help please
- afif mansib ch
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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
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
Re: help please
Have you tried using induction?
"Everything should be made as simple as possible, but not simpler." - Albert Einstein
- afif mansib ch
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Re: help please
i tried with induction but can't find any nice solution .will you help please?
- Tahmid Hasan
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Re: help please
please use LaTeX or equation editor.i can't understand the equation quite well.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
বড় ভালবাসি তোমায়,মা
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Re: help please
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.
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......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- afif mansib ch
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Re: help please
thanks a lot.
Re: help please
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...
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...
- Abdul Muntakim Rafi
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