if $(2^p-1)$ is mersenne prime then $2^{(p-1)}.(2^p-1)$ is perfect number. prove they are sum of odd number's cube. if it sum of 1~nth odd number's cube then $n=2^{\frac{p-1}{2}}$
Hint: calculate sum of n odd numbers where $n=2^{\frac{p-1}{2}}$ then show result as $2^{(p-1)}.(2^p-1)$
Search found 56 matches
- Sat May 07, 2011 12:25 am
- Forum: Number Theory
- Topic: Perfect Number Is Sum Of Cube
- Replies: 9
- Views: 5493
- Thu May 05, 2011 10:30 pm
- Forum: Geometry
- Topic: A QUESTION ABOUT VECTOR
- Replies: 4
- Views: 3252
Re: A QUESTION ABOUT VECTOR
Happy Birth Day Mahi !
multiplications of vector is two kind of.both scalar and vector answer can found.
multiplications of vector is two kind of.both scalar and vector answer can found.
- Wed May 04, 2011 3:11 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 2007-5
- Replies: 15
- Views: 9519
Re: Imo 2007-5
@Masum, i don't assume $(4ab-1)$ divisor of $4a^2-1$. i say if $4ab-1$ divisor of $(4a^2-1)^2$ then $4ab-1$ and $4a^2-1$ have common divisor suppose $d$. then $d$ divisor of $(4ab-1)-(4a^2-1)$. then it makes a contradiction. which only can ignore if we accept $a=b$. u also can see my reply of Tanvir...
- Wed May 04, 2011 2:54 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 2007-5
- Replies: 15
- Views: 9519
Re: Imo 2007-5
if $B$ divide $A^2$ then $A,B$ must have common divisor. let $A=P1^(x1).P2^(x2)..Pn^(xn)$ $A^2=P1^(2x1)P2^(2x2)...Pn^(2xn)$ if $A^2=B.m$ then sqr root of $B.m=P1^(x1).P2^(x2)...Pn^(xn)$. since $B.m$ integer and they sharing same prime factors of $A$ then it seems $A,B,m$ have some common divisors. r...
- Tue May 03, 2011 7:46 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 2007-5
- Replies: 15
- Views: 9519
Re: Imo 2007-5
$(4a^2-1)=A$ $(4ab-1)=B$ now $B$ factor of $A^2$ so $A,B$ have one or more common factors. suppose $d$ factor of $A$ and $B$. then $d/A,B=d$ or, $d/B-A=d$ so, $d/4ab-4a^2$. $d$ not equal to $4$,bcz $A=-1(mod)4$ so $d$ factor of $a$. suppose $a=nd$. now $A=4(nd)^2-1$ $B=4ndb-1$. now $A=-1(mod)d$ $B=-...
- Tue May 03, 2011 2:59 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 1999-2
- Replies: 6
- Views: 4878
Re: APMO 1999-2
cube root of 420 not an ineger is not it a problem ?
- Mon May 02, 2011 1:16 am
- Forum: Number Theory
- Topic: Secondary Special Camp 2011: NT P 1
- Replies: 20
- Views: 11356
Re: Secondary Special Camp 2011: NT P 1
that's what i say when 2n+1=prime then it restrict answer bcz a prime must has no factor so i say in this case d=2n+1 and i don't think about 1. that's different what u say and i said. nothing to be anxious much. but ur discussion about this solution is very helpful. also ur replies on my reply clar...
- Sun May 01, 2011 9:46 pm
- Forum: Number Theory
- Topic: Secondary Special Camp 2011: NT P 1
- Replies: 20
- Views: 11356
Re: Secondary Special Camp 2011: NT P 1
now u messed up, if (2n+1) is a prime then how d could be any divisor of (2n+1)? a prime has no divisor.
thanks for ur clear solution.
thanks for ur clear solution.
- Sun May 01, 2011 9:37 pm
- Forum: Secondary Level
- Topic: Keo ki proman korte parbe - Por por 5 ti kromic sonkhar gunf
- Replies: 2
- Views: 2770
Re: Keo ki proman korte parbe - Por por 5 ti kromic sonkhar
among 1~n numbers one number is divisible by n. bcz if u divide a number suppose a by n and maximum (n-1) left as reminder then a+(n-1) is divisible by n that means every time between 1~n numbers one has n as a factor. thus every (n-1) numbers one number is divisible by (n-1). so every 1~(n-2) numbe...
- Sun May 01, 2011 1:42 am
- Forum: Geometry
- Topic: Homothetic Triangles!!
- Replies: 8
- Views: 5434
Re: Homothetic Triangles!!
if A`C` extend meet C`` with BC then angle C=C``=C` bcz B`C` parallel BC and A`C` parallel AC. thus angle A=A` B=B`. so why it said not congruent=similar ?