Search found 244 matches
- Wed Dec 28, 2011 3:08 pm
- Forum: Number Theory
- Topic: Preparation Marathon
- Replies: 112
- Views: 48795
Re: Preparation Marathon
solution No 2 : $\binom{2012+4-1}{4-1} =\binom{2015}{3}$ No 3 : $\sum_{k=0}^{7} k\binom{7}{k}$ No 5 : $1^2+2^2+...+9^2=285$ No 7 : $0$ No 8 : $(1+2+...+8)(1+2+...+9)=1620$ No 9 : $\binom{4012}{2011} \equiv 1 mod 2011$ No 10 : 2 3 6 14 30 1 3 8 16 2 5 8 3 3 $n^{th}$ term is= 2+(n-1)+(n-1)(n-2)+{(n-1)...
- Wed Dec 28, 2011 1:04 pm
- Forum: Number Theory
- Topic: Preparation Marathon
- Replies: 112
- Views: 48795
Re: Preparation Marathon
If you select m person's from a set of n persons then you can assign captain in m ways.Abdul Muntakim Rafi wrote: 3.$2^7-1$
So that the answer will be
$\sum_{k=0}^{7}k \binom{7}{k}=7.2^6$
- Wed Dec 28, 2011 12:52 pm
- Forum: Number Theory
- Topic: Preparation Marathon
- Replies: 112
- Views: 48795
Re: Preparation Marathon
No 1 : LEt S(n) donate the sum of digits of n . If the number 0f digits be $\geq$ 4 then $S(n)_{max} =7.9.4=252 $ which is not a four digit number. So number of digits will be $\leq$3 If the number 0f digits be 3 then $S(n)_{max} =7.9.3=189 $ so that $n=100+a.10+b=7+7a+7b$ or $93+3a=6b$ that cant be...
- Wed Dec 28, 2011 10:27 am
- Forum: Number Theory
- Topic: Preparation Marathon
- Replies: 112
- Views: 48795
Re: Preparation Marathon
No 9 :
- Sat Nov 05, 2011 3:40 pm
- Forum: National Math Camp
- Topic: Camp exam problem 1
- Replies: 4
- Views: 3508
Re: Camp exam problem 1
one eleven
- Thu Nov 03, 2011 11:56 am
- Forum: National Math Camp
- Topic: Exam - Bangladesh Online Math Camp
- Replies: 7
- Views: 5689
Re: Exam - Bangladesh Online Math Camp
$hiks ^{hiks}$
- Tue Nov 01, 2011 10:11 pm
- Forum: National Math Camp
- Topic: A helpful tools :(n+1)^n > n!
- Replies: 15
- Views: 7776
Re: A helpful tools :(n+1)^n > n!
with bernaulli's inequality we have
$(1+1)^{n}\geq(1+n)$
so
$2^{n}>n$
$(1+1)^{n}\geq(1+n)$
so
$2^{n}>n$
- Tue Nov 01, 2011 9:40 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Algebra: Inequalities
- Replies: 14
- Views: 8090
Re: Algebra: Inequalities
here is a)
$\frac{x^{2}+y^{2}}{2} \geq xy$*
$\frac{y^{2}+z^{2}}{2} \geq yz$*
$\frac{z^{2}+x^{2}}{2} \geq zx$*
added three stars
$\frac{x^{2}+y^{2}}{2} \geq xy$*
$\frac{y^{2}+z^{2}}{2} \geq yz$*
$\frac{z^{2}+x^{2}}{2} \geq zx$*
added three stars
- Tue Nov 01, 2011 9:29 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Algebra: Inequalities
- Replies: 14
- Views: 8090
Re: Algebra: Inequalities
a) rearrangement .
b)
$a^{4}+b^{4}+c^{4}=3[4,0,0]$
$abc(a+b+c)=3[2,1,1]$
$[4,0,0] \geq 3[2,1,1]$
so proved
b)
$a^{4}+b^{4}+c^{4}=3[4,0,0]$
$abc(a+b+c)=3[2,1,1]$
$[4,0,0] \geq 3[2,1,1]$
so proved
- Tue Nov 01, 2011 9:04 pm
- Forum: National Math Camp
- Topic: help ! help ! help !
- Replies: 8
- Views: 4457
Re: help ! help ! help !
oh shourav da .. josss
$x^{3}-x^{2}z-x^{2}y+xyz=x(x-y)(x-z)$
$y^{3}-y^{2}x-y^{2}z+xyz=y(y-x)(y-z)$
$z^{3}-z^{2}x-z^{2}y+xyz=z(z-y)(z-x)
$
put n=1 in scaur's Inequality ...
$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)
\geq0 $
so Its proved ...
That means IMO 2000-02 is solved too
$x^{3}-x^{2}z-x^{2}y+xyz=x(x-y)(x-z)$
$y^{3}-y^{2}x-y^{2}z+xyz=y(y-x)(y-z)$
$z^{3}-z^{2}x-z^{2}y+xyz=z(z-y)(z-x)
$
put n=1 in scaur's Inequality ...
$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)
\geq0 $
so Its proved ...
That means IMO 2000-02 is solved too