#proofinatoot that a number \(a\)

is divisible by \(2^n\) if the \(n^{th}\) last digit of \(a\) are divisible by \(2^n\).

We note \(a_i\) the \(i^{th}\) digit of \(a\). Then we get

\[S = \frac{10^wa_w}{2^n} + \ldots + \frac{10^0a_0}{2^n}.\]

yet, \(10^x \equiv 0\pmod{2^n} \iff x - n\)

\[\ge 0 \iff x \ge n\]

Then, only the \(n^{th}\) first digit of \(a\) need to be divisible by \(2^{n}\).

#proofinatoot that if the sum of the digits of a number \(a\) is divisible by 9, then this number is divisible by 9.

Let \(\sum_{i=0}^{n}10^{i}a_i = a, a_i \in {0,...,9}\).

When we divide it by 9, we get

\[ \frac{a}{9} = \frac{a_n10^n}{9} + \ldots + \frac{a_010^0}{9}\]

\[= \frac{a_n(10^n - 1) + a_n}{9} + \ldots + \frac{a_0}{9}\]

\[= a_n \times \frac{10^n - 1}{9} + \ldots + \frac{\sum_{i=0}^{n}a_i}{9}\]

So since all the \(10^k - 1\) factors are divisible, it remains the last term, our sum.

A odd number \(x\) is noted \(2k\). We are looking at \(\sum_{i=1}^{n}(2i) = S\).

\begin{align}

S &= \sum_{i=1}^{n}(2i) \\

&= 2 \times \frac{n(n + 1)}{2}\\

&= n(n + 1).

\end{align}

#proofinatoot that \[\sum^{n}_{k=1}k^2 = \frac{n(n + 1)(2n + 1)}{6}.\]

For that, we will need to imagine a regular staking of ball forming a rectangular pyramid. The first floor will contain \(n^2\) ball, etc.

Then, instead of counting the ball vertically, we will count then horizontally, witch give us:

\[\sum^{n}_{k=1}k^2 = n \times \frac{n(n + 1)}{2} - \sum^{n - 1}_{k=1}\frac{k(k + 1)}{2}.\]

(only 500 caracters...)

\[=\frac{2n^3 + 3n^2 + n}{6}.\]

And after factoring, we have our formula.

#proofinatoot that the derivate of \(\sqrt{u(x)}' = \frac{u(x)'}{2\sqrt{u(x)}}\).

\[\sqrt{u(a)}' = \lim_{h \to 0} \frac{\sqrt{u(a + h)} - \sqrt{u(a)}}{h}\]

\[= \lim_{h \to 0} \frac{u(a + h) - u(a)}{h \times (\sqrt{u(a + h)} + \sqrt{u(a)})}.\]

Then we isolate \(u(a)'\):

\[\sqrt{u(a)}' = \lim_{h \to 0} \frac{u(a)'}{\sqrt{u(a + h)} + \sqrt{u(a)}}\]

\[= \frac{u(a)'}{2\sqrt{u(a)}}.\]

e.g.: The derivate of \(f(x) = \sqrt{1 - x^2}\) is \(f'(x) = -\frac{x}{\sqrt{1 - x^2}}\).

#proofinatoot that the equation \(f(x)=ax^2+bx+c\) is always tangent to \(g(x)=-ax^2+bx+c\) with \((a,c,b) \in \mathbb{R}^3\).

We get:

\(ax^2+bx+c=-ax^2+bx+c\)

\(2ax^2=0\)

Since b and c are canceled, the discriminant equals 0. There is only one solution, i.e. only one point I of intersection.

\(x_I = 0; y_I = f(0) = c\).

#proofinatoot that the perpendicular to the line of equation \(y = mx\) is \(y = -\frac{1}{m}x\).

The angle between the abcsissa axis is \(\theta = \arctan(m)\).

So we are looking at a slope \(m'\) (where \(\alpha = \arctan(m')\)), such that \(\theta - \alpha = \frac{\pi}{2}\). Indeed, we will use radians.

By rearranging, we are getting,

\[m' = \tan(\arctan(m) - \frac{\pi}{2})\]

\[\iff m' = -\frac{1}{m}\]

because \(\tan = \frac{sin}{cos}\).

Finally, the slope \(m'\) is \(-\frac{1}{m}\).

#proofinatoot that there are infinitely many solutions to the equation \(xy = x + y\).

i.e. \[1 = \frac{1}{x} + \frac{1}{y}\]

Let \(x = c/a\) and \(y = c/b\). We have \(1 = \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\)

\(\iff c = a + b\).

So \(xy = x + y\) is satisfied when \(x = c/a\), \(y = c/b\) where \(c = a + b\), for any value of a and b.

e.g.: \(\frac{37}{15} + \frac{37}{22} = \frac{37}{15} \times \frac{37}{22} = \frac{1369}{330}\).

A odd number \(x\) can be noted \(2k-1\). We are looking at \(\sum_{i=1}^{n}(2i - 1)\).

And we have \(\sum_{i=1}^{n}(-1) = -n\).

So \(\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n\)

\(= 2 \times \frac{n^2 + n}{2} - n = n^2\).

Its palindrome is \(\sum_{i=1}^{n}10^{n-(i-1)}a_i = \overline{a}\).

Thus \(a - \overline{a} = \sum_{i=1}^{n}a_i(10^{i-1} - 10^{n-(i-1)})\), or we know that \(\forall a,b \in \mathbb{N}, 10^a - 10^b \equiv 0\pmod 9\).

Consequently, \(a - \overline{a} \equiv 0\pmod 9\).

Joined May 2017