Messages posted by ®µD®A

you mean 3 right ??? :)

Let $Ax_1+By_1+C_1=0\ and\ Ax_2+By_2+C_2=0$ be two parallel lines.. And d is the perpendicular distance between them... Then.

$d=\frac{|C_2-C_1|}{\sqrt{A^2+B^2}}$

http://www.goiit.com/posts/show/1103211.htm

http://en.wikipedia.org/wiki/Euler%27s_theorem

and my answer was not a guess.. try renaming the file before uploading it :D

sikkim

From Euler's totient theorem

$3^{phi{100}}equiv01mod100$

$3^{40}equiv01mod100$

So, $3^{4000}equiv01mod100$

rational roots theorem can be a help if the equation has rational roots. If it has irrational roots the you can not solve without using a computer.

These are quite easy if you know congruency...

7¹⁰³ is divided by 5

From Fermat Little Theorem,

$7^4equiv 1mod5Rightarrow 7^{103}equiv343equiv3mod5$

32^{32power 32} is divided by 7

$32^6equiv 1mod7Rightarrow 32^{32^{32}}equiv2^{32}equiv2^{30}cdot2^2equiv4mod7$

So leaves remainder 4 when divided by 7..

32³² is divided by 3

$32^2\equiv1\mod3\Rightarrow 32^{32}\equiv1\mod3$

2222⁵⁵⁵⁵ + 5555²²²² is divided by 7

$2222^6\equiv1\mod7\\\Rightarrow2222^{6\times925}\equiv1\mod7\\\Rightarrow 2222^{5555}\equiv2222^5\equiv3^5\equiv5\mod7$

$5555^6\equiv1\mod7\\\Rightarrow5555^{6\times370}\equiv1\mod7\\\Rightarrow 5555^{2222}\equiv5555^2\equiv4^2\equiv2\mod7$

So, 2222⁵⁵⁵⁵ + 5555²²²² leaves remainder 0 when divided by 7

$r=sqrt{x^2+y^2}=2$

Let

arg z

Now lies in IIIrd quadrant..

$\arg z=\theta=-\pi+\frac\pi{3}$

$z=re^{i heta}$

$z=2\left(\cos\left(-\pi+\frac\pi{3}\right)+i\sin\left(-\pi+\frac\pi{3}\right)\right)=2\left(\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right)=2\left(\cos\left(\frac{2\pi}{3}\right)-i\sin\left(\frac{2\pi}{3}\right)\right)$

Hint :- $2\sqrt{n}-2<1+\frac{1}{\sqrt{2}}+\frac1{\sqrt{3}}+\dots+\frac1{\sqrt{n}}<2\sqrt{n}-1$

Prove the above inequality..

in integral calculus ??????????

in number theory $a^{\phi (n)} \equiv 1 \mod{n}$

no answers till now !! I can't wait :D

i am telling the answer.. 197

oops...

sin nx ? i consider that to be a typo..

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