MaxWong

An interesting theorem: If $z + \dfrac1z = 2\cos \alpha$, then $z^n + \dfrac1{z^n} = 2\cos n\alpha$ for positive integers n and real $\alpha$.

(If you don't like to read proofs, then you can jump to the point where I marked "END OF PROOF".)

Proof of the theorem:

We use strong induction.

Base case #1 : n = 1.

Substituting n = 1 and checking shows that base case #1 is true.

Base case #2 : n = 2.

$z^2 + \dfrac1{z^2} = \left(z + \dfrac1z\right)^2 - 2 = (2\cos \alpha)^2 - 2=2(2\cos^2 \alpha - 1) = 2\cos 2\alpha$

So base case #2 is true.

Inductive step: Assume $z^t + \dfrac1{z^t} = 2\cos t\alpha$ and $z^{t - 1} + \dfrac1{z^{t - 1}} = 2\cos(t -1)\alpha$ for some positive integer $t$.

$\quad z^{t + 1} + \dfrac1{z^{t + 1}}\\ = \left(z^t + \dfrac1{z^t}\right)\left(z + \dfrac1z\right) - \left(z^{t - 1} + \dfrac1{z^{t - 1}}\right)\\ = 2\cos t\alpha \cdot 2\cos \alpha - 2\cos (t - 1)\alpha\\ \text{By product to sum formula:}\\ = 2 \left(\cos (t + 1)\alpha + \cos(t - 1)\alpha\right) - 2\cos(t-1)\alpha\\ =2\cos(t + 1)\alpha$

Therefore by principles of mathematical induction, the proposition is true.

-------------------------------------------------------------------------(END OF PROOF)-------------------------------------------------------------------------

Back to the problem. Using the theorem I mentioned above, and substituting $\alpha = 60^\circ$ into the formula,

$\quad z^{10} + \dfrac1{z^{10}}\\ = 2\cos\left(10\cdot 60^\circ\right)\\ = 2 \cos\left(600^\circ\right)\\ = 2 \cos 240^\circ\\ = \boxed{-1}$

We need 8 digits, but we only had 6, which is 001101. 

When we don't have enough digits, we add zeroes in front of the number. This is because doing this does not affect the value of the number.

We add 2 zeroes in front, so it gives 00001101.

It is good if we can express everything in terms of x. Because it will become a linear equation in one variable. 

In fact, we can. Manipulating the first equation:

3x = 5y

y = 3x/5 = 0.6 x

3x = 8z

z = 3x/8 = 0.375 x

Then, we plug these into the equation:

$x-0.6x-0.375x=-3$

We simplify the left hand side:

$0.025x = -3$

Changing that 0.025 into a fraction gives

$\dfrac x{40} =-3\\ $

We multiply 40 to both sides to get

x = -120.

Try to expand and you will get a linear equation. After that you can solve it by moving terms around and simplifying it.

(x + A)(x + B) = (x + C)(x + D)

x² + (A + B)x + AB = x² + (C + D)x + CD

(A + B)x + AB = (C + D)x + CD

(A + B)x - (C + D)x = CD - AB

(A + B - C - D)x = CD - AB

x = ____________________

One step left for you to complete.

A story:

Imagine a wall being placed on the right of a binary number.

Whenever a number gets left shifted, a zero comes out of the wall to fill the gap between the wall and the number.

Whenever a number gets right shifted, the last digit just goes through the wall and never come back again.

Now, let's try left shifting and right shifting some numbers.

right shift of two places on 00110100.

The bar denotes the "wall", and blue digits means the digit is gone.

$00110100\Big|$

right shift once

$0011010\Big|\color{blue}0$

right shift once more

$001101\Big|\color{blue}00$

The result should be $001101$, but because you needs 8 bits for a byte, it becomes $00001101$. We add zeroes in front when we don't have enough digits.

Hints for the others:

Q2) Think of what a left shift and a right shift really does on the value of the binary number by trying out few examples.

Q3) Similar to Q1.

EDIT: Italicized text is what I changed.

See my answer here:
https://web2.0calc.com/questions/please-help-question-not-answered#r2

Group the terms in pairs. Using the formula a² - b² = (a - b)(a + b), we have

$\quad(a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\\ =\left((a +(2n+1)d)^2- (a + (2n)d)^2\right) +\left((a + (2n-1)d)^2 - (a+(2n-2)d)^2\right) + \cdots + \left((a+d)^2 - a^2\right)\\ =(\color{red}(a + (2n + 1)d)\color{black} - \color{blue}(a + (2n)d)\color{black})(\color{red}(a + (2n + 1)d)\color{black} + \color{blue}(a + (2n)d)\color{black}) \\\quad\quad+ (\color{red}(a + (2n-1)d)\color{black} - \color{blue}(a + (2n-2)d)\color{black})(\color{red}(a + (2n-1)d)\color{black} + \color{blue}(a + (2n-2)d)\color{black}) + \cdots \\\quad\quad\qquad+(\color{red}(a + d)\color{black} - \color{blue}a\color{black})(\color{red}(a + d)\color{black} + \color{blue}a\color{black})$

The last line may be a bit long, but if we simplify it:

$= d(2a + \color{red}(4n + 1)\color{black}d) + d(2a + \color{red}(4n - 3)\color{black}d) + \cdots + d(2a + \color{red}1\color{black}d)$

Now we see that the original expression is nothing other than the sum of A.S. in disguise.

Factorising out a common factor for a more easy-to-spot A.S.:

$= d((2a + (4n + 1)d) + (2a + (4n - 3)d) + \cdots + (2a + d))$

We can now see the expression inside the bracket is an A.S. with first term = (2a + (4n + 1)d), common difference = -4d, number of terms = (n + 1).

We can plug those into the formula:

$\quad\text{sum of A.S.} \\= \dfrac{\text{number of terms}}{2} \left(2\times \text{first term} + (\text{number of terms} - 1)\times \text{common difference}\right) \\= \dfrac{n + 1}{2}\left(2(2a + (4n + 1)d) + n(-4d)\right) \\= \dfrac{n + 1}2\left(4a + (4n + 2)d\right) \\= (n + 1)(2a+(2n + 1)d)$

The details are left to you.

Notice that the complex numbers are $\omega = \dfrac{-1 + i\sqrt 3}{2}$ and $\omega^2 = \dfrac{-1 - i\sqrt 3}{2}$ respectively, where $\omega$ is the cube root of unity.

The original expression equates to $\omega^{2021} + \left(\omega^2\right)^{2021} = \omega^{2021} + \omega^{4042}$.

Noticing that $\omega^3 = 1$, $\omega^{2021} + \omega^{4042} = \omega^2 + \omega = \dfrac{-1-i\sqrt 3}2+ \dfrac{-1+i\sqrt 3}2 = \boxed{-1}$.

This means that 

$\begin{cases}\text{sum of roots} = \dfrac k3\\\text{product of roots} = \dfrac{14}3\end{cases}$

Let the roots be 7p and 6p. Using the second formula,

$(7p)(6p) = \dfrac{14}3\\ p^2 = \dfrac19\\ p = \pm \dfrac13$

Using the first formula,

$13p = \dfrac k3\\ k = 39p = \pm13$