leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

leehuan · Jan 30, 2016

$\\ $The modular equation ${2}^{n} \mod 9 \, \forall n \in \mathbb {Z}^{+} $ forms an interesting pattern as such:$ \\ {2}^{1} \mod 9 = {2}^{7} \mod 9 = \dots = 2 \\ {2}^{2} \mod 9 = {2}^{8} \mod 9 = \dots = 4 \\ {2}^{3} \mod 9 = {2}^{9} \mod 9 = \dots = 8 \\ {2}^{4} \mod 9 = {2}^{10} \mod 9 = \dots = 7 \\ {2}^{5} \mod 9 = {2}^{11} \mod 9 = \dots = 5 \\ {2}^{6} \mod 9 = {2}^{12} \mod 9 = \dots = 1$

Is there a mathematical method to prove this pattern?

InteGrand · Jan 30, 2016

leehuan said:
$\\ $The modular equation ${2}^{n} \mod 9 \, \forall n \in \mathbb {Z}^{+} $ forms an interesting pattern as such:$ \\ {2}^{1} \mod 9 = {2}^{7} \mod 9 = \dots = 2 \\ {2}^{2} \mod 9 = {2}^{8} \mod 9 = \dots = 4 \\ {2}^{3} \mod 9 = {2}^{9} \mod 9 = \dots = 8 \\ {2}^{4} \mod 9 = {2}^{10} \mod 9 = \dots = 7 \\ {2}^{5} \mod 9 = {2}^{11} \mod 9 = \dots = 5 \\ {2}^{6} \mod 9 = {2}^{12} \mod 9 = \dots = 1$

Is there a mathematical method to prove this pattern?

$\noindent Yeah, this is just because $2^6$ is congruent to 1 modulo 9.$

InteGrand · Jan 30, 2016

Paradoxica · Jan 30, 2016

Euler's Theorem and Fermat's Little Theorem. You should learn those, for number theory.

leehuan · Jan 30, 2016

Paradoxica said:
Euler's Theorem and Fermat's Little Theorem. You should learn those, for number theory.

Noted for later

leehuan · Jan 31, 2016

$\\ $Find all $n \in \mathbb N $ such that $\sqrt{n} + \sqrt{n+253} $is a natural number.$

With a hint: To use difference of two squares.

InteGrand · Jan 31, 2016

leehuan said:
$\\ $Find all $n \in \mathbb N $ such that $\sqrt{n} + \sqrt{n+253} $is a natural number.$

With a hint: To use difference of two squares.

$\noindent If we rationalise the numerator and apply difference of two squares, it becomes $\frac{-253}{\sqrt{n}-\sqrt{n+253}}=\frac{253}{\sqrt{n+253}-\sqrt{n}}$. Note that the denominator is monotone decreasing with limit 0. So for this to be a natural number, we need the denominator here to be a factor of 253. The factors of 253 are: $1,11,23,253$. So we can set the denominator equal to each of these and then solve for $n$ and just take the ones that have a natural number solution for $n$.$

leehuan · Jan 31, 2016

InteGrand said:
$\noindent If we rationalise the numerator and apply difference of two squares, it becomes $\frac{-253}{\sqrt{n}-\sqrt{n+253}}=\frac{253}{\sqrt{n+253}-\sqrt{n}}$. Note that the denominator is monotone decreasing with limit 0. So for this to be a natural number, we need the denominator here to be a factor of 253. The factors of 253 are: $1,11,23,253$. So we can set the denominator equal to each of these and then solve for $n$ and just take the ones that have a natural number solution for $n$.$

Ah ok yep. That makes better sense than these answers. But mind interpreting them for me?

My brain isn't really working right now.

(Note: These answers appear as though they assume not even much prelim maths has been covered either)

Drsoccerball · Jan 31, 2016

leehuan · Jan 31, 2016

Drsoccerball said:
$$ Using $ : \int{\frac{\sin h(x)}{\cos h(2x)}}dx $ and $ \int{\frac{\cos h(x)}{\cos h(2x)}}dx$

$$ Find: $ \int_0^1{\frac{1}{1+x^4}}dx$

._. ?

Drsoccerball · Jan 31, 2016

leehuan said:
._. ?

Can't I post questions here aswell

?

Speed6 · Jan 31, 2016

Is this a one-stop place where I can ask any maths question I like, without making a new thread each time I do?

leehuan · Jan 31, 2016

Wow guys this is my SOS thread for when I need help

Speed6 · Jan 31, 2016

leehuan said:
Wow guys this is my SOS thread for when I need help

Ok no problem, I just thought everyone posts their maths problems here and you just help out whenever.

InteGrand · Jan 31, 2016

Drsoccerball said:
Can't I post questions here aswell ?

Haha I think this was an SOS thread. And in your question, did you mean sinh and cosh? These have LaTeX codes too, just write like \sinh and \cosh.

Drsoccerball · Jan 31, 2016

InteGrand said:
Haha I think this was an SOS thread. And in your question, did you mean sinh and cosh? These have LaTeX codes too, just write like \sinh and \cosh.

I tried that it didn't work? Anyways how would you do the question? $\sinh{x}$

leehuan · Jan 31, 2016

Drsoccerball said:
I tried that it didn't work? Anyways how would you do the question? $\sinh{x}$

Just write the hyperbolic functions in the exponential form.

turntaker · Jan 31, 2016

Drsoccerball said:
$$ Using $ : \int{\frac{\sin h(x)}{\cos h(2x)}}dx $ and $ \int{\frac{\cos h(x)}{\cos h(2x)}}dx$

$$ Find: $ \int_0^1{\frac{1}{1+x^4}}dx$

calculus shit tHat’s sum cAlcuLUς $#!t thats∰some good∰∭integration right∳∰there right⨌there⨍find the limitfind thelimitthat’s not showing your work right tᴴᵉᴿe show w0rk(chorus: ᵍᴵᵛᴱn ⁽ˣ⁾)mMMMMMM∰∯∰ НO0ОଠOOOOOОଠଠOoooᵒᵒᵒᵒᵒᵒᵒᵒᵒ∰

CAlcUlUs $#!t

InteGrand · Jan 31, 2016

Drsoccerball said:
I tried that it didn't work? Anyways how would you do the question? $\sinh{x}$

Haha it worked this time. And you should probably post that question in the extracurricular integration thread so that we don't derail this thread.

Drsoccerball · Jan 31, 2016

leehuan said:
Just write the hyperbolic functions in the exponential form.

The first part is easy but the second how can we get it in that form we can't multiply integrals.

leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

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