Search found 16 matches
- Tue Feb 08, 2011 4:15 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2007/4
- Replies: 5
- Views: 4349
Re: BdMO National Higher Secondary 2007/4
n, (n+1), (n+2) are back to back integers. If n is even, then (n+1) is odd and (n+2) is even. And if n is odd, then (n+1) is even and (n+2) is odd. So, (n+1) is coprime to both n and (n+2). So.... a) (n,n+1)=1 b) (n+1,n+2)=1 c) (n+1, n(n+2))=1 Now, as there is no comon factor between (n+1) and n(n+2...
- Tue Feb 08, 2011 3:54 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2009/9
- Replies: 3
- Views: 3433
Re: BdMO National Higher Secondary 2009/9
Though it was a problem of combinatorics, most of us solved it as a sequence that year.... :p
- Mon Feb 07, 2011 11:58 pm
- Forum: Combinatorics
- Topic: good problems
- Replies: 3
- Views: 3572
Re: good problems
According to my solution, the answer should be nC4....photon wrote:2nd prob.
A point can't intersect its 2 sides' points.
my ans . (n-2)!ways
- Sun Feb 06, 2011 1:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: remainder
- Replies: 2
- Views: 2784
Re: remainder
$$F(x)=x^{n}+x^{n-1}+.................+x^2+x+2$$ find the remainder when $$F(x^{n+1})$$ is divided by $$F(x)$$. \[f(x)=x^n+x^{n-1}+........+x^2+x+2\] \[\therefore f(x^{n+1})=(x^{n+1})^n+(x^{n+1})^{n-1}+........+(x^{n+1})^2+x^{n+1}+2\] Now, \[f(x)=\, \frac{x(x^n-1)}{x-1}+2\]\[=\, {P}x(x-1)+2\] And, ...
- Sun Feb 06, 2011 12:13 am
- Forum: Number Theory
- Topic: Find the remainder
- Replies: 7
- Views: 5108
Re: Find the remainder
What is the remainder when 2^1990 divided by 1990? \[1990=2\times 5\times 199\] \[\Rightarrow 995=5\times 199\]: \[So,\: 2\: is\: a\: coprime\: to\: 995.\: According\: to\: Fermat's\: theorem,\] \[\Rightarrow 2^{995-1}\equiv 1(mod\: 995)\] \[\Rightarrow (2^{994})^2\equiv 1^2(mod\: 995)\] \[\Rightar...
- Sun Feb 06, 2011 12:09 am
- Forum: Number Theory
- Topic: Find the remainder
- Replies: 7
- Views: 5108
Re: Find the remainder
\[1990=2\times 5\times 199\] \[\Rightarrow 995=5\times 199\]: \[So,\: 2\: is\: a\: coprime\: to\: 995.\: According\: to\: Fermat's\: theorem,\] \[\Rightarrow 2^{995-1}\equiv 1(mod\: 995)\] \[\Rightarrow (2^{994})^2\equiv 1^2(mod\: 995)\] \[\Rightarrow 2^{1988}\equiv 1(mod\: 995)\] \[\Rightarrow 2^{1...