Page 2 of 2
Re: dhaka div 2010
Posted: Sat Jan 01, 2011 6:34 pm
by tushar7
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 $5^4=625\equiv 4\bmod (9)$
Re: dhaka div 2010
Posted: Sun Jan 02, 2011 2:56 am
by Moon
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.
Dhaka Higher Secondary 2010/6
Posted: Tue Jan 18, 2011 2:05 pm
by BdMO
What is the remainder when $2^{1024} + 5^{1024}$ is divided by $3$?
Re: Dhaka Higher Secondary 2010/6
Posted: Tue Jan 18, 2011 6:46 pm
by leonardo shawon
the remainder is 2.
Power of 2 and 5 is even. So i checked for power 2 and 4.
Re: Dhaka Higher Secondary 2010/6
Posted: Tue Jan 18, 2011 7:05 pm
by Moon
You are right. Actually $a^{2k} \equiv 1 \pmod{3}\quad \forall \; a,k \in \mathbb{N}$ with $gcd(a,3)=1$.
Re: Dhaka Higher Secondary 2010/6
Posted: Tue Jan 18, 2011 8:57 pm
by tushar7
fermat's little theorem can be useful ..........but this is too trivial