help please

For discussing Olympiad Level Number Theory problems
User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment
help please

Unread post by afif mansib ch » Sun Aug 07, 2011 10:31 pm

\[\]
i am having problem with solving a few simple problems of the ELEMENTARY NUMBER THEORY book.please show me the solution of one of them:
proof that :
5^2n+3.2^5n-2is divisable by 7

User avatar
nayel
Posts:268
Joined:Tue Dec 07, 2010 7:38 pm
Location:Dhaka, Bangladesh or Cambridge, UK

Re: help please

Unread post by nayel » Mon Aug 08, 2011 3:24 pm

Have you tried using induction?
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: help please

Unread post by afif mansib ch » Mon Aug 08, 2011 8:13 pm

i tried with induction but can't find any nice solution .will you help please? :?:

User avatar
Tahmid Hasan
Posts:665
Joined:Thu Dec 09, 2010 5:34 pm
Location:Khulna,Bangladesh.

Re: help please

Unread post by Tahmid Hasan » Mon Aug 08, 2011 10:26 pm

afif mansib ch wrote:\[\]
i am having problem with solving a few simple problems of the ELEMENTARY NUMBER THEORY book.please show me the solution of one of them:
proof that :
5^2n+3.2^5n-2is divisable by 7
please use LaTeX or equation editor.i can't understand the equation quite well. :shock:
বড় ভালবাসি তোমায়,মা

sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
Location:Dhaka
Contact:

Re: help please

Unread post by sourav das » Mon Aug 08, 2011 11:24 pm

For $5^{2n} + 3.2^{5n} - 2$ you get contradiction for $n = 2$.

For $5^{2n} + 3.2^{5n - 2}$
$2^{5n}\equiv 32^{n}\equiv 4^{n}\ (mod \ 7 ) $
$g.c.d(2^2,7)=1$
So,
$2^{5n-2}\equiv 4^{n-1}\ (mod \ 7 ) $
$5^{2n}\equiv 25^{n}\equiv 4^{n}\ (mod \ 7 ) $
and so,
$5^{2n} + 3.2^{5n - 2} \equiv 4^{n-1} (3 + 4 ) \equiv 0 ( mod \ 7 )$
And we are done.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: help please

Unread post by afif mansib ch » Tue Aug 09, 2011 3:09 am

thanks a lot.

Shifat
Posts:53
Joined:Sun Jul 31, 2011 12:21 pm
Location:Dhaka, Bangladesh

Re: help please

Unread post by Shifat » Tue Aug 23, 2011 1:19 pm

I am posting a solution with induction:-Nine ten er higher math book follow kore dicchi.
first as usual when n=1 then $5^{2n} + 3\times 2^{5n-2}$ is dividable by 7.......(1)
now let for n=m
$5^{2m} + 3\times 2^{5m-2}=7k$........(2)
then induction law goes, if for n=m+1, $5^{2n} + 3\times 2^{5n-2}$ is dividable by 7 then the equation is true
(1) LHS n= m+1 implies
$5^{2m+2} + 3\times 2^{5m-2+5}
=5^{2m}\times 5^2 + 3\times 2^{5m-2}\times 2^5$
we can write $2^5= 5^2 +7$
so the expression is $ 5^2 (5^{2m}+3\times 2^{5m-2}) + 7(3\times 2^{5m-2})$
which is dividable by 7, so
PROVED... :)

User avatar
Abdul Muntakim Rafi
Posts:173
Joined:Tue Mar 29, 2011 10:07 pm
Location:bangladesh,the earth,milkyway,local group.

Re: help please

Unread post by Abdul Muntakim Rafi » Wed Aug 24, 2011 5:22 am

Bravo Shifat :)
Man himself is the master of his fate...

Post Reply