Dhaka Higher Secondary 2010/6

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
tushar7
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Re: dhaka div 2010

Unread post by tushar7 » Sat Jan 01, 2011 6:34 pm

tushar7 wrote:what is the remainder when $2^{1024} +5^{1024} +1$ is divided by $9$?
the answer given in KMC website is 0 . but i am having the same answer $3$
$2^3\equiv -1\bmod{9}$ $\Rightarrow 2^{1024}\equiv -2\bmod{9}$
$5^6\equiv 1\bmod{9}$ $\Rightarrow 5^{1024}\equiv 4\bmod{9}$
so $-2+4+1=3$ , where is the faulty?
i think the part with just $2$ you did understand(if not plz reply)
and i used euler's toient theorem to figure out $5^6\equiv 1\bmod (9)$ and $1020$ is divisible by $9$ and $5^4=625\equiv 4\bmod (9)$

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Moon
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Re: dhaka div 2010

Unread post by Moon » Sun Jan 02, 2011 2:56 am

I'll try to correct the solutions when I get time. Actually there might me mistakes; however, as Avik vi has already noted, those problems are checked (and rechecked during the examining period) to make sure that the solutions are correct.
So you don't need to worry about this. :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

BdMO
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Dhaka Higher Secondary 2010/6

Unread post by BdMO » Tue Jan 18, 2011 2:05 pm

What is the remainder when $2^{1024} + 5^{1024}$ is divided by $3$?

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leonardo shawon
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Re: Dhaka Higher Secondary 2010/6

Unread post by leonardo shawon » Tue Jan 18, 2011 6:46 pm

the remainder is 2.

Power of 2 and 5 is even. So i checked for power 2 and 4.
Last edited by leonardo shawon on Wed Jan 19, 2011 2:38 am, edited 1 time in total.
Ibtehaz Shawon
BRAC University.

long way to go .....

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Moon
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Re: Dhaka Higher Secondary 2010/6

Unread post by Moon » Tue Jan 18, 2011 7:05 pm

You are right. Actually $a^{2k} \equiv 1 \pmod{3}\quad \forall \; a,k \in \mathbb{N}$ with $gcd(a,3)=1$.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

tushar7
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Re: Dhaka Higher Secondary 2010/6

Unread post by tushar7 » Tue Jan 18, 2011 8:57 pm

fermat's little theorem can be useful ..........but this is too trivial

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