YES,$p_{d}$ means $p_{d_{i}}$.let $d$ be adivisor of n.Then take the value of the divisors of n for $d_{i}$ which are divisible by $d$.

Moderation Note: $L^AT_EX$ed correctly.

YES,$p_{d}$ means $p_{d_{i}}$.let $d$ be adivisor of n.Then take the value of the divisors of n for $d_{i}$ which are divisible by $d$.

Moderation Note: $L^AT_EX$ed correctly.

Moderation Note: $L^AT_EX$ed correctly.

- Sun Jan 22, 2012 6:17 am
- Forum: Number Theory
- Topic: Sum Challenge
- Replies:
**6** - Views:
**828**

Yes, i found it latter.Actually there's been a mistake while posting it!!!!!

- Sun Jan 22, 2012 5:57 am
- Forum: National Math Olympiad (BdMO)
- Topic: INEQUALITY-2(OWN!!)
- Replies:
**2** - Views:
**562**

Let $a,b,c$ be positive numbers such that $a^{2}+b^{2}-ab=c^{2}$.prove that,

$(a-c)(b-c)=<0$.

$(a-c)(b-c)=<0$.

- Sun Jan 22, 2012 5:55 am
- Forum: Algebra
- Topic: INEQUALITY
- Replies:
**2** - Views:
**506**

find floor

Let $a_{0}=$ 1996 and $a_{n+1}=a_{n}/[a_{n}^{2}+1]$ for $n=1,2,3,...,$.prove that $[a_{n}]=1996-n$ for $n=1,2,3,...,999$.Here $[x]$ denotes the greatest positive integer less than or equal to $x$.

Let $a_{0}=$ 1996 and $a_{n+1}=a_{n}/[a_{n}^{2}+1]$ for $n=1,2,3,...,$.prove that $[a_{n}]=1996-n$ for $n=1,2,3,...,999$.Here $[x]$ denotes the greatest positive integer less than or equal to $x$.

- Sun Jan 22, 2012 5:39 am
- Forum: Number Theory
- Topic: Determinining the floor
- Replies:
**5** - Views:
**650**

SECONDARY GEOMETRY (OWN) Let a circle $[_{1}$ be drawn through the vertices $A,B$ of $triangle_ ABC$ touching $BC$ at $B$. Similarly drtaw the circle $[_{2}$ passing through $A,C$ touching $BC$ at $C$.Cord $AB$ produces an angle of $45^0$ at the center of $[_{1}$. Cord $AC$ produces an angle of $60...

- Wed Jan 18, 2012 11:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: GEOMETRY
- Replies:
**2** - Views:
**591**

GENERAL SUMMATION

LET $s(n) = \sum_{k=1}^{n}(p-1+k)Pk$ where p is a given positive integer and nis a natural number.then find a nice simple formula for $s(n)$ involving $n,p$.

LET $s(n) = \sum_{k=1}^{n}(p-1+k)Pk$ where p is a given positive integer and nis a natural number.then find a nice simple formula for $s(n)$ involving $n,p$.

- Wed Jan 18, 2012 11:37 pm
- Forum: National Math Olympiad (BdMO)
- Topic: SUMMATION
- Replies:
**2** - Views:
**596**

INEQUALITY-2(OWN!!)

Let $a,b,c$ be positive real numbers such that $\sum_{cyclic}a = 6$

prove that $2abc +\sum_{cyclic}ab(a+b)$<36.

Let $a,b,c$ be positive real numbers such that $\sum_{cyclic}a = 6$

prove that $2abc +\sum_{cyclic}ab(a+b)$<36.

- Wed Jan 18, 2012 11:22 pm
- Forum: National Math Olympiad (BdMO)
- Topic: INEQUALITY-2(OWN!!)
- Replies:
**2** - Views:
**562**

It's open ended because the equality is only my conjecture.

- Tue Jan 17, 2012 6:32 pm
- Forum: Number Theory
- Topic: Inequality(open ended)[made by shanzeed anwar]
- Replies:
**2** - Views:
**462**

Sum Challenge 001

Let the divisors of n be A={d1,d2,d3,….,dn} where

d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.

Remember that σ(x)=sum of the divisors of x.

Prove that, I

∑d/n μ(d)Pd=1.

Let the divisors of n be A={d1,d2,d3,….,dn} where

d1=1 and dn =n. Let Pd =∑i σ(di), d Є A, d/di.

Remember that σ(x)=sum of the divisors of x.

Prove that, I

∑d/n μ(d)Pd=1.

- Fri Dec 30, 2011 9:29 am
- Forum: Number Theory
- Topic: Sum Challenge
- Replies:
**6** - Views:
**828**

(n+1)^n (1) =n^n+C(n,1)xn^(n-1)+C(n,20)xn^(n-2)+.....+1 =n^n+nxn^(n-1)+.....+1 =n^n+n^n+C(n,2)xn^(n-2)+.....+1 (2) From (1) and (2),we can say, (n+1)^n>n^n+n^n [For n>1] or,(n+1)^n>2xn^n or,((n+1)^n)/(n^n)>2 or,((n+1)/n)^n>2 or,(n+1)/n>2^(1/n) or,1+(1/n)>2^(1/n) or,1/n>2^(1/n)-1 or, 1/( 2^(1/n)-1)>n...

- Wed Dec 28, 2011 10:14 pm
- Forum: Junior Level
- Topic: Inequality !!! [self-made]
- Replies:
**2** - Views:
**621**