Search found 101 matches
- Wed Dec 29, 2010 1:30 pm
- Forum: Divisional Math Olympiad
- Topic: number theory problem
- Replies: 10
- Views: 6643
Re: number theory problem
from $5!$ to $99!$ is divisible bt 30 .
- Wed Dec 29, 2010 1:44 am
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/6
- Replies: 15
- Views: 10867
Re: dhaka div 2010
oh thanks for the clarification . i was stumped to the answer .
- Wed Dec 29, 2010 1:15 am
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/6
- Replies: 15
- Views: 10867
Re: dhaka div 2010
that was for the H.secondary .Avik Roy wrote:The question in Dhaka was:
What is the remainder when $2^{1024} + 5^{1024}$ is divided by $3$?
The answer is $2$
the problem i gave was from the secondary group .
- Tue Dec 28, 2010 11:28 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/6
- Replies: 15
- Views: 10867
Re: dhaka div 2010
i am not understanding .... . Going to PM moon bhaya
- Tue Dec 28, 2010 11:09 pm
- Forum: Secondary Level
- Topic: Combi probs from Moon vai's note...
- Replies: 13
- Views: 8069
Re: Combi probs from Moon vai's note...
repetetion allowed or not?!
- Tue Dec 28, 2010 9:06 pm
- Forum: Divisional Math Olympiad
- Topic: plz quick reply.................
- Replies: 2
- Views: 2345
Re: plz quick reply.................
the sum of the all the angles , let $n$ be the number of sides of the polygon .
then ..
$\frac{(n-2)180}{n}=160$
so n=18
then ..
$\frac{(n-2)180}{n}=160$
so n=18
- Tue Dec 28, 2010 4:26 pm
- Forum: Secondary Level
- Topic: having problem in congruence
- Replies: 11
- Views: 8138
Re: having problem in congruence
thats what i have been trying to say
- Tue Dec 28, 2010 2:43 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/6
- Replies: 15
- Views: 10867
dhaka div 2010
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 fa...
- Tue Dec 28, 2010 1:08 pm
- Forum: Secondary Level
- Topic: Combi probs from Moon vai's note...
- Replies: 13
- Views: 8069
Re: Combi probs from Moon vai's note...
counting for 4 digits:: fixing 5 as unit digit, tens digit and hundreds digit can be in $10^2$ ways and thousands digit can be in 4 ways(as it's less than 5000)... considering same for fixing 0 as unit digit, total permutation is $4\cdot10\cdot10\cdot2$ when we put $5$ or $0$ in the last digit . an...
- Tue Dec 28, 2010 12:58 pm
- Forum: Secondary Level
- Topic: having problem in congruence
- Replies: 11
- Views: 8138
Re: having problem in congruence
this comment is right .....?tushar7 wrote:i know it was for the 1st property . so $m$ has to be the lcm of $k_1$ and $k_2$HandaramTheGreat wrote:HandaramTheGreat wrote:if $m=k_1\cdot k_2$ and $gcd\left(k_1, k_2\right)=1$...