Search found 73 matches

by Tasnood
Mon Feb 19, 2018 9:49 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2018 News
Replies: 4
Views: 681

Re: BdMO 2018 News

Neither BdMO nor APMO has a specific syllabus like school/college syllabus. And before the registration, add will be published on Prothom Alo.
by Tasnood
Mon Feb 19, 2018 9:12 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2018 News
Replies: 4
Views: 681

Re: BdMO 2018 News

Nothing to do. We need to attend both BdMO and APMO. So good luck! :)
by Tasnood
Mon Feb 19, 2018 8:47 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2018 News
Replies: 4
Views: 681

BdMO 2018 News

Dear participants of BdMO 2018 national round, the program is going to be held on 9-10 March, 2018 at the same location. Get more information here:
http://matholympiad.org.bd/
by Tasnood
Sun Feb 18, 2018 11:00 am
Forum: Number Theory
Topic: 13 -11
Replies: 6
Views: 3922

Re: 13 -11

:oops: I missed the middle stamp!
However a simple change will make the solution correct.
by Tasnood
Sun Feb 18, 2018 10:08 am
Forum: Number Theory
Topic: 13 -11
Replies: 6
Views: 3922

Re: 13 -11

Easy to prove by Modular Arithmetic $11|13^N-1$ We write: $13^N-1 \equiv 0 \Rightarrow 13^N \equiv 1 \Rightarrow (11+2)^N \equiv 1$ mod $(11)$ We can write the expression $(11+2)^N=11^N+{{N} \choose {1}} 11^{N-1}.2+{{N} \choose {2}} 11^{N-2}.2^2+...+{{N} \choose {N-1}} 11.2^{N-1}+2^N$ Here, all term...
by Tasnood
Fri Feb 16, 2018 7:45 pm
Forum: Secondary Level
Topic: secondary regional 2017
Replies: 9
Views: 1575

Re: secondary regional 2017

You didn't write that in the last question. :?:
by Tasnood
Fri Feb 16, 2018 7:42 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Junior 2016/1
Replies: 7
Views: 1448

Re: BdMO National Junior 2016/1

We can write the expression in this form: $1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$. [We will call "Brackets" a pair of starting and ending brackets] If we take the whole expression under brackets we will get the number $2017$. If we encounter the first brackets, the resul...
by Tasnood
Fri Feb 16, 2018 6:56 pm
Forum: Secondary Level
Topic: secondary regional 2017
Replies: 9
Views: 1575

Re: secondary regional 2017

$34x+51y=6z \Rightarrow 34x=51y-6z=0$ $34x$ is even for all values of $x$. $6z$ is even too. Then, $51y$ must be even , $y=2$ then. We can write: $6z-34x=102$ [Applying the value of $y$] $ \Rightarrow 3z-17x=51$ $3$ divides RS . So, $3$ must divide $(3z-51)$. $3$ divides $3z$. Then $17x$. $3$ doesn'...
by Tasnood
Thu Feb 15, 2018 11:22 pm
Forum: Secondary Level
Topic: secondary regional 2017
Replies: 9
Views: 1575

Re: secondary regional 2017

Another clear-cut solution $34x+51y=6z \Rightarrow 34x+51y-6z=0$ Here, $34x$ and, $6z$ are even . So, $34x-6z$ is even . As LS is even $(0)$, $51y$ must be even . $51$ is not even . So, $y$ is obviously even and, $y=2$ We can write: $6z-34x=51 \times 2 \Rightarrow 6z-34x=102 \Rightarrow 3z-17x=51$ ...
by Tasnood
Thu Feb 15, 2018 4:16 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Primary 2016/2
Replies: 2
Views: 627

Re: BdMO National Primary 2016/2

$(11+11) \times (12+12) \times (13+13) \times...\times (18+18) \times (19+19)=22 \times 24 \times 26 \times...\times 38$
Here, the fifth term will be: $(15+15)=30$ which is divisible by $10$
As $x=30 \times y$
So, $10|x$
Remainder=$0$
[Can't simplify more for primary :D ]