heureka

here is the picture:

A stick has a length of \(5\) units. The stick is then broken at two points, chosen at random.

What is the probability that all three resulting pieces are longer than \(1\) unit?

The probability that all three resulting pieces are longer than \(1\) unit = \(\dfrac{ A_{\color{darkorange}orange} } {A} \)

\(\begin{array}{|rcll|} \hline H^2 + \left(\dfrac{S}{2}\right)^2 &=& S^2 \\ H^2 &=& S^2 -\dfrac{S^2}{4} \\ H^2 &=& \dfrac{3S^2}{4} \qquad \mathbf{H=\dfrac{S\sqrt{3}}{2}} \qquad (1) \\ S^2 &=& \dfrac{4H^2}{3} \\ S &=& \dfrac{2H}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{S} &=& \mathbf{\dfrac{2H\sqrt{3}}{3}} \qquad (2) \\ \hline S &=& \dfrac{2H\sqrt{3}}{3} \quad | \quad H = 5 \\ \mathbf{S} &=& \mathbf{\dfrac{10\sqrt{3}}{3}} \\ \hline 2A &=& H*S \quad | \quad H=5,\ \mathbf{S=\dfrac{10\sqrt{3}}{3}} \\ \mathbf{2A} &=& \mathbf{\dfrac{50\sqrt{3}}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(30^\circ) &=& \dfrac{1}{x} \quad | \quad \tan(30^\circ) = \dfrac{1}{\sqrt{3}}\\ \dfrac{1}{\sqrt{3}} &=& \dfrac{1}{x} \\ \mathbf{ x } &=& \mathbf{\sqrt{3}} \\ \hline \mathbf{s} &=& \mathbf{S-2x} \quad | \quad \mathbf{S=\dfrac{10\sqrt{3}}{3}},\ \mathbf{ x =\sqrt{3}} \\ s &=& \dfrac{10\sqrt{3}}{3}-2\sqrt{3} \\ s &=& \dfrac{10\sqrt{3}}{3}-\dfrac{6\sqrt{3}}{3} \\ \mathbf{ s } &=& \mathbf{\dfrac{4\sqrt{3}}{3}} \\ \hline \mathbf{h^2 + \left(\dfrac{s}{2}\right)^2} &=& \mathbf{s^2} \\ h^2 &=& s^2 -\dfrac{s^2}{4} \\ h^2 &=& \dfrac{3s^2}{4} \\ \mathbf{h} &=& \mathbf{\dfrac{s\sqrt{3}}{2}}\quad | \quad \mathbf{ s =\dfrac{4\sqrt{3}}{3}} \\ h &=& \mathbf{\dfrac{\dfrac{4\sqrt{3}}{3}\sqrt{3}}{2}} \\ h &=& \dfrac{4}{2} \\ \mathbf{h} &=& \mathbf{2} \\ \hline 2A_{\color{darkorange}orange}&=& h*s \quad | \quad h=2,\ \mathbf{ s=\dfrac{4\sqrt{3}}{3}} \\ 2A_{\color{darkorange}orange} &=& 2*\dfrac{4\sqrt{3}}{3} \\ \mathbf{2A_{\color{darkorange}orange}} &=& \mathbf{\dfrac{8\sqrt{3}}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{ A_{\color{darkorange}orange} } {A} &=& \dfrac{ 2A_{\color{darkorange}orange} } {2A} \\\\ &=& \dfrac{ \dfrac{8\sqrt{3}}{3} } {\dfrac{50\sqrt{3}}{3}} \\\\ &=& \dfrac{8} {50} \\\\ \mathbf{ \dfrac{ A_{\color{darkorange}orange} } {A} } &=& \mathbf{ \dfrac{4}{25} } \\ \hline \end{array}\)

You have linear functions \(p(x)\) and \(q(x)\). You know \(p(2)=3\), and \(p\Big(q(x)\Big)=4x+7\) for all x.
Find \(q(-1)\).

\(\text{Let $p(x)=ax+b$}\)

\(\begin{array}{|rcll|} \hline p(x) &=& ax+b \quad | \quad x=2 \\ p(2) &=& 2a+b \quad | \quad p(2) = 3 \\ 3 &=& 2a+b \\ \mathbf{2a} &=& \mathbf{3-b} \\ \hline \end{array} \begin{array}{|rcll|} \hline p\Big(q(x)\Big) &=& aq(x)+b \quad | \quad p\Big(q(x)\Big) = 4x+7 \\ 4x+7 &=& aq(x)+b \quad | \quad x=-1 \\ 4(-1)+7 &=& aq(-1)+b \\ 3 &=& aq(-1)+b \\ 3-b &=& aq(-1) \quad | \quad * 2 \\ 2(3-b) &=& 2aq(-1) \quad | \quad \mathbf{2a=3-b} \\ 2(3-b) &=& (3-b)q(-1) \quad | :( 3-b) \\ 2 &=& q(-1) \\ \mathbf{q(-1)} &=& \mathbf{2} \\ \hline \end{array}\)

How do you work out what y and z are if x=y (mod z), where 7x+1=0 mod 3, x=0 mod 2,
without guessing and checking, and finding the LCM?

\(\begin{array}{|lrcll|} \hline & 7x+1 &\equiv& 0 \pmod{ 3 } & \text{or} \\ & 7x+1 &=& 0 + 3n,\ n\in \mathbb{Z} \\ & 7x+1 &=& 3n \\ (1) & \mathbf{7x} &=& \mathbf{3n-1} \\ \hline & x &\equiv& 0 \pmod{ 2 } & \text{or}\\ & x &=& 0 + 2m,\ m\in \mathbb{Z} \\ & x &=& 2m \quad | \quad *7 \\ (2) & \mathbf{7x} &=& \mathbf{14m} \\ \hline & \mathbf{7x} = 3n-1 &=& 14m \\ & 3n-1 &=& 14m \\ (3) & \mathbf{3n-14m} &=& \mathbf{1} \\ \hline \end{array} \)

What is the remainder of \(N=1\times 3\times5\times7\times...\times101\)  when it is divided by 8?

So
\(\begin{array}{rcll} N &\equiv& x \pmod{8} \\ 1\times 3\times5\times7\times\ldots\times101 &\equiv& x \pmod{8} \\ \end{array}\)

\(\begin{array}{|rcl|rcl|rcl|rcl|} \hline \mathbf{1} &\equiv& 1 \pmod{8} & \mathbf{3} &\equiv& 3 \pmod{8}& \mathbf{5} &\equiv& 5 \pmod{8}& \mathbf{7} &\equiv& 7 \pmod{8} \\ \mathbf{9} &\equiv& 1 \pmod{8} & \mathbf{11} &\equiv& 3 \pmod{8}& \mathbf{13} &\equiv& 5 \pmod{8}& \mathbf{15} &\equiv& 7 \pmod{8} \\ \mathbf{17} &\equiv& 1 \pmod{8} & \mathbf{19} &\equiv& 3 \pmod{8}& \mathbf{21} &\equiv& 5 \pmod{8}& \mathbf{23} &\equiv& 7 \pmod{8} \\ \mathbf{25} &\equiv& 1 \pmod{8} & \mathbf{27} &\equiv& 3 \pmod{8}& \mathbf{29} &\equiv& 5 \pmod{8}& \mathbf{31} &\equiv& 7 \pmod{8} \\ \vdots &&&\vdots&&&\vdots&&&\vdots \\ \mathbf{89} &\equiv& 1 \pmod{8} & \mathbf{91} &\equiv& 3 \pmod{8}& \mathbf{93} &\equiv& 5 \pmod{8}& \mathbf{95} &\equiv& 7 \pmod{8} \\ \mathbf{87} &\equiv& 1 \pmod{8} & \mathbf{99} &\equiv& 3 \pmod{8}& \mathbf{101} &\equiv& 5 \pmod{8} \\ \hline \end{array} \)

\(\small{ \begin{array}{|rcll|} \hline (1\times 3\times5\times7)\times (9\times11\times13\times 15)\times (17\times19\times21\times 23)\times \ldots \times (89\times91\times93\times 95)\times(97 \times99 \times101) &\equiv& x \pmod{8} \\ (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times (1\times 3\times5\times7)\times \ldots \times (1\times 3\times5\times7)\times(1 \times3 \times5) &\equiv& x \pmod{8} \\ \boxed{ (1\times 3\times5\times7) = 105 \equiv 1\pmod{8}\\ (1\times 3\times5) = 15 \equiv 7\pmod{8} } \\ (1)\times (1)\times (1)\times \ldots \times (1)\times(7) &\equiv& {\color{red}7} \pmod{8} \\ \hline \end{array} }\)

The remainder of \(N=1\times 3\times5\times7\times\cdots\times101\)  when it is divided by 8 is \(\mathbf{\color{red}7}\)

In triangle ABC, AB = BC = 25 and AC = 40. What is \(\sin( \angle ACB)\) ?

Heron's formula states that the area of a triangle whose sides have lengths \(a \), \(b\), and \(c\) is

\(\qquad A = \sqrt{s(s-a)(s-b)(s-c)}\),

where s is the semi-perimeter of the triangle; that is,

\(\qquad s=\dfrac{a+b+c}{2}\)

\(\begin{array}{rcll} \text{Let $BC=a=25$} \\ \text{Let $AC=b=40$} \\ \text{Let $AB=c=25$} \\ \end{array}\qquad \begin{array}{|rcll|} \hline \mathbf{ s } &=& \mathbf{ \dfrac{a+b+c}{2} } \\ s &=& \dfrac{25+40+25}{2} \\ \mathbf{ s } &=& \mathbf{45 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ A } &=& \mathbf{\sqrt{s(s-a)(s-b)(s-c)} } \\ A &=& \sqrt{45(45-25)(45-40)(45-25)} \\ A &=& \sqrt{45*20*5*20} \\ A &=& \sqrt{15^2*20^2} \\ \mathbf{ A } &=& \mathbf{15 * 20} \\ \hline \end{array}\)

Formula: \(2A = ab\sin(\angle ACB)\)

\(\begin{array}{|rcll|} \hline 2A &=& ab\sin(\angle ACB) \quad | \quad \mathbf{ A = 15 * 20},\ a=25,\ b=40 \\ 2*15 * 20 &=& 25*40 \sin(\angle ACB) \\ 15 * 40 &=& 25*40 \sin(\angle ACB) \quad | \quad : 40 \\ 15 &=& 25 \sin(\angle ACB) \\ \sin(\angle ACB) &=& \dfrac{15}{25} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{\dfrac{3}{5}} \\ \mathbf{ \sin(\angle ACB) } &=& \mathbf{0.6} \\ \hline \end{array} \)

A, B and C each represent a single digit with no two digits being the same.
Find the values of A, B and C so that the following is true: \(AB + C=38\) and \(BC+A=29\)

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{AB + C} &=& \mathbf{38} \\ 10A+B+C &=& 38 \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{BC+A} &=& \mathbf{29} \\ 10B+C+A &=& 29 \qquad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2): & 10A+B+C -(10B+C+A) &=& 38-29 \\ & 10A+B+C - 10B -C -A &=& 38-29 \\ & 10A+B - 10B -A &=& 9 \\ & 9A-9B &=& 9 \quad | \quad :9 \\ & A-B &=& 1 \\ & \mathbf{B} &=& \mathbf{A-1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{10A+B+C} &=& \mathbf{38} \quad | \quad \mathbf{B=A-1} \\ & 10A+A-1+C &=& 38 \\ & 11A-1+C &=& 38 \\ & 11A +C &=& 39 \\ & \mathbf{ 11A } &=& \mathbf{39-C} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|} \hline C & 39-C & A = \dfrac{39-C}{11} \\ \hline 0 & 39 & \text{not integer} \\ \hline 1 & 38 & \text{not integer} \\ \hline 2 & 37 & \text{not integer} \\ \hline 3 & 36 & \text{not integer} \\ \hline 4 & 35 & \text{not integer} \\ \hline 5 & 34 & \text{not integer} \\ \hline \color{red}6 & 33 & \mathbf{A =3} \\ \hline 7 & 32 & \text{not integer} \\ \hline 8 & 31 & \text{not integer} \\ \hline 9 & 30 & \text{not integer} \\ \hline \end{array}\), so \(\mathbf{C=6}\)

\(\begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{A-1} \quad | \quad \mathbf{A =3} \\ B &=& 3-1 \\ \mathbf{B} &=& \mathbf{2} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline 32+6 &=& 38 \qquad ( AB+C = 38) \\ 26+3 &=& 29 \qquad ( BC+A = 29) \\ \hline \end{array}\)

Hello Alan,

here is your pic: https://web2.0calc.com/img/upload/c037cd1f330c754341269173/g1.jpg

If a 7-digit number 13ab45c is divisible by 792, what is b?

\(\mathbf{792=2^3*3^2*11}\)

1.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{2}\), so \(c=\{2,4,6,8\}\)

2.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{2^3=8}\), so \(45c \equiv 0 \pmod{8}\)

\(\begin{array}{|c|r|c|} \hline c & 45c & 45c \pmod{8} \equiv 0\ ?\\ \hline 2 & 452 & \text{no} \\ \hline 4 & 454 & \text{no} \\ \hline \mathbf{\color{red}6} & 456 & \text{yes} \\ \hline 8 & 458 & \text{no} \\ \hline \end{array} \), so \(\mathbf{c=6}\)

3.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{3^2=9}\), so \(13ab456 \equiv 0 \pmod{9}\)

\(\begin{array}{|rcll|} \hline 1+3+a+b+4+5+6 &\equiv& 0 \pmod{9} \\ a+b+19 &\equiv& 0 \pmod{9} \quad | \quad 19 \equiv 1 \pmod{9} \\ a+b+1 &\equiv& 0 \pmod{9} \quad | \quad - 1 \\ a+b &\equiv& -1 \pmod{9} \\ a+b &\equiv& -1+9 \pmod{9} \\ \mathbf{a+b} &\equiv& \mathbf{ 8 \pmod{9} } \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|} \hline \mathbf{a+b}\pmod{9}= 8 \ ? \\ b \rightarrow & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9}\\ a \downarrow \\ \hline \mathbf{0} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 \\ \hline \mathbf{1} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 \\ \hline \mathbf{2} & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 \\ \hline \mathbf{3} & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 \\ \hline \mathbf{4} & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 \\ \hline \mathbf{5} & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathbf{6} & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathbf{7} & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathbf{8} & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7& \color{red}8 \\ \hline \mathbf{9} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7& \color{red}8 & 0 \\ \hline \end{array}\)

, so \((a,b)=\{(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(8,0),(8,9),(9,8)\}\)

4.
\(\mathbf{13ab45c}\) is divisible by \(\mathbf{11}\), so \(13ab456 \equiv 0 \pmod{11}\)

\(\begin{array}{|rcll|} \hline 1-3+a-b+4-5+6 &\equiv& 0 \pmod{11} \\ a-b+3 &\equiv& 0 \pmod{11} \quad | \quad - 3 \\ a-b &\equiv& -3 \pmod{11} \\ a-b &\equiv& -3+11 \pmod{11} \\ \mathbf{a-b} &\equiv& \mathbf{ 8 \pmod{11} } \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|} \hline \mathbf{a-b}\pmod{11}= 8 \ ? \\ b \rightarrow & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9}\\ a \downarrow \\ \hline \mathbf{0} & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 \\ \hline \mathbf{1} & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 \\ \hline \mathbf{2} & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 \\ \hline \mathbf{3} & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 \\ \hline \mathbf{4} & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 \\ \hline \mathbf{5} & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 \\ \hline \mathbf{6} & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 \\ \hline \mathbf{7} & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 \\ \hline \mathbf{8} & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 \\ \hline \mathbf{9} & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \end{array}\)

, so \((a,b)=\{(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9),(8,0),(9,1)\}\)

compare:

\(\mathbf{a+b}\pmod{9}= 8 : \quad (a,b)=\{(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),{\color{red}(8,0)},(8,9),(9,8)\} \\ \mathbf{a-b}\pmod{11}= 8 :\quad (a,b)=\{(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9),{\color{red}(8,0)},(9,1)\}\)

,so \(a=8\) and \(b=0\)

\(\begin{array}{|lr|} \hline & \mathbf{13ab45c} = \mathbf{1380456} \\ \hline \text{check}: \\ & 1380456 : 792 = 1743 \\ \hline \end{array}\)