Search found 194 matches
- Sat May 22, 2021 10:39 pm
- Forum: Social Lounge
- Topic: Struggling with math
- Replies: 6
- Views: 13899
Re: Struggling with math
Please somebody answer ;-;
- Fri May 21, 2021 3:05 pm
- Forum: Social Lounge
- Topic: Struggling with math
- Replies: 6
- Views: 13899
Struggling with math
I am a very noob and inexperienced problem solver and just started taking prep for MO after SSC vacation.(although after 3 or 4 months again quited for a long time and just got active before 2021 BdMO again and worth mentioning that i mostly invested that 3,4 months on Geo :oops: especially studying...
- Thu May 20, 2021 9:50 pm
- Forum: College / University Level
- Topic: Maybe not as hard as you see
- Replies: 8
- Views: 24386
Maybe not as hard as you see
Prove that any closed loop is inscribed in a square .(It is not the unsolved topology problem rather a brain teaser) Disclaimer: I don't know how to prove it formally but the intuitive proof(key) is enough. Hint: It is easy to prove that every closed loop is inscribed in a rectangle then try to mes...
- Thu May 13, 2021 10:06 pm
- Forum: Social Lounge
- Topic: Eid mubarak
- Replies: 1
- Views: 6963
Eid mubarak
Eid mubarak may Allah SWT accept our deeds and make the Eid blessed for us. ( Also pray for Palestine )
- Tue May 11, 2021 11:00 pm
- Forum: Algebra
- Topic: A question about FE
- Replies: 9
- Views: 15134
Re: A question about FE
Bumpty Bumpty Bump
- Mon May 10, 2021 11:44 pm
- Forum: College / University Level
- Topic: A question about derivatives
- Replies: 2
- Views: 9145
Re: A question about derivatives
Consider this following proof of a combinatorial identity i.e $n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ Proof: $(1+x)^n=\sum_{k=1}^{n} \binom{n}{k} x^k$ Differentiating both sides wrt x gives $n(1+x)^{n-1}=\sum_{k=1}^{n}k \binom{n}{k} x^{k-1}$ $x:=1 \Rightarrow n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ ...
- Mon May 10, 2021 10:20 pm
- Forum: College / University Level
- Topic: A question about derivatives
- Replies: 2
- Views: 9145
A question about derivatives
Consider this following proof of a combinatorial identity i.e $n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ Proof: $(1+x)^n=\sum_{k=1}^{n} \binom{n}{k} x^k$ Differentiating both sides wrt x gives $n(1+x)^{n-1}=\sum_{k=1}^{n}k \binom{n}{k} x^{k-1}$ $x:=1 \Rightarrow n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$ H...
- Sun May 09, 2021 1:10 pm
- Forum: Physics
- Topic: A Problem of Secondary Physics Olympiad 2021
- Replies: 3
- Views: 11942
Re: A Problem of Secondary Physics Olympiad 2021
here is the solution(maybe do check the calculation) consider kg,m,s as variables then you should get $m^{a+2b+3c}kg^{b+c}s^{-a-2b-2c}=kgm^{-1}$ Comparing the power you get the following set of equations $a+2b+3c=-1,b+c=1,-a-2b-2c=0$ solving it yields $(a,b,c)=(-\frac{5}{2},\frac{3}{2},-\frac{2}{1})...
- Fri May 07, 2021 2:50 pm
- Forum: Number Theory
- Topic: An OLD NT
- Replies: 1
- Views: 6540
Re: An OLD NT
Find all positive integers $x,y,z$ such that $x^2 +y^2 =3z^2$. Here is a elegant solution (saw in thanic bhaiya's book): It is easy to check that $3|x,y$ write $x=3a$ and $y=3b$ now it follows that $3|z$ writing $z=3c$ implies $3a^2 +3b^2= 9c^2 \Rightarrow a^2+b^2=3c^2$ so if $x,y,z$ is a solution ...
- Thu May 06, 2021 10:42 pm
- Forum: Number Theory
- Topic: An OLD NT
- Replies: 1
- Views: 6540
An OLD NT
Find all positive integers $x,y,z$ such that $x^2 +y^2 =3z^2$.