Problem 5:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Source: অলিম্পিয়াড সমগ্র বই
Search found 138 matches
- Thu Nov 15, 2012 6:47 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 307590
- Tue Nov 13, 2012 10:31 am
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 307590
Re: Secondary and Higher Secondary Marathon
Phlembac Adib Hasan wrote: Problem 4: (Posted before)
If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are integers.
Source: Self-made.
Is it true ?
- Sun Nov 11, 2012 11:32 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 307590
Re: Secondary and Higher Secondary Marathon
Problem 2 Solution nona.jpg :arrow: $BC$ ও $FE$ কে বর্ধিত করি; তারা $M$ বিন্দুতে ছেদ করে। $EC$ and $FB$ are parallel with $HI$. We can observe that $CE=BD$ and $BF=BG$. Each angle of the nonagon is $(180-\frac{360}{9}) = 140^o = \angle EDC$ As $ED=CD$, $\angle DEC = \angle DCE = 20^o$ $\angle FEM$ ...
- Sat Nov 10, 2012 6:16 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 307590
Re: Secondary and Higher Secondary Marathon
Problem 2
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$
$ABCDEFGHI$ is a regular nonagon . show that $BG = BC + BD$
- Sat Nov 10, 2012 5:26 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 307590
Re: Secondary and Higher Secondary Marathon
Problem 1 Solution: Simplifying the expression we get, $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$ $=(m+n)(m-n)(m+2n)(m-2n)(m+3n)$ It has five factors. On the other hand, 33 has only two factors. If one factor or the product of two factors of the expression is 33, the other factors will be 1. But then...
- Sun Sep 16, 2012 4:14 pm
- Forum: Junior Level
- Topic: Equilateral Triangle
- Replies: 3
- Views: 3792
Equilateral Triangle
$ABCD$ is a square. $P$ is a point inside the square.
$\angle PAB=\angle PBA=15º$
Prove that $∆CPD$ is an equilateral triangle.
$\angle PAB=\angle PBA=15º$
Prove that $∆CPD$ is an equilateral triangle.
- Sun Sep 16, 2012 4:07 pm
- Forum: Combinatorics
- Topic: Fantasy Cricket
- Replies: 3
- Views: 3927
Re: Fantasy Cricket
I should mention it earlier.
I chose this serial. --Batsmen,Bowler,A.R , W.K and at last $12^{th}$ man--
I chose this serial. --Batsmen,Bowler,A.R , W.K and at last $12^{th}$ man--
- Fri Sep 14, 2012 9:52 pm
- Forum: Geometry
- Topic: SAMO 2012 Problem 2
- Replies: 4
- Views: 3108
Re: SAMO 2012 Problem 2
photon wrote: by the way what is SAMO?
SAMO means "South African Math Olympiad". It's a problem of SAMO, Senior Round 3.
- Fri Sep 14, 2012 4:09 pm
- Forum: Combinatorics
- Topic: Fantasy Cricket
- Replies: 3
- Views: 3927
Fantasy Cricket
You are playing a game on T20 World Cup. There are $12$ teams and each team has $5$ Batsmen, $5$ Bowlers, $4$ All-rounders and $1$ wicket keeper in their squad. You have to create a team of $12$ players; consists of $4$ Batsmen, $3$ Bowlers, $3$ A.R , $1$ W.K and a $12^{th}$ man.[The 12^{th} man can...
- Fri Sep 14, 2012 12:03 pm
- Forum: Secondary Level
- Topic: Number Theory Problems
- Replies: 4
- Views: 3453
Re: Number Theory Problems
Problem 2 can be solved in 2 ways. 1st way $n^2+3n+5$ is divisible by $11$, only when $n≡4(mod 11)$. We can write it as $11k+4$. Now $(11k+4)^2+3(11k+4)+5$ $\rightarrow 121k^2+88k+16+33k+12+5$ $\rightarrow 121k^2+121k+33$ Here $121$ doesn't divide $33$. So $n^2+3n+5$ is not divisible by $121$. 2nd w...