sakibtanvir wrote:I think it will be 30 as well...........(Really they r friendly )
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- Sat Jan 14, 2012 1:52 pm
- Forum: Junior Level
- Topic: Friendly Triangles[self-made]
- Replies: 4
- Views: 3982
Re: Friendly Triangles[self-made]
- Fri Jan 13, 2012 10:14 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problem Set 3
- Replies: 25
- Views: 12499
Re: Problem Set 3
No 8: $2^{2a}\equiv1(modb)$ $2^{2b}\equiv1(modc)$ $2^{2c}\equiv1(moda)$ a,b,c wont be even . So that If b \geq 3 $2^{(2a,\phi(b))} \equiv 2^2 mod \equiv 1(modb)$ (i) else $b=1$ (ii) let $a=2n+1$ then for (i) $2^{2n+1}+1=2^a+1 \equiv 3 \equiv 0( mod b)$ or $b=3$ Thus we can say $c=1,3$,$a=1,3$ are p...
- Fri Jan 13, 2012 10:04 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problem Set 3
- Replies: 25
- Views: 12499
Re: Problem Set 3
Thank you.I missed the case.Now the proof has been edited.sourav via wrote:What if $x=2k$ where $(2,k)=1$ , $2|2$, $k|a$. Are you considering $x$ a prime?I didn't get your solution...Phlembac Adib Hasan wrote:As $2$ and $a$ are co-primes, $x|2$ or $x|a$
- Fri Jan 13, 2012 12:01 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problem Set 3
- Replies: 25
- Views: 12499
Re: Problem Set 3
Actually you should say that, let $p^a|m$ for a prime $p$ such that $p^{a+1}$ does not divide $m$. If we can prove that $p^a|n!$ then we are done since distinct primes are always co-prime. Actually I didn't thought in this way.As every composite number can be written as multiplication of prime powe...
- Thu Jan 12, 2012 10:02 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Problem Set 3
- Replies: 25
- Views: 12499
Re: Problem Set 3
Problem 7: $m \leq \frac {n^2} {4}$ implies that $\sqrt{m} \leq \frac{n}{2}$.So it's clear that if we prove the statement for prime power $m$s, it will be proved for every $m$.Let $m=p^a$ Then $ 2p^{\frac {a} {2} } \leq n$ Now it's enough to show that $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \rig...
- Thu Jan 12, 2012 8:35 pm
- Forum: News / Announcements
- Topic: Prothom Alo 4/1/2012 page 6 column 5
- Replies: 13
- Views: 10464
Re: Prothom Alo 4/1/2012 page 6 column 5
I also agree with you.Actually I needed only 10 minutes.FahimFerdous wrote:The problem is actually an exercise and it's super easy. I did it just when I saw it. I don't know what Mr. Alamgir thinks of himself.
- Thu Jan 12, 2012 10:26 am
- Forum: News / Announcements
- Topic: Prothom Alo 4/1/2012 page 6 column 5
- Replies: 13
- Views: 10464
Re: Prothom Alo 4/1/2012 page 6 column 5
Thanks.Sorry,I can't give you its link because the online copy of Prothom Alo doesn't publish all the advertisements.sourav das wrote:Solution:
- Wed Jan 11, 2012 9:34 pm
- Forum: News / Announcements
- Topic: Prothom Alo 4/1/2012 page 6 column 5
- Replies: 13
- Views: 10464
Prothom Alo 4/1/2012 page 6 column 5
In an advertisement titled "A difficult sum" Mr. Alamgeer Hossain gave a math problem and told that it may be included in Guiness Book of World. He also claimed nobody was able to solve that till then and if anyone can solve it ,then he or she must announce it before 31/1/2012.The problem and my sol...
- Wed Jan 11, 2012 7:51 pm
- Forum: Geometry
- Topic: Angle Construction [self-discovered]
- Replies: 0
- Views: 1835
Angle Construction [self-discovered]
Suppose $k\in \mathbb{N}$. Prove that we can construct all the angles of the form $\dfrac {\pi k} {60} $ using only a straitedge and a compass. Hint: No calculation, no difficult theorem; just logic.I discovered this beautiful thing about two and a half years ago when I was in class 6 and knew nothi...
- Wed Jan 11, 2012 6:51 pm
- Forum: Junior Level
- Topic: Friendly Triangles[self-made]
- Replies: 4
- Views: 3982
Friendly Triangles[self-made]
$ABCD$ is a parallelogram. Extend $AD$ and take a point $P$ on its extension. $BP$ intersect $CD$ at $Q$.$\bigtriangleup ADQ=30$,Then $\bigtriangleup CPQ=?$