relation among radiuses
- Replies: 0
- Views: 1578
relation among radiuses
The incircle of triangle $ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively . The circle passing through $A$ and touching the line $BC$ at $D$ intersects $BF$ and $CE$ at points $K$ and $L$ respectively . The line passing through $E$ and parallel to $DL$ and the line passing through $F$ and parallel t...
- Sat Oct 12, 2013 5:26 pm
- Forum: Higher Secondary Level
- Topic: dividing $x^{2013}-1$
- Replies: 7
- Views: 6874
dividing $x^{2013}-1$
Find the remainder on dividing $x^{2013} - 1$ by $(x^2+1)(x^2+x+1)$ .
- Sun Sep 29, 2013 10:23 pm
- Forum: Geometry
- Topic: Where I visualize cyclic ness?
- Replies: 4
- Views: 4421
Re: Where I visualize cyclic ness?
$\angle RQE = \angle QEP = \angle AEF = \angle ABC$ $\angle RQC = \angle RBC$ , $B,R,C,Q$ are cycic . $\therefore RD.DQ=BD.DC$ Let M be the midpoint of $BC$ . $M,D,E,F$ lie in nine-point circle . so , $PE.PF=PD.PM$ $\displaystyle \Rightarrow PC.PB = PD.PM$ [$PE.PF=PC.PB$ as $B,F,E,C$ are cyclic] $\d...