Search found 1015 matches
- Fri Dec 30, 2011 11:54 am
- Forum: Number Theory
- Topic: Sum Challenge
- Replies: 6
- Views: 4247
Re: Sum Challenge
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. The problem statement is confusing. Please clearly define $P_d$.Is it $P_{d_i} $? Otherwise it seems that $ P_...
- Fri Dec 30, 2011 8:43 am
- Forum: Number Theory
- Topic: Round 3
- Replies: 31
- Views: 15363
Re: Round 3
I don't understand what prob 1 says.It's not clear what informations are given .Can anyone explain it?
- Thu Dec 29, 2011 11:55 am
- Forum: Junior Level
- Topic: Inequality !!! [self-made]
- Replies: 2
- Views: 2752
Re: Inequality !!! [self-made]
(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)>...
- Wed Dec 28, 2011 8:16 am
- Forum: Secondary Level
- Topic: Helpful Inequality [self-made]
- Replies: 4
- Views: 3182
Helpful Inequality [self-made]
Let $n$ is an integer and $n\ge 3$.Prove that \[ n^{n+1}> (n+1)^n \].
[Many regional and national problems can be solved using it.]
[Many regional and national problems can be solved using it.]
- Wed Dec 28, 2011 8:10 am
- Forum: Junior Level
- Topic: Inequality !!! [self-made]
- Replies: 2
- Views: 2752
Inequality !!! [self-made]
$n$ is a real number and $n\ge1$.Prove that \[ \frac {1}{2^{\frac {1}{n} }-1} \ge n\]
with equality if and only if $n=1$.
A JUNIOR TYPE PROBLEM.NOT FOR OTHERS.
with equality if and only if $n=1$.
A JUNIOR TYPE PROBLEM.NOT FOR OTHERS.
- Sun Dec 18, 2011 11:34 am
- Forum: Junior Level
- Topic: Tom's Mysterious Quadrilateral [Self-Made]
- Replies: 6
- Views: 4770
Re: Tom's Mysterious Quadrilateral [Self-Made]
Nope, I think. $EFGH$ is a rectangle. But it implies $ABCD$ a kite, not necessarily a rhombus.
- Sun Dec 18, 2011 10:56 am
- Forum: Number Theory
- Topic: PRIME BEAUTY {SELF-MADE}
- Replies: 9
- Views: 5086
Re: PRIME BEAUTY {SELF-MADE}
Sorry, I was an idiot; used $\sum$ notation instead of $\Pi$. However, my problem was correct!
- Sat Dec 17, 2011 7:25 pm
- Forum: Number Theory
- Topic: PRIME BEAUTY {SELF-MADE}
- Replies: 9
- Views: 5086
PRIME BEAUTY {SELF-MADE}
$P$ is a positive integer.Prove that $p=1,4$ or any odd prime if and only if
\[({\prod_{k=0} ^{p-1}{^{p-1}}C_k})^2\equiv1(mod p)\]
\[({\prod_{k=0} ^{p-1}{^{p-1}}C_k})^2\equiv1(mod p)\]
- Sat Dec 17, 2011 7:06 pm
- Forum: Higher Secondary Level
- Topic: PROPERTY OF 3 [Self-Made]
- Replies: 1
- Views: 2542
PROPERTY OF 3 [Self-Made]
For $n\ge 1$ prove that ${\sum^{2n}_{k=1}} ^{2n+1}C_k\equiv 0(mod3)$
Just Exercise.
Just Exercise.
Re: SUM [OWN]
I have used the phrase ''sum of'' to mean $\sum$.well I'll try my best to look it better.