An exercise from Math StackExchange

Problem:

Give all the positive whole number solutions to the equation $ x^3−y^3 = 602 $

Solution:

Here is really just a replication of my answer on math.stackexchange.com:

Note that $ (x - y)^2 = x^2 - 2xy + y^2 $, we can write

$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x - y)((x - y)^2 + 3xy) $

For simplicity, let $ z = x - y $, we have

$ 602 = z(z^2 + 3xy) $

Suppose for a moment that $ z $ is known, now we can calculate

$ z^2 + 3xy = \frac{602}{z} $

$ 3xy = \frac{602}{z} - z^2 $

$ 3(x - y + y)y = \frac{602}{z} - z^2 $

$ 3(z + y)y = \frac{602}{z} - z^2 $

$ 3zy + 3y^2 = \frac{602}{z} - z^2 $

$ 3y^2 + 3zy + z^2 - \frac{602}{z} = 0 $

Despite the deceiving complexity, since $ z $ is assumed to be known, we can easily find $ y $ using the quadratic formula.

Now we have $ 602 = 2 \times 7 \times 43 $, so $ z $ can only be these options

* 1
* 2
* 7
* 43
* $ 2 \times 7 $
* $ 2 \times 43 $
* $ 7 \times 43 $
* $ 2 \times 7 \times 43 $

And the negative of these values

Out of these 16 choices, we can easily enumerate the solutions. Of course, many of these choices does not generate integer solution, just ignore them.

For example, if I choose $ z = 2 $, we get $ 11^3 - 9^3 = 602 $ and also $ (-9)^3 - (-11)^3 = 602 $.

Exercise from Quora

Problem:

Please find the problem here.

Solution:

$\begin{eqnarray*} (0! + 0! + 0!)! &=& 6 \\ (1 + 1 + 1)! &=& 6 \\ 2 + 2 + 2 &=& 6 \\ (3 + 3 - 3)! &=& 6 \\ (4 - 4 / 4)! &=& 6 \\ 5 + 5 / 5 &=& 6 \\ 6 + 6 - 6 &=& 6 \\ 7 - 7 / 7 &=& 6 \\ (\sqrt{8 + 8 / 8})! &=& 6 \\ (\sqrt{9 + 9 - 9})! &=& 6 \end{eqnarray*} $

Exercise from 9gag.com?

Problem:

http://9gag.com/gag/aVDxn1O

Solution:

Denote bottle by $ b $

$ b + b + b = 30 $, therefore $ b = 10 $.

Denote the hamburger by $ h $

$ b + h + h = 20 $, therefore $ h = 5 $

Denote the beer by $ e $

$ h + e + e = 9 $, therefore $ e = 2 $.

Now here is the fun

$ \int\limits_{2h - b}^{\infty}{\frac{b\sin x}{ex}dx} = \int\limits_{0}^{\infty}{\frac{10\sin x}{2x}dx} = 5\int\limits_{0}^{\infty}{\frac{\sin x}{x}dx} = \frac{5\pi}{2} $

The last integral is well known, for example, see:
http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2

This video explains the Feynman trick!
https://www.youtube.com/watch?v=3LsXWPzlOhQ

Have fun!

Mathematical Analysis - Exercise 1.3

Problem:

Solution:

Suppose $ n = ab $ is not prime, then we can write $ 2^{ab} = (2^{a})^b $. Applying the identity we just proved in the previous problem, we have:

$ 2^{ab} - 1 =  (2^{a})^b - 1 =  (2^{a})^b - 1^b = (2^a - 1)(\cdots) $.

Therefore if $ a \ne 1 $, we have a non-trivial factor for $ 2^{ab} - 1 $.

The contradiction show $ n $ has to be prime.

Mathematical Analysis - Exercise 1.2

Problem:

Solution:

The right hand side sounds like something we can telescope, let's see:

$ \begin{eqnarray*} & & (a - b)\sum\limits_{k = 0}^{n-1}{a^{k}b^{n-1-k}} \\ &=& a\sum\limits_{k = 0}^{n-1}{a^{k}b^{n-1-k}} - b\sum\limits_{k = 0}^{n-1}{a^{k}b^{n-1-k}} \\ &=& \sum\limits_{k = 0}^{n-1}{a^{k+1}b^{n-1-k}} - \sum\limits_{k = 0}^{n-1}{a^{k}b^{n-k}} \\ &=& (a^n + \sum\limits_{k = 0}^{n-2}{a^{k+1}b^{n-1-k}}) - (\sum\limits_{k = 1}^{n-1}{a^{k}b^{n-k}} + b^n ) \\ &=& (a^n + \sum\limits_{k = 1}^{n-1}{a^{k}b^{n-k}}) - (\sum\limits_{k = 1}^{n-1}{a^{k}b^{n-k}} + b^n ) \\ &=& a^n - b^n \end{eqnarray*} $

Mathematical Analysis - Exercise 1.1

Problem:

Solution:

Suppose there exists a largest prime, that means there is only finite number of primes. Consider the product of them plus 1. This number cannot be a prime number because it is larger than the largest prime.

Now consider its prime factorization. Note that when this number is divided by any prime, the remainder 1, therefore, there is just no way of prime factorizing it, contradicting the fundamental theorem of arithmetic, therefore there is no largest prime!

An exercise about Lagrange mean value theorem

Problem:

Solution:

This is a really bored after lunch hour, so I decided let's do a simple calculus exercise to wake my brain up.

The $ \frac{1}{1 + u^2} $ reminded me this has something to do with $ \tan $. So let's do this simple integration.

$ \int{\frac{du}{1+u^2}} = \int{\frac{\sec^2\theta d\theta}{1+\tan^2 \theta}} = \int{d\theta} = \theta = \tan^{-1}u $.

That's refresh my memory! $ \frac{1}{1 + u^2} $ is the derivative of $ \tan^{-1} u $.

Now we apply the Lagrange's mean value theorem to get:

$ \frac{\tan^{-1}v - \tan^{-1}u}{v - u} = \frac{1}{1 + \delta^2} $.

Where $ u < \delta < v $.

$ \tan^{-1}v - \tan^{-1}u= \frac{v - u}{1 + \delta^2} $.

Therefore we conclude (remember when you increase the denominator, the value decrease, and vice versa):

$ \frac{v-u}{1 + v^2} < \tan^{-1}v - \tan^{-1}u< \frac{v - u}{1 + u^2} $.

Last but not least, set $ u = 1 $ and $ v = \frac{4}{3} $ gives

$ \frac{\frac{4}{3}-1}{1 + \left(\frac{4}{3}\right)^2} < \tan^{-1}\frac{4}{3} - \tan^{-1}1< \frac{\frac{4}{3}-1}{1 + 1^2} $.

$ \frac{3}{25} < \tan^{-1}\frac{4}{3} - \tan^{-1}1< \frac{1}{6} $.

$ \frac{\pi}{4} + \frac{3}{25} < \tan^{-1}\frac{4}{3} < \frac{\pi}{4} + \frac{1}{6} $.