heureka

Find polynomial f(n) such that for all integers $n \ge 1$ we have
$3\left( 1\cdot2 + 2\cdot3 + \ldots + n(n+1) \right) = f(n).$

$\begin{array}{|rcll|} \hline f(n) &=& 3\Big( 1\cdot2 + 2\cdot3 + 3\cdot4 + \ldots + n(n+1) \Big) \quad & | \quad n(n+1) = n^2+n \\ &=& 3 ( 1^2+1+2^2+2+3^2+3 + \ldots + n^2+n ) \\ &=& 3 ( \underbrace{1+2+3+ \ldots + n}_{=\dfrac{(n+1)n}{2}} + \underbrace{1^2 +2^2 +3^2 + \ldots + n^2}_{=\dfrac{(n+1)n(2n+1)}{6}} ) \\\\ &=& 3 \left( \dfrac{(n+1)n}{2} + \dfrac{(n+1)n(2n+1)}{6} \right) \\\\ &=& 3 \left[ \dfrac{(n+1)n}{2}\left(1+ \dfrac{2n+1}{3}\right) \right] \\\\ &=& \dfrac{(n+1)n}{2}\left(3+ 2n+1\right) \\\\ &=& \dfrac{(n+1)n}{2}\left(2n+4\right) \\\\ &=& (n+1)n(n+2) \\\\ &=& n(n+1)(n+2) \\\\ &\mathbf{=}& \mathbf{ n^3+3n^2+2n } \\\\ \hline \end{array}$

Triangle AHI is equilateral. We know BC, DE, and FG are all parallel to HI and AB = BD = DF = FH.
What is the ratio of the area of the trapezoid FGIH to the area of triangle AHI?
Express your answer as a common fraction.

$\text{Let $FG = s $} \\ \text{Let $HI = c $} \\ \text{Area of the triangle $AHI = A_{AHI}$ } \\ \text{Area of the triangle $AGF = A_{AGF}$ } \\ \text{Area of the trapezoid $FGIH = A_{FGIH}$ } \\ \text{Let $AK = h$ (height of the triangle$_{AGF}$) } \\ \text{Let $AL = H$ (height of the triangle$_{AHI}$) } \\ \text{Let $AF = \frac34 c $}$

1.

$\begin{array}{|rcll|} \hline A_{AHI} &=& A_{AGF} + A_{FGIH} \\\\ \dfrac{cH}{2} &=& \dfrac{sh}{2} + \left( \dfrac{s+c}{2}\right)(H-h) \quad & | \quad \times 2 \\\\ cH &=& sh + (s+c)(H-h) \\\\ \not{cH} &=& \not{sh} + sH-\not{sh}+\not{cH}-ch \\\\ \mathbf{ch} &\mathbf{=}& \mathbf{ sH } \qquad (1) \\\\ &\text{or}& \\\\ \mathbf{\dfrac{s}{c}} &\mathbf{=}& \mathbf{\dfrac{h}{H}} \qquad (2) \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline \text{ratio} &=& \dfrac{ A_{FGIH} } {A_{AHI}} \\\\ &=& \dfrac{ \left( \dfrac{s+c}{2}\right)(H-h) } {\dfrac{cH}{2}} \\\\ &=& \dfrac{(s+c)(H-h)}{cH} \\\\ &=& \dfrac{sH-sh+cH-ch}{cH} \quad & | \quad ch=sH \qquad (1) \\\\ &=& \dfrac{cH-sh}{cH} \\\\ &=& 1-\dfrac{sh}{cH} \quad & | \quad \dfrac{s}{c} = \dfrac{h}{H} \qquad (2) \\\\ &=& 1-\dfrac{h^2}{H^2} \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\dfrac{h}{H}\right)^2} \\\\ && \boxed{\mathbf{3.}\\ \dfrac{h}{\frac34 c} = \dfrac{H}{c} \\ h = \frac34 c \cdot \dfrac{H}{c} \\ h = \frac34 \cdot H \\ \mathbf{\dfrac{h}{H} = \frac34} } \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\frac34 \right)^2} \\\\ &=& 1- \frac{9}{16} \\\\ &=& \frac{16-9}{16} \\\\ &\mathbf{=}& \mathbf{\dfrac{7}{16}} \\ \hline \end{array}$

So  the ratio  of areas is $\mathbf{\dfrac{7}{16}}$

Solve and find the domain of the equation:

$\huge{ x^{\log_\sqrt{x} (x-2)}=9 }$

1. Domain:

$x-2 > 0 \\ x > 2$

$\text{ $\{x \in R : x>2\}$ (assuming a function from reals to reals) }$

2. Solve:

$\begin{array}{|rcll|} \hline \large{ x^{\log_\sqrt{x} (x-2)} }& \large{=}& \large{9} \\ && \boxed{ \log_\sqrt{x} (x-2) = \dfrac{\ln(x-2)}{\ln(\sqrt{x})} \\ = \dfrac{\ln(x-2)}{\ln(x^{\frac12})}\\ = \dfrac{\ln(x-2)}{\frac12\ln(x)}\\ = \dfrac{2\ln(x-2)}{\ln(x)}} \\ \large{ x^{\dfrac{2\ln(x-2)}{\ln(x)}}}& \large{=}& \large{9} \\ && \boxed{\text{Formula: } a^b = e^{\ln(a^b)} = e^{b\ln(a)} } \\ && \boxed{ x^{\dfrac{2\ln(x-2)}{\ln(x)}} = e^{ \dfrac{2\ln(x-2)}{\ln(x)} \cdot \ln(x) } =e^{2\ln(x-2)} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \large{ e^{2\ln(x-2)} } & \large{=} & \large{9} \\ \large{ e^{\ln\left((x-2)^2\right)} } & \large{=} & \large{9} \quad & | \quad e^{\ln\left((x-2)^2\right)} = (x-2)^2 \\ \large{ (x-2)^2 } & \large{=} & \large{9} \quad & | \quad \sqrt{} \text{ both sides} \\ \large{ x-2 } & \large{=} & \large{\pm 3} \\\\ x_1 -2 &=& 3 \\ x_1 &=& 3+2 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{5} \\\\ x_2 -2 &=& -3 \\ x_2 &=& -3+2 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-1} \quad & | \quad \text{no solution, see domain } x> 2 \\ \hline \end{array}$

I assume the question is:

$\mathbf{\text{A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed $\\$ so that the first three terms are in geometric progression, $\\$ the second, third, and fourth terms are in arithmetic progression, $\\$ and, in general, for all $n\ge1,$ $\\$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, $\\$ and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. $\\$ Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $.}}$

Let AP = arithmetic progression,

Let GP = geometric progression

$\text{AP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i(j-k) + t_j(k-i) + t_k(i-j) = 0 \\\\ & i = 2n & t_i = a_{2n} \\ & j = 2n+1 & t_j = a_{2n+1} \\ & k = 2n+2 & t_k = a_{2n+2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline t_i(j-k) + t_j(k-i) + t_k(i-j) &=& 0 \\ a_{2n}(2n+1-(2n+2)) + a_{2n+1}(2n+2-2n) + a_{2n+2}(2n-(2n+1)) &=& 0 \\ \mathbf{-a_{2n} + 2a_{2n+1} - a_{2n+2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}$

$\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ & i = 2n-1 & t_i = a_{2n-1} \\ & j = 2n & t_j = a_{2n} \\ & k = 2n+1 & t_k = a_{2n+1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ a_{2n-1}^{2n-(2n+1)} \times a_{2n}^{2n+1-(2n-1)} \times a_{2n+1}^{2n-1-(2n)} = 1 \\\\ \mathbf{a_{2n-1}^{-1} \times 2a_{2n}^{2} \times a_{2n+1}^{-1} } &\mathbf{=}& \mathbf{1} \\ \hline \end{array}$

So we have:

$\begin{array}{|lrcll|} \hline (1) & \mathbf{a_{2n+2}} &\mathbf{=}& \mathbf{2a_{2n+1} -a_{2n} } \\ (2) & \mathbf{a_{2n+1} } &\mathbf{=}& \mathbf{\dfrac{ a_{2n}^{2} }{a_{2n-1} } } \\ \hline \end{array}$

$\mathbf{a_1=1 }\\ \begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_2^2}{a_1} \quad a_1 = 1 & | \quad \text{ Formula $(2)$} \\ \mathbf{a_3} & \mathbf{=} & \mathbf{ a_2^2 } \\ \hline a_4 &=& 2a_3-a_2 \quad a_3 = a_2^2 & | \quad \text{ Formula $(1)$} \\ &=& 2a_2^2-a_2 \\ \mathbf{a_4} & \mathbf{=} & \mathbf{ a_2(2a_2-1) } \\ \hline a_5 &=& \dfrac{a_4^2}{a_3} & | \quad \text{ Formula $(2)$} \\ a_5 &=& \dfrac{a_2^2(2a_2-1)^2}{a_2^2} \\ \mathbf{a_5} & \mathbf{=} & \mathbf{ (2a_2-1)^2 } \\ \hline a_6 &=& 2a_5-a_4 & | \quad \text{ Formula $(1)$} \\ &=& 2(2a_2-1)^2-a_2(2a_2-1) \\ \mathbf{a_6} & \mathbf{=} & \mathbf{ (2a_2-1)(3a_2-2) } \\ \hline a_7 &=& \dfrac{a_6^2}{a_5} & | \quad \text{ Formula $(2)$} \\ a_7 &=& \dfrac{(2a_2-1)^2(3a_2-2)^2}{(2a_2-1)^2} \\ \mathbf{a_7} & \mathbf{=} & \mathbf{ (3a_2-2)^2 } \\ \hline a_8 &=& 2a_7-a_6 & | \quad \text{ Formula $(1)$} \\ &=& 2(3a_2-2)^2-(2a_2-1)(3a_2-2) \\ \mathbf{a_8} & \mathbf{=} & \mathbf{ (3a_2-2)(4a_2-3) } \\ \hline a_9 &=& \dfrac{a_8^2}{a_7} & | \quad \text{ Formula $(2)$} \\ a_9 &=& \dfrac{(3a_2-2)^2(4a_2-3)^2}{(3a_2-2)^2} \\ \mathbf{a_9} & \mathbf{=} & \mathbf{ (4a_2-3)^2 } \\ \hline a_{10} &=& 2a_9-a_8 & | \quad \text{ Formula $(1)$} \\ &=& 2(4a_2-3)^2-(3a_2-2)(4a_2-3) \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{ (4a_2-3)(5a_2-4) } \\ \hline \end{array}$

$\mathbf{a_9+a_{10}=646}\\ \begin{array}{|rcll|} \hline \mathbf{a_9+a_{10}} & \mathbf{=}& \mathbf{646} \\ (4a_2-3)^2+(4a_2-3)(5a_2-4) &=& 646 \\ (4a_2-3) (4a_2-3+5a_2-4) &=& 646 \\ (4a_2-3) (9a_2-7) &=& 646 \\ \ldots \\ 36a_2^2-55a_2-625 &=& 0 \\\\ a_2 &=& \dfrac{55\pm\sqrt{55^2-4\cdot36\cdot(-625)} }{2\cdot 36} \\\\ &=& \dfrac{55\pm\sqrt{93025} }{72} \\\\ &=& \dfrac{55\pm305 }{72} \\\\ a_2 &=& \dfrac{55+305 }{72} \\\\ \mathbf{a_2 } & \mathbf{=} & \mathbf{5} \\\\ && \text{or} \\\\ a_2 &=& \dfrac{55-305 }{72} \\\\ a_2 & = & -3.47222222222 \\ && \text{no solution, because $a_n$ is a sequence}\\ && \text{of positive integers!}\\ \hline \end{array}$

$\text{Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $}$

$\begin{array}{|lclcl|} \hline a_1 && &=& 1 \\ a_2 && &=& 5 \\ a_3 &=& 5^2 &=& 25 \\ a_4 &=& 5\cdot 9 &=& 45 \\ a_5 &=& 9^2 &=& 81 \\ a_6 &=& 9\cdot 13 &=& 117 \\ a_7 &=& 13^2 &=& 169 \\ a_8 &=& 13\cdot 17 &=& 221 \\ a_9 &=& 17^2 &=& 289 \\ a_{10} &=& 17\cdot 21 &=& 357 \\ a_{11} &=& 21^2 &=& 441 \\ a_{12} &=& 21\cdot 25 &=& 525 \\ a_{13} &=& 25^2 &=& 625 \\ a_{14} &=& 25\cdot 29 &=& 725 \\ a_{15} &=& 29^2 &=& 841 \\ \color{red}a_{16} &=& 29\cdot 33 &\color{red}=& \color{red}957 \qquad a_n < 1000\\ a_{17} &=& 33^2 &=& 1089 \qquad a_n > 1000 \\ \hline \end{array}$

$\mathbf{\text{$a_n$ is $ 957$}}$

$\text{GP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{ratio} & a_{2n-1} & a_{2n} & a_{2n+1} \\ \hline 1 &5 &1 & 5 & 25 \\ 2&1.8 &25 & 45 & 81 \\ 3&1.4\ldots &81 & 117 & 169 \\ 4&1.30769230769 &169 & 221 & 289 \\ 5&1.23529411765 &289 & 357 & 441 \\ 6&1.19047619048 &441 & 525 & 625 \\ 7&1.16 &625 & 725 & 841 \\ 8&1.13793103448 &841 & 957 & 1089 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}$

$\text{AP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{common difference} & a_{2n} & a_{2n+1} & a_{2n+2} \\ \hline 1 &20 & 5 & 25 & 45 \\ 2 &36 &45 & 81 & 117 \\ 3 &52 &117 & 169 & 221 \\ 4 &68 & 221 & 289 & 357 \\ 5 &84 &357 & 441 & 525 \\ 6 &100& 525 & 625 & 725 \\ 7 &116& 725 & 841 & 957 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}$

Find the sum of the terms from the (n + 1) th to the m th term inclusive of an arithmetical progression whose

first term is a and whose second term is b.

If m = 13, n =   3 and the sum is 12a, find the ratio b: a.

Answer (m-n){a+1/2(m+n-1)(b-a)} ; 77/75

$\text{Formula arithmetical progression: }\\ \begin{array}{|rcll|} \hline a_1 &=& a \\ a_2 &=& a+d \\ \hline \end{array}$

$\text{second term is b: }\\ \begin{array}{|rcll|} \hline a_2 &=& a+d \\ a_2 &=& b \\ \hline a+d &=& b \\ \mathbf{d} & \mathbf{=} & \mathbf{b-a} \\ \hline \end{array}$

The common difference is $\mathbf{b-a}$

$\text{The sum of the member is: }\\ \begin{array}{|rcll|} \hline s_n = n\cdot a + \dfrac{n(n-1)}{2}\cdot d \\ \hline \end{array}$

$\text{The sum of the terms from the (n + 1) th to the m th term is: }\\ \begin{array}{|rcll|} \hline s_m-s_n &=& \underbrace{m\cdot a + \dfrac{m(m-1)}{2}\cdot (b-a)}_{=s_m} - \left[ \underbrace{n\cdot a + \dfrac{n(n-1)}{2}\cdot (b-a)}_{=s_n} \right] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m(m-1)-n(n-1) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-m-n^2+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-n^2-m+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m^2-n^2-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ (m-n)(m+n)-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)(m-n)( m+n-1 ) \\\\ \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\ \hline \end{array}$

$\small{ \text{If $m = 13, n = 3$ and the sum is $12a$, find the ratio $b: a$ }\\ \begin{array}{|rcll|} \hline \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\\\ 12a & = & (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b}{2}\right)( m+n-1 )- \left(\dfrac{a}{2}\right)( m+n-1 ) \\\\ a\left( \dfrac{12}{m-n} - 1+ \dfrac{m+n-1}{2} \right) & = & b\left( \dfrac{m+n-1}{2} \right) \qquad m = 13 \qquad n = 3 \\\\ a\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) & = & b\left( \dfrac{15}{2} \right) \\\\ \dfrac{b}{a} &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) \\\\ &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12-10+75}{10} \right) \\\\ &=& \dfrac{2\cdot 77}{150} \\\\ &=& \dfrac{2\cdot 77}{2\cdot 75} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{ =}& \mathbf{ \dfrac{77}{75}} \\ \hline \end{array} }$

In triangle
$ABC, \overline{AB}=5, \overline{AC}=6,$
and
$\overline{BC}=7.$
Circles are drawn with centers $A, B,$ and $C,$
so that any two circles are externally tangent.
Find the sum of the areas of the circles.

$\text{Let $r_a =$ radius of circle $A$ } \\ \text{Let $r_b =$ radius of circle $B$ } \\ \text{Let $r_c =$ radius of circle $C$ }$

$\begin{array}{|lrcll|} \hline (1) & r_a + r_b &=& 5 \\ (2) & r_b + r_c &=& 7 \\ (3) & r_c + r_a &=& 6 \\ \hline (1)+(2)+(3): \\ & 2(r_a+r_b+r_c) &=& 5+6+7 \\ & 2(r_a+r_b+r_c) &=& 18 \quad & | \quad : 2 \\ & r_a+r_b+r_c &=& 9 \\ \hline \end{array}$

$\mathbf{r_c = \ ?}$

$\begin{array}{|rcll|} \hline \underbrace{r_a+r_b}_{=5}+r_c &=& 9 \\ 5+r_c &=& 9 \\ r_c &=& 9-5 \\ \mathbf{r_c} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$

$\mathbf{r_a = \ ?}$

$\begin{array}{|rcll|} \hline r_a+\underbrace{r_b+r_c}_{=7} &=& 9 \\ r_a+7 &=& 9 \\ r_a &=& 9-7 \\ \mathbf{r_a} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}$

$\mathbf{r_b = \ ?}$

$\begin{array}{|rcll|} \hline r_a+r_b+r_c &=& 9 \quad & | \quad r_a+r_c = 6\\ 6 + r_b &=& 9 \\ r_b &=& 9-6 \\ \mathbf{r_b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}$

The sum of the areas of the circles:

$\begin{array}{|rcll|} \hline && \pi r_a^2 + \pi r_b^2 + \pi r_c^2 \\ &=& \pi \cdot( r_a^2 + r_b^2 + r_c^2 ) \\ &=& \pi \cdot( 2^2 + 3^2 + 4^2 ) \\ &=& \pi \cdot( 4+9+16 ) \\ &\mathbf{=}& \mathbf{29\pi} \\ \hline \end{array}$

Thank you, Melody

When the minute hand of a clock points exactly at a full minute, the hour hand is exactly two minutes away.

Give all possible times in a 12-hour period that satisfy condition.

$\text{ angular velocity minute hand: $ \omega_m^{\circ} =\dfrac{360^{\circ}}{1~ h} $ }\\ \text{ angular velocity hour hand: $ \omega_h^{\circ} =\dfrac{360^{\circ}}{12~ h} $ }\\ \boxed{\text{angle = angular velocity }\times \text{ time}}\\ \text{ angle minute hand: $ \alpha_m^{\circ} = \omega_m^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle hour hand: $ \alpha_h^{\circ} = \omega_h^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle difference minute hand-hour hand: $ \alpha_m^{\circ}-\alpha_h^{\circ} = \Delta\alpha^{\circ}$ } \\ \Delta\alpha^{\circ} = \alpha_m^{\circ}-\alpha_h^{\circ} \\ \Delta\alpha^{\circ} = \omega_m^{\circ} \times t^h - \omega_h^{\circ} \times t^h \\ \Delta\alpha^{\circ} = \left(\omega_m^{\circ} - \omega_h^{\circ} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(\dfrac{360^{\circ}}{1~ h} - \dfrac{360^{\circ}}{12~ h} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(360\cdot \dfrac{11}{12} \right) \times t^h \\ \Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} \\$

$\mathbf{\boxed{\Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} }}$

Example:

$t^h = 6\ h \\ \Delta\alpha^{\circ} = 330 \times 6 \pmod{360^{\circ}} \\ \Delta\alpha^{\circ} = 1980^{\circ} \pmod{ 360^{\circ} } \\ \Delta\alpha^{\circ} = 1980^{\circ} - 5\cdot 360^{\circ} \\ \Delta\alpha^{\circ} = 180^{\circ}$

1.

$\text{The hour hand of a clock points exactly at a full minute:} \\ \begin{array}{rcll} \dfrac{360^{\circ}}{60~\min.} &=& \dfrac{x}{1~\min.} \\\\ x&=&\dfrac{1~ \min.}{60~ \min.}\cdot 360^{\circ} \\\\ \mathbf{x}&\mathbf{=}&\mathbf{6^{\circ}} \\ \end{array}$

1 minute on the clock conforms 6 degrees.

$\text{The hour hand of a clock points exactly at a full minute, so} \\ \text{ $\alpha_h^{\circ} = 6^{\circ} \cdot n \qquad | \qquad n$ is an integer}$

2. $\mathbf{t^h =\ ?}$

$\begin{array}{rcll} \alpha_h^{\circ} &=& \omega_h^{\circ} \times t^h \\\\ t^h &=& \dfrac{ \alpha_h^{\circ} } {\omega_h^{\circ}} \quad & | \quad \alpha_h^{\circ} = 6^{\circ} \cdot n \qquad \omega_h^{\circ} = \dfrac{360^{\circ}}{12~ h} \\\\ t^h &=& \dfrac{ 6^{\circ} \cdot n } { \dfrac{360^{\circ}}{12~ h}} \\\\ \mathbf{t^h} &\mathbf{=}& \mathbf{0.2n} \\ \end{array}$

3.

$\text{The hour hand is exactly two minutes away: $\Delta\alpha^{\circ} = \pm 12^{\circ}$ } \\ \begin{array}{rcll} \Delta\alpha^{\circ} &=& 330 \times t^h \pmod{360^{\circ}} \quad & | \quad \mathbf{t^h=0.2n} \qquad \Delta\alpha^{\circ} = \pm 12^{\circ} \\ \pm 12^{\circ} &=& 330 \times 0.2n \pmod{360^{\circ}} \\ \pm 12^{\circ} &=& 66n \pmod{360^{\circ}} \\ \pm 12^{\circ} -66n &=& 360^{\circ}m \qquad & n \in N,\ m \in N \\ \pm 12^{\circ} &=& 66n + 360^{\circ}m \quad & | \quad : 6 \\ \pm 2^{\circ} &=& 11n + 60^{\circ}m \\ &&\boxed{1. \text{ Diophantine equation: } 11n + 60^{\circ}m = 2} \\ &&\boxed{2. \text{ Diophantine equation: } 11n + 60^{\circ}m = -2} \\ \end{array}$

4. Solution diophantine equation 11n + 60m = 2:

$\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& 2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \\ &=& \dfrac{ 2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m + 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ 2 - 5m } {11} \\ n &=&-5m+ \dfrac{ 2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 - 5m } {11} \\ & 11a &=& 2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& 2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \\ &=& \dfrac{ 2 - 10a-a } {5} \\ &=& \dfrac{ - 10a + 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ 2 -a } {5} \\ m &=& -2a+ \dfrac{ 2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ 2 - a } {5} \\ & 5b &=& 2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2 - 5b } \\ \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \quad & | \quad \mathbf{a = 2 - 5b }\\ & = & \dfrac{ 2 - 11 (2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{-4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = -4 + 11b }\\ & = & \dfrac{ 2 - 60(-4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{22 - 60b } \\ \end{array}$

$\boxed{ \text{1. Diophantine equation: }\\ \mathbf{n = 22 - 60b} \\\mathbf{m=-4 + 11b} \qquad b \in Z }$$\begin{array}{|lrcll|} \hline b = 0: & n &=& 22 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 22 \\ & &=& 4.4\quad (4:24) \\ \hline \end{array}$

5. Solution diophantine equation 11n + 60m = -2:

$\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& -2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ -2 - 60m } {11}} \\ &=& \dfrac{ -2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m - 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ -2 - 5m } {11} \\ n &=&-5m+ \dfrac{ -2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ -2 - 5m } {11} \\ & 11a &=& -2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& -2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \\ &=& \dfrac{ -2 - 10a-a } {5} \\ &=& \dfrac{ - 10a - 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ -2 -a } {5} \\ m &=& -2a+ \dfrac{ -2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ -2 - a } {5} \\ & 5b &=& -2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ -2 - 5b } \\ \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \quad & | \quad \mathbf{a = -2 - 5b }\\ & = & \dfrac{ -2 - 11 (-2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = 4 + 11b }\\ & = & \dfrac{ -2 - 60(4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{-22 - 60b } \\ \end{array}$

$\boxed{ \text{2. Diophantine equation: }\\ \mathbf{n = -22 - 60b} \\\mathbf{m=4 + 11b} \qquad b \in Z }$$\begin{array}{|lrcll|} \hline b = -1: & n &=& -22+60 \\ & &=& 38 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 38 \\ & &=& 7.6\quad (7:36) \\ \hline \end{array}$

The solutions are: 4:24 and 7:36

How many two-digit positive integers are congruent to 1 (mod 3)?

$\begin{array}{|rcll|} \hline \text{ $x \equiv 1 \pmod 3$ $\\$ or $ \\ x-1 = n\cdot 3 $ } \\ \hline \end{array}$

$\begin{array}{lrcll} \text{If $x = 99$} & 99-1 &=& n\cdot 3 \\ & 98 &=& n\cdot 3 \\ & n &=& \dfrac{98}{3} \\ & n &=& 32.7 \\ & \boxed{ n = 0\ldots 32 \qquad x = 1\ldots 99 } \\ \end{array}$

$\begin{array}{lrcll} \text{If $x = 9$} & 9-1 &=& m\cdot 3 \\ & 8 &=& m\cdot 3 \\ & m &=& \dfrac{8}{3} \\ & m &=& 2.7 \\ & \boxed{ m = 0\ldots 2 \qquad x = 1\ldots 9 } \\ \end{array}$

$\begin{array}{|rcll|} \hline x = 10\ldots 99 \\ n-m &=& 33 -3 = 30 \\ \hline \end{array}$

30 two-digit positive integers are congruent to 1 (mod 3)

$\begin{array}{|l|rcll|} \hline 1. & 10 \\ 2. & 13 \\ 3. & 16 \\ 4. & 19 \\ 5. & 22 \\ 6. & 25 \\ 7. & 28 \\ 8. & 31 \\ 9. & 34 \\ 10. & 37 \\ 11. & 40 \\ 12. & 43 \\ 13. & 46 \\ 14. & 49 \\ 15. & 52 \\ 16. & 55 \\ 17. & 58 \\ 18. & 61 \\ 19. & 64 \\ 20. & 67 \\ 21. & 70 \\ 22. & 73 \\ 23. & 76 \\ 24. & 79 \\ 25. & 82 \\ 26. & 85 \\ 27. & 88 \\ 28. & 91 \\ 29. & 94 \\ 30. & 97 \\ \hline \end{array}$

A triangle with vertices A(5,3), B(1,-4) and C(-3,1) is rotated 90 degrees counterclockwise about vertex B.

Find the co-ordinates of the image of vertex A.

$\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \vec{A} \text{ before rotation} \\ & \vec{A}' \text{ after rotation} \\ & \text{$\vec{B_{rotation\ axis}}$ at $\binom{1}{-4} $ } \\ & \overset{\curvearrowleft}{D}_{\varphi} = \begin{pmatrix} \cos(\varphi) & \sin(\varphi) \\ -\sin(\varphi) & \cos(\varphi) \end{pmatrix} \ \text{ Matrix of rotation counterclockwise}\\ & \overset{\curvearrowleft}{D}_{90^{\circ}} =\begin{pmatrix} \cos(90^{\circ}) & \sin(90^{\circ}) \\-\sin(90^{\circ}) & -\cos(90^{\circ}) \end{pmatrix} =\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\ \end{array}$

$\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \begin{array}{|rcll|} \hline \vec{A}' &=& \left(\dbinom{5}{3}-\dbinom{1}{-4} \right)\cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{4}{7} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{-7}{4}+\dbinom{1}{-4} \\ &=& \dbinom{-6}{0} \\ \hline \end{array} \end{array}$