heureka

It took Lara five days to read a novel. Each day after the first day,
Lara read half as many pages as the day before.
If the novel was 248 pages long,
how many pages did she read on the first day?

Sequence is a geometric sequence:

Formula:
\(a_n = a\cdot r^{n-1} \\ sum~ s_n = \dfrac{a\left(1-r^{n+1}\right)} {1-r} \)

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{a\left(1-r^{n}\right)} {1-r} \quad & | \quad n=5,~ r=\dfrac12,~ s_n = 248 \\\\ 248 &=& \dfrac{a\left(1-\left(\dfrac12 \right)^{5}\right)} {1-\dfrac12} \\\\ 248 &=& \dfrac{a\left(1-\dfrac{1}{2^5}\right)} {\dfrac12} \\\\ 248 &=& 2a\left(1-\dfrac{1}{32}\right) \quad & | \quad :2 \\\\ 124 &=& a\left(1-\dfrac{1}{32}\right) \\\\ a\left(1-\dfrac{1}{32}\right) &=& 124 \\\\ a\left(\dfrac{32-1}{32}\right) &=& 124 \\\\ a\left(\dfrac{31}{32}\right) &=& 124 \quad & | \quad \cdot \dfrac{32}{31} \\\\ a &=& 124\cdot \dfrac{32}{31} \\\\ \mathbf{a} & \mathbf{=} & \mathbf{128} \\ \hline \end{array}\)

Lara read 128 pages on the first day.

Ochsengespann

Herr Hofer kehrte von der Stadt nach Hause zurück.

Zwei Drittel des Weges fuhr er mit dem Fahrrad viermal schneller, als wenn er zu Fuss gegangen wäre.

Dann stieg er auf sein wartendes Ochsengespann um, was die Fahrt wesentlich verlangsamte.

Am Ende kam er zur gleichen Zeit nach Hause, wie wenn er zu Fuss gegangen wäre.

Zu Fuss legt Herr Hofer 5 km in einer Stunde zurück.

Wie schnell fährt sein Ochsengespann?

\(\text{$\mathbf{s}$ ist der Weg von der Stadt nach Hause.}\\ \text{$\mathbf{v_{f}}$ ist die Geschwindigkeit die Herr Hofer mit dem Fahrrad fährt.}\\ \text{$\mathbf{t_{f}}$ ist die Zeit die Herr Hofer mit dem Fahrrad fährt.}\\ \text{$\mathbf{v_{O}}$ ist die Geschwindigkeit die Herr Hofer mit dem Ochsengespann fährt.}\\ \text{$\mathbf{t_{O}}$ ist die Zeit die Herr Hofer mit dem Ochsengespann fährt.}\\ \text{$\mathbf{v =5\ \frac{\text{km}}{\text{h}}}$ ist Herrn Hofers Geschwindigkeit, wenn er zu Fuss geht.}\\ \text{$\mathbf{t}$ ist die Zeit, wenn er zu Fuss von der Stadt nach Hause geht.}\)

\(\begin{array}{|lrcll|} \hline & \frac23s&=&v_f\cdot t_f \quad & | \quad v_f = 4v \\ & \frac23s&=&4v\cdot t_f \\ (1) & t_f &\mathbf{=}& \frac{s}{6v} \\\\ & \frac13s &=& v_O\cdot t_O \\ (2) & t_O & \mathbf{=}& \frac{s}{3v_O} \\\\ & s &=& v\cdot t \quad & | \quad t = t_f+t_O \\ & s &=& v\cdot (t_f+t_O) \\ & s &=& v\cdot (\frac{s}{6v}+\frac{s}{3v_O}) \\ & s &=& s\cdot v\cdot (\frac{1}{6v}+\frac{1}{3v_O}) \\ & 1 &=& v\cdot (\frac{1}{6v}+\frac{1}{3v_O}) \\ & \frac1v &=& \frac{1}{6v}+\frac{1}{3v_O} \\ & \frac1v - \frac{1}{6v} &=& \frac{1}{3v_O} \\ & \frac{1}{3v_O} &=& \frac1v - \frac{1}{6v} \\ & \frac{1}{3v_O} &=& \frac1v(1- \frac{1}{6} ) \\ & \frac{1}{3v_O} &=& \frac1v \cdot \frac{5}{6} \\ & \frac{1}{3v_O} &=& \frac{5}{6v} \\ & \frac{3v_O}{1} &=& \frac{6v}{5} \\ & v_O &=& \frac{6v}{3\cdot 5} \\ & v_O &=& \frac{6\cdot 5\ \frac{\text{km}}{\text{h}}}{15} \quad & | \quad v =5\ \frac{\text{km}}{\text{h}} \\\\ & v_O &=& \mathbf{2\ \frac{\text{km}}{\text{h}} }\\ \hline \end{array}\)

Sein Ochsengespann fährt \(\mathbf{2\ \frac{\text{km}}{\text{h}} } \) schnell.

3x+4y=72

I need the x and y intercepts.

\(\begin{array}{|rcll|} \hline 3x+4y &=& 72 \quad & | \quad :72 \\\\ \dfrac{3x+4y}{72} &=& 1 \\\\ \dfrac{3x}{72} + \dfrac{4y}{72} &=& 1 \\\\ \dfrac{x}{\dfrac{72}{3}} + \dfrac{y}{\dfrac{72}{4}} &=& 1 \\\\ \dfrac{x}{24} + \dfrac{y}{18} &=& 1 \\\\ \text{x-intercept}~ = 24, \text{if }~y=0\\ \text{y-intercept}~ = 18, \text{if }~x=0\\ \hline \end{array}\)

Part (b):

Find all pairs of positive integers \((a, n)\)  such that \(n \ge 2\) and 

\(a + (a + 1) + (a + 2) + \ldots + (a + n - 1) = 100\).

\(\begin{array}{|rcll|} \hline a + (a + 1) + (a + 2) + \dots + (a + n - 1) &=& 100 \\\\ \underbrace{(a + a + a + \ldots + a)}^{n ~\text{times}}_{\text{sum}=n\cdot a} + \underbrace{( 1 + 2 + \ldots + n - 1)}_{\text{sum}=\frac{1+(n-1)}{2} \cdot(n-1) }&=& 100 \\\\ n\cdot a + \dfrac{1+(n-1)}{2} \cdot(n-1) &=& 100 \\\\ n\cdot a + \dfrac{n(n-1)}{2} &=& 100 \quad & | \quad \cdot 2 \\\\ 2n\cdot a + n(n-1) &=& 200 \\ \mathbf{n\cdot (\underbrace{2a+n-1)}_{=b}} &\mathbf{=}& \mathbf{\underbrace{200}_{=n\cdot b}} \\ \hline \end{array} \)

Divisors of 200:

\(\begin{array}{|rcll|} \hline 200 = n &\cdot& b = \\ 1 &\cdot &200 \\ 2 &\cdot &100 \\ 4 &\cdot &50 \\ 5 &\cdot &40 \\ 8 &\cdot &25 \\ 10& \cdot & 20 \\ \hline \end{array} \)

\(\mathbf{(a,n) =\ ?}\)

\(\begin{array}{|r|r|r|c|} \hline n \ge 2 & b=2a+n-1 & a = \frac{b-n+1}{2} & a ~ \text{is integer} & (a,n) \\ \hline 2 & 100 \\ \hline 4 & 50 \\ \hline \color{red}5 & 40 & \color{red}18 & \checkmark & (18,5) \\ \hline \color{red}8 & 25 & \color{red}9 & \checkmark & (9,8) \\ \hline 10 & 20 \\ \hline \end{array} \)

\(\begin{array}{llcr} (18,5) :~ 18 + (18+1)+ (18+2)+ (18+3)+ (18+4)&=&100 \\ (9,8) :~ 9 + (9+1)+ (9+2)+ (9+3)+ (9+4)+ (9+5)+ (9+6)+ (9+7)&=&100 \\ \end{array}\)

A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies.

What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?

Formula:

\(\begin{array}{|rcll|} \hline \gcd(a,b,c) &=& \gcd( \gcd(a,b), c ) \\ \gcd(a,b ) &=& \gcd (a-b,b ) & a \ge b \\ \gcd(a,a) &=& a & a \ge 0 \\ \gcd(a,b) &=& \gcd(b,a) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ && \gcd(30,24) \\ && = \gcd(30-24,24) \\ && = \gcd(6,24) \\ && = \gcd(24,6) \\ && = \gcd(24-6,6) \\ && = \gcd(18,6) \\ && = \gcd(18-6,6) \\ && = \gcd(12,6) \\ && = \gcd(12-6,6) \\ && = \gcd(6,6) \\ && = 6 \\\\ \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ &=& \gcd( 6, 12 ) \\ &=& \gcd( 12, 6 ) \\ &=& \gcd( 12-6, 6 ) \\ &=& \gcd( 6, 6 ) \\ &=& 6 \\ \hline \end{array}\)

The greatest common divisor of two integers is \((x+2)\)
and their least common multiple is \(x(x+2)\),
where \(x\) is a positive integer.

If one of the integers is \(24\),
what is the smallest possible value of the other one ?

Let \(b\) is the value of the other one

greatest common divisor\((24,b) = x+2 \text{ and } x \in N\)

least common multiple\((24,b) = x(x+2) \text{ and } x \in N\)

Formula:

\(\begin{array}{|rcll|} \hline \text{greatest common divisor}(a,b) \times \text{least common multiple}(a,b) = a\times b \\ \hline \end{array}\)

\(\mathbf{b =\ ?}\)

\(\begin{array}{|rcll|} \hline 24\times b &=& (x+2)\times x(x+2) \\\\ \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{x(x+2)^2}{24}} \qquad x \in N, ~ x \gt 0 \\ \hline \end{array} \)

\(\begin{array}{|r|c|c|} \hline x \in N & \mathbf{b=\dfrac{x(x+2)^2}{24}} & \text{integer}\ ? \\ \hline 1 & \dfrac{3}{8} \\\\ 2 & \dfrac{4}{3} \\\\ 3 & \dfrac{25}{8} \\\\ 4 & \color{red}6 & \checkmark \\\\ \cdots & \cdots \\\\ \hline \end{array}\)

The smallest possible value of the other one is \(\mathbf{6}\) and \(x = 4\)

Check:

\(\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}\)

Define
\(f(x)=\dfrac{1+x}{1-x}\)
and
\(g(x)=\dfrac{-2}{x+1}\).
Find the value of
\(g(f(g(f(\ldots g(f(12)) \ldots ))))\)
where there are 16 compositions of the functions g, and f,
alternating between the two.

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{x+1}{1-x} \\\\ g(f(x)) = \dfrac{-2}{\dfrac{x+1}{1-x}+1} &&=& x-1 \\\\ f(g(f(x))) = \dfrac{x-1+1}{1-(x-1)} &=& \dfrac{x+0}{2-x} \\\\ g(f(g(f(x)))) = \dfrac{-2}{\dfrac{x}{2-x}+1}& &=&x-2 \\\\ f(g(f(g(f(x))))) = \dfrac{x-2+1}{1-(x-2)} &=& \dfrac{x-1}{3-x} \\\\ g(f(g(f(g(f(x))))))=\dfrac{-2}{\dfrac{x-1}{3-x}+1} & &=& x-3 \\\\ f(g(f(g(f(g(f(x))))))) = \dfrac{x-3+1}{1-(x-3)} &=& \dfrac{x-2}{4-x} \\\\ g(f(g(f(g(f(g(f(x))))))))=\dfrac{-2}{\dfrac{x-2}{4-x}+1} & &=& x-4 \\\\ \ldots \\ \hline \end{array}\)

If \(h(x) = g(f(x))\) 8 times, then:

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-8\quad & | \quad x = 12 \\ &=& 12-8 \\ &=& 4 \\ \hline \end{array} \)

Find

\(\large{\left(\dfrac{1+i}{\sqrt{2}}\right)^{46}}\)

\left(\frac{1+i}{\sqrt{2}}\right)^{46}.

\(\begin{array}{|rcll|} \hline && \left(\dfrac{1+i}{\sqrt{2}}\right)^{46} \\\\ &=& \dfrac{ \left(1+i \right)^{46} }{ \left(\sqrt{2} \right)^{46} } \\\\ &=& \dfrac{ \left( \left(1+i \right)^{2}\right)^{23} }{ 2^{\frac{46}{2}} } \quad & | \quad \left(1+i \right)^{2} = 2i \\\\ &=& \dfrac{ \left( 2i \right)^{23} } { 2^{23} } \\\\ &=& \dfrac{ 2^{23}i^{23} } { 2^{23} } \\\\ &=& i^{23} \\\\ &=& \left(i^{2}\right)^{11}i \quad & | \quad i^2 = -1 \\\\ &=& \left(-1 \right)^{11}i \\\\ &\mathbf{=}& \mathbf{-i} \\ \hline \end{array}\)

When working modulo m, the notation \(a^{-1}\) is used to denote the residue \(b\) for which \(ab\equiv 1\pmod{m}\),
if any exists.
For how many integers a satisfying \(0 < a < 100\) is it true that \(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\) ?

\(\begin{array}{|rcll|} \hline a(a-1)^{-1} &\equiv& 4a^{-1} \pmod{20} \quad & | \quad \cdot a \\ a^2(a-1)^{-1} &\equiv& 4a^{-1}\cdot a \pmod{20} \quad & | \quad a^{-1}\cdot a = \dfrac{a}{a} = 1 \\ a^2(a-1)^{-1} &\equiv& 4 \pmod{20} \quad & | \quad a^2(a-1)^{-1} = \dfrac{a^2}{a-1} = a+1 + \dfrac{1}{a-1} \\ a+1 + \underbrace{\dfrac{1}{a-1}}_{\text{is integer, only if }a=2} &\equiv& 4 \pmod{20} \quad & | \quad a = 2 \\\\ 2+1 + \dfrac{1}{2-1} &\equiv& 4 \pmod{20} \quad & | \quad \\ 3 + 1 &\equiv& 4 \pmod{20} \\ 4 &\equiv& 4 \pmod{20} \ \checkmark \\ \hline \end{array}\)

One Integer solution \(\mathbf{a=2}\)

Betty goes to the store to get flour and sugar.
The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar,
and is no more than twice the amount of sugar.
Find the least number of pounds of sugar that Betty could buy.

Let \(y\) = flour
Let \(x\) = sugar

\(\begin{array}{|rcll|} \hline y &\ge& 6+\dfrac x2 \\ y &\le& 2x \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && & y \ge 6+\dfrac x2 \\ && & y \le 2x \\\\ \underline{x_{\text{min}}:} \\ 2x &=& 6+\dfrac x2 \\\\ 2x-\dfrac x2 &=& 6 \\\\ \dfrac32\cdot x &=& 6 \\\\ x &=& \dfrac23 \cdot 6 \\\\ \mathbf{ x} & \mathbf{=} & \mathbf{4\ \text{pounds}} \\ \hline \end{array}\)

The least number of pounds of sugar that Betty could buy are \(\mathbf{4\ \text{pounds}}\).