Search found 73 matches
- Fri Feb 09, 2018 10:23 pm
- Forum: Number Theory
- Topic: Prime numbers
- Replies: 4
- Views: 6870
Re: Prime numbers
Basically the solution may be this kind of: $n$ is a natural prime number. So, it can be either an odd number or the only even number ,$2$ For, $n=2$, we get: $(n-6)^2+1=(2-6)^2+1=(-4)^2+1=17$; which is a prime. But for, $n$ =odd , $n-6$= odd , $(n-6)^2$= odd , but $(n-6)^2+1$= even The only even pr...
- Fri Feb 09, 2018 8:53 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2008/1
- Replies: 2
- Views: 2974
Re: BdMO National Higher Secondary 2008/1
Sum of the first $2008$ even positive integers=$2+4+6+...+n$ Sum of the first $2008$ odd positive integers=$1+3+5+7+...+n$ [Where $n$ denotes the $2008$-th term of the sequence] Difference between the first terms of two sequence=$1$ Difference between the second terms of two sequence=$1$ So,Differen...
- Thu Feb 08, 2018 8:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO Secondary P8
- Replies: 4
- Views: 4255
Re: National BDMO Secondary P8
Please help to solve: Secondary-10,2016.PNG In $\triangle ABC$, $AD$ is an angle bisector. So, $\frac{BD}{CD}=\frac{AB}{AC} \Rightarrow \frac{BD}{CD}=\frac{5}{3} \Rightarrow \frac{BC}{CD}=\frac{8}{3} \Rightarrow CD=\frac{3 \times 7}{8} \Rightarrow CD=\frac{21}{8}$ So, $BD=BC-CD=7-\frac{21}{8}=\frac{...
- Wed Feb 07, 2018 9:02 pm
- Forum: Junior Level
- Topic: Cylinder probability
- Replies: 1
- Views: 5456
Re: Cylinder probability
Total surface area of the cylinder=$2\pi r(r+h)$=$2\pi \times 5(5+10)=500\pi$
Area of the plane round surface=$2\pi r^2$=$2\pi 5^2$=$50\pi$
Probability=$\frac{500\pi}{50\pi}$=$10$
Area of the plane round surface=$2\pi r^2$=$2\pi 5^2$=$50\pi$
Probability=$\frac{500\pi}{50\pi}$=$10$
- Mon Feb 05, 2018 10:39 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies: 17
- Views: 11119
Re: BdMO National Secondary: Problem Collection(2016)
5(b) We can set the Earthlings in $n!$ ways if they are in straight line. But considering rotation, there exists a group of $n$ arrangements, that are actually same if we rotate them. If $n=3$, rearrangements are: $ABC,BCA,CAB,ACB,CBA,BAC$, where the first three are same and so are the last three. ...
- Mon Feb 05, 2018 10:18 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 3
- Replies: 6
- Views: 4498
Re: BDMO 2017 National round Secondary 3
Stuck at last step. Please help if anyone can solve Let $r_1$ and $r_2$ be the roots of the equation $x^2+3x-1=0$. So, we can write: $r_1+r_2=\frac{-b}{a}=\frac{-3}{1}=-3$... 1|) $r_1r_2=\frac{c}{a}=\frac{-1}{1}=-1$... (2) We can write the quartic equation in this way: $x^4+dx^3+ax^2+bx+c=0$ where,...
- Sun Feb 04, 2018 7:30 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies: 19
- Views: 17400
Re: BDMO 2017 National round Secondary 1
Can anyone give the correct solution?
- Sun Feb 04, 2018 12:05 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary: Problem Collection(2016)
- Replies: 17
- Views: 11119
Re: BdMO National Secondary: Problem Collection(2016)
5(a) When there are $m$ Martians and $n$ Earthlings. If we consider the Earthlings set, there are $(n+1)$ places for the Martians. So, the place can be chosen in ${n+1 \choose m}$. But the Earthlings can change their place in $n!$ ways where, the Martians can in $m$ ways. So, total combination =$m!...
- Sat Feb 03, 2018 12:19 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies: 19
- Views: 17400
Re: BDMO 2017 National round Secondary 1
For $5$ match, result may be of $20$ ways in total. (Excluding unimportant results when series ends in 3/4 matches) (a) If Bangladesh wins in 3 games, none but the sequence must be:$WWW$. So, the probability=$\frac{1}{20}$ (b) If Bangladesh wins in 4 matches, the sequence may be following: $WWLW, WL...
- Fri Feb 02, 2018 11:11 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 6
- Replies: 9
- Views: 8298
Re: BDMO 2017 National round Secondary 6
We can rotate the octahedron by 8\times3 or 24 ways(to prove it,just assign one side of the octahedron as A and one of it's neighbour as B and observe it's rotation). So,the total number of distinguishable octahedron is \frac{8!}{24} or 1680. He wants to mean: The octahedron has $8$ surfaces. If we...