Find the remainder

For discussing Olympiad Level Number Theory problems
protik
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Find the remainder

Unread post by protik » Wed Dec 29, 2010 8:10 pm

What is the remainder when 2^1990 divided by 1990?

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Labib
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Re: Find the remainder

Unread post by Labib » Wed Dec 29, 2010 11:53 pm

Protik, could ya verify my proof posting it over here?? plzzzz....
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Labib
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Re: Find the remainder

Unread post by Labib » Wed Dec 29, 2010 11:56 pm

My soln was
$1024$
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Zzzz
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Re: Find the remainder

Unread post by Zzzz » Fri Jan 07, 2011 7:18 am

Labib wrote:My soln was
$1024$
Please explain your solution.
Every logical solution to a problem has its own beauty.
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jagdish
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Re: Find the remainder

Unread post by jagdish » Thu Jan 27, 2011 10:59 pm

Can we solve using this by binomial theorem........
jagdish

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Cryptic.shohag
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Re: Find the remainder

Unread post by Cryptic.shohag » Sun Feb 06, 2011 12:09 am

\[1990=2\times 5\times 199\]
\[\Rightarrow 995=5\times 199\]:

\[So,\: 2\: is\: a\: coprime\: to\: 995.\: According\: to\: Fermat's\: theorem,\]

\[\Rightarrow 2^{995-1}\equiv 1(mod\: 995)\]
\[\Rightarrow (2^{994})^2\equiv 1^2(mod\: 995)\]
\[\Rightarrow 2^{1988}\equiv 1(mod\: 995)\]
\[\Rightarrow 2^{1988}\times 4\equiv 1\times 4(mod\: 995\times 4)\]
\[\Rightarrow 2^{1990}\equiv 4(mod\: 1990\times 2)\]
\[\Rightarrow 2^{1990}\equiv 4(mod\: 1990)\]

So, the remainder is 4. :)
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein

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Cryptic.shohag
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Re: Find the remainder

Unread post by Cryptic.shohag » Sun Feb 06, 2011 12:13 am

protik wrote:What is the remainder when 2^1990 divided by 1990?
\[1990=2\times 5\times 199\]
\[\Rightarrow 995=5\times 199\]:

\[So,\: 2\: is\: a\: coprime\: to\: 995.\: According\: to\: Fermat's\: theorem,\]

\[\Rightarrow 2^{995-1}\equiv 1(mod\: 995)\]
\[\Rightarrow (2^{994})^2\equiv 1^2(mod\: 995)\]
\[\Rightarrow 2^{1988}\equiv 1(mod\: 995)\]
\[\Rightarrow 2^{1988}\times 4\equiv 1\times 4(mod\: 995\times 4)\]
\[\Rightarrow 2^{1990}\equiv 4(mod\: 1990\times 2)\]
\[\Rightarrow 2^{1990}\equiv 4(mod\: 1990)\]

So, the remainder is 4. :)
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein

AntiviruShahriar
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Re: Find the remainder

Unread post by AntiviruShahriar » Thu Mar 10, 2011 12:22 pm

Cryptic.shohag wrote: \[1990=2\times 5\times 199\]
\[\Rightarrow 995=5\times 199\]:

\[So,\: 2\: is\: a\: coprime\: to\: 995.\: According\: to\: Fermat's\: theorem,\]

\[\Rightarrow 2^{995-1}\equiv 1(mod\: 995)\]
Fermat's theorem \[\Rightarrow\]
$ a^{p-1} \equiv 1 (mod p)$ কিন্তু 995 prime না।
এক্ষেত্রে, $a^{\phi(n)} \equiv 1 (mod n)$ $a \perp n$
যদি এইটা বযবহার করতে হয় তাইলে, $ \phi(1990) = 792$

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