Dhaka Secondary 2010/2

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2010/2

Unread post by BdMO » Fri Jan 21, 2011 7:04 pm

What is the remainder when $2^{1024} + 5^{1024} +1$ is divided by $9$?

protik
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Re: Dhaka Secondary 2010/2

Unread post by protik » Sat Jan 22, 2011 10:53 pm

6. I am not sure about my answer..

AntiviruShahriar
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Re: Dhaka Secondary 2010/2

Unread post by AntiviruShahriar » Sun Jan 23, 2011 11:10 am

$5^{ \phi(9)} \equiv 2^{ \phi(9)} \equiv 1$(mod 9)
$5^{1020} \equiv 2^{1020} \equiv 1$(mod 9)
$5^{1024} \equiv -5 \equiv 4$(mod 9)
$2^{1024} \equiv -2 \equiv 7$(mod 9)
$7+4+1=12 \equiv 3$(mod 9)
Ans:3

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leonardo shawon
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Re: Dhaka Secondary 2010/2

Unread post by leonardo shawon » Sun Jan 30, 2011 11:55 pm

Cant we use it??
as 1024 is evan so
$5^2+2^2+1=30$ remainder 3.....

[[Actually i dont know Mod. :(]]
Ibtehaz Shawon
BRAC University.

long way to go .....

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