heureka

If  \(x + \dfrac{1}{x} = 1\),  then find the value of  \(x^{1000} + \dfrac{1}{x^{1000}}\)

\(\text{Let $x=e^{i\varphi} $} \\ \text{Let $\dfrac{1}{x}=x^{-1}=e^{-i\varphi}=e^{i(-\varphi)} $} \\ \text{Let $x^{1000}=e^{i(1000\varphi)} $} \\ \text{Let $\dfrac{1}{x^{1000}}=x^{-1000}=e^{-i\varphi 1000}=e^{i(-1000\varphi)} $}\)

\(\begin{array}{|rcll|} \hline x + \dfrac{1}{x} &=& 1 \\ x + x^{-1} &=& 1 \\ e^{i\varphi} + e^{i(-\varphi)} &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(-\varphi)+i\sin(-\varphi) &=& 1 \\ \cos(\varphi)+i\sin(\varphi)+\cos(\varphi)-i\sin(\varphi) &=& 1 \\ 2\cos(\varphi) &=& 1 \\ \cos(\varphi) &=& \dfrac{1}{2} \\ \varphi &=& \arccos\left( \dfrac{1}{2} \right) \\ \mathbf{ \varphi } &=& \mathbf{60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{x^{1000} + \dfrac{1}{x^{1000}}} \\ &=& e^{i(1000\varphi)} + e^{i(-1000\varphi)} \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(-1000\varphi)+i\sin(-1000\varphi) \\ &=& \cos(1000\varphi)+i\sin(1000\varphi)+\cos(1000\varphi)-i\sin(1000\varphi) \\ &=& 2\cos(1000\varphi) \quad | \quad \mathbf{ \varphi=60^\circ} \\ &=& 2\cos(1000*60^\circ) \\ &=& 2\cos(60000^\circ) \\ &=& 2\cos(240^\circ) \\ &=& 2(-0.5) \\ &=& \mathbf{-1} \\ \hline \end{array}\)

If we divide a 2-digit positive integer by the sum of its digits,
we get the quotient and remainder of 4 and 3, respectively.
If we divide the same 2-digit positive by the product of its digits,
we get quotient and remainder of 3 and 5, respectively.
What is the 2-digit integer?

The 2-digit integer \(ab\) is \(10a + b\)

\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the sum of its digits} \\ \end{array} \\ \hline (1): & \mathbf{\dfrac{10a + b}{a+b}} &=& \mathbf{4 + \dfrac{3}{a+b}} \quad | \quad \times (a+b) \\\\ & 10a + b &=& 4(a+b) + 3 \\ & 10a + b &=& 4a+4b + 3 \\ & 10a-4a + b-4b &=& 3 \\ & 6a-3b &=& 3 \quad | \quad : 3 \\ & 2a-b &=& 1 \\ & \mathbf{b} &=& \mathbf{2a-1} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & \begin{array}{rcll} \text{We divide a 2-digit positive integer}\\ \text{by the product of its digits} \\ \end{array} \\ \hline (2): & \mathbf{\dfrac{10a + b}{ab}} &=& \mathbf{3 + \dfrac{5}{ab}} \quad | \quad \times (ab) \\\\ & 10a + b &=& 3ab+5 \quad | \quad \mathbf{b=2a-1} \\ & 10a + 2a-1 &=& 3a(2a-1) + 5 \\ & 12a-1 &=& 6a^2-3a + 5 \\ & 6a^2 -15a +6 &=& 0 \quad | \quad : 3 \\ & \mathbf{2a^2 -5a +2} &=&\mathbf{ 0 } \\\\ & a &=& \dfrac{5\pm \sqrt{5^2-4*2*2} }{2*2} \\ & a &=& \dfrac{5\pm \sqrt{9}}{4} \\ & a &=& \dfrac{5\pm 3}{4} \\\\ & a &=& \dfrac{5+ 3}{4} \\\\ & a &=& \dfrac{8}{4} \\\\ & \mathbf{a} &=& \mathbf{2} \\\\ \text{or} & a &=& \dfrac{5- 3}{4} \\\\ & a &=& \dfrac{2}{4} \qquad \text{no integer }! \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{2a-1} \quad | \quad \mathbf{a=2} \\\\ b &=& 2*2-1 \\ \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}\)

The 2-digit integer ab is \(\mathbf{23}\)

In a 100 meter race, Allen and Bruce reached the finish line in 12 seconds and 15 seconds, respectively.
When Allen finished the race, how far was Bruce from the finish line?
Assume that they each move at a constant rate.

\(v_{\text{Bruce}} = \dfrac{100\ m}{15\ s} \)

\(\begin{array}{|rcll|} \hline s_{\text{Bruce}} &=& \dfrac{100\ m}{15\ s} \times (15\ s -12\ s) \\\\ s_{\text{Bruce}} &=& \dfrac{100\ m}{15\ s} \times (3\ s) \\\\ \mathbf{s_{\text{Bruce}}} &=& \mathbf{20\ \text{meter}} \\ \hline \end{array}\)

Bruce was \(\mathbf{20\ \text{meter}}\) from the finish line.

When a natural number x is divided by 5, the remainder is 2.
When a natural number y is divided by 5, the remainder is 4.
The remainder is z when x+y is divided by 5.
Find the value of \(\dfrac{2z - 5}{3}\).

\(\begin{array}{|lrcll|} \hline & x & \equiv & 2\pmod{5} \qquad (1) \\ & y & \equiv & 4\pmod{5} \qquad (2) \\ \hline (1)+(2): & x+y & \equiv & 2+4\pmod{5} \\ & x+y & \equiv & 6 \pmod{5} \\ & x+y & \equiv & 6-5 \pmod{5} \\ & x+y & \equiv & {\color{red}1} \pmod{5} \\ \hline & z &=& 1 \\\\ & && \mathbf{\dfrac{2z - 5}{3}} \\\\ & &=& \dfrac{2*1 - 5}{3} \\\\ & &=& \dfrac{-3}{3} \\\\ & &=& \mathbf{-1} \\ \hline \end{array}\)

Find all complex numbers z z^2=i

1)  \(i=\sqrt {-1}\)

2) Euler:

 \(e^{i\pi}=-1\\ \sqrt{e^{i\pi}}=\sqrt{-1 } \\ e^{i \frac{\pi}{ 2}}=i\)

\(\sqrt{ e^{ i \frac{\pi}{2} } } =\pm \sqrt{i} \\ e^{ i\frac{\pi}{4} } =\pm \sqrt{i} \\ \sqrt{i} = \pm e^{i\frac{\pi}{4} } \)

\(z^2=i \\ z= \sqrt{i}\)

\(z_1= + e^{i\frac{\pi}{4}} \\= \cos(45^\circ) + i \sin(45^\circ) \\ = \frac{\sqrt{2}} {2}+i \frac{\sqrt2} {2} \)

\(z_2= -e^{i\frac{\pi}{4}} \\=- \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\)

Find the value of the sum \(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98}\)

\((1+i)^{99}=\dbinom{99}{0}+i\dbinom{99}{1}-\dbinom{99}{2}-i\dbinom{99}{3}+\dbinom{99}{4}+i\dbinom{99}{5}- \ldots -\dbinom{99}{98}-i\dbinom{99}{99} \qquad (1)\)

Note here that \(\binom{99}{0} - \binom{99}{2} + \binom{99}{4}- \binom{99}{6} + \dots - \binom{99}{98}\) is the real part of (1).

\(\begin{array}{|rcll|} \hline 1+i &=& \sqrt{1^2+1^2} \Bigg(\cos\Big( \arctan(\frac{1}{1} )\Big) +i\sin\Big( \arctan(\frac{1}{1} )\Big) \Bigg) \\ 1+i &=& \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \\ \hline (1+i)^{99} &=& \Bigg( \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \Bigg)^{99} \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(99*45^\circ) +i\sin(99*45^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(4455^\circ) +i\sin(4455^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(135^\circ) +i\sin(135^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-\dfrac{\sqrt{2}}{2} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{\frac{1}{2}-1} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{-\frac{1}{2}} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}}*(-2^{-\frac{1}{2}}) +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}}2^{-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{98}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& \underbrace{\color{red}-2^{49}}_{\text{the real part}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ \hline \end{array} \)

\(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98} = \mathbf{-2^{49}}\)

Let a < b < c < d be four consecutive positive integers such that     

- a is divisible by 5,     
- b is divisible by 7,    
- c is divisible by 9, and      
- d is divisible by 11. 

Find the minimum value of a+b+c+d. 

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ b=a+1 & \equiv & 0 \pmod{7} \\ c=a+2 & \equiv & 0 \pmod{9} \\ d=a+3 & \equiv & 0 \pmod{11} \end{array}\right\} \begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ a & \equiv & -1 \pmod{7} \\ a & \equiv & -2 \pmod{9} \\ a & \equiv & -3 \pmod{11} \end{array} \\\\ a+b+c+d = 4a+6 \\ \end{array} \)

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ a & \equiv & -1 \pmod{7} \\ \end{array}\right\} \begin{array}{rcl} a & \equiv & -15 \pmod{35} \\ \end{array} \\ \left.\begin{array}{rcl} a & \equiv & -2 \pmod{9} \\ a & \equiv & -3 \pmod{11} \end{array}\right\} \begin{array}{rcl} a & \equiv & -47 \pmod{99} \\ \end{array} \end{array} \)

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & -15 \pmod{35} \\ a & \equiv & -47 \pmod{99} \\ \end{array}\right\} \begin{array}{rcl} a & \equiv & 1735 \pmod{3465} \\ a &=& 1735 + 3465n,\ n\in \mathbb{Z} \\ \end{array} \\ \end{array} \)

\(\text{The minimum value of $\mathbf{a}$ is $\mathbf{1735}$} \\ \text{The minimum value of $\mathbf{a+b+c+d} = 6*1735 + 4 \mathbf{=6946}$ } \)

Check:

\(\begin{array}{|rcll|} \hline 1735 & \equiv & 0 \pmod{5} \\ 1736 & \equiv & 0 \pmod{7} \\ 1737 & \equiv & 0 \pmod{9} \\ 1738 & \equiv & 0 \pmod{11} \\ \hline \end{array}\)

There are three consecutive multiples of five. The sum of the first two numbers is 1,945.
What are the integers?

\(\begin{array}{rcll} x~ \text{is the multiply of}~ 5 \\ \text{First} &=& x \\ 2\text{nd} &=& x+5 \\ 3\text{rd} &=& (x+5)+5 \\ \end{array} \)

\(\begin{array}{|rcll|} \hline x+x+5 &=& 1945 \\ 2x+5 &=& 1945 \\ 2x &=& 1945-5 \\ 2x &=& 1940 \\ \mathbf{x} &=& \mathbf{970} \\ \hline \end{array}\)

\(\begin{array}{rcll} \text{First} &=& 970 \\ 2\text{nd} &=& 970+5 \\ &=& 975 \\ 3\text{rd} &=& 975+5 \\ &=& 980 \\ \end{array}\)

Compute

\(S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +............+ 500^2/(999.1001)\)

Find the smallest positive integer n such that 135n is divisible by 360.

\(\begin{array}{|rcll|} \hline \text{lcm}(360,135) &=& 1080 \\ 135n &=& 1080 \\ n &=& \dfrac{1080}{135} \\ \mathbf{n} &=& \mathbf{8} \\ \hline \end{array}\)