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 $$2^{n}$$.

Does anyone know a method similar to the vectorial product, which allows to create a plan from three plan? (in space) thank you !

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.

that the sum of the first $$n^{th}$$ even numbers is equal to $$n(n + 1)$$.
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}

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.

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 $$f'(x) = -\frac{x}{\sqrt{1 - x^2}}$$.

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.

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

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 $$-\frac{1}{m}$$.

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.: $$\frac{37}{15} + \frac{37}{22} = \frac{37}{15} \times \frac{37}{22} = \frac{1369}{330}$$.

that the sum of the `$$n^{th}$$ first odd numbers is equal to $$n^2$$.
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 $$\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n$$
$$= 2 \times \frac{n^2 + n}{2} - n = n^2$$.

that every natural number minus its palindrome is divisble by 9. Let $$\sum_{i=1}^{n}10^{i-1}a_i = a, a_i \in {0,...,9}$$.
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, $$a - \overline{a} \equiv 0\pmod 9$$.

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