heureka

2.)

An other ex. is this: \(\sigma=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)\)
The answer is \((1,11,3,2,4,6,7,5,8,9)(10) \Rightarrow (1,11,3,2,4,6,7,5,8,9)\)

Start with \(x = 1 \text{ in } k(x)\)

\(\small{ \begin{array}{|ccccccccccc|c|cccccc|} \hline f &\circ& g &\circ& h &\circ& i &\circ& j &\circ& k & x & k(x) & j(k(x)) & i(j(k(x))) & \tiny{h(i(j(k(x))))} & \tiny{g(h(i(j(k(x)))))} & \tiny{f(g(h(i(j(k(x))))))} \\ \hline (1,3,5)&\circ&(2,7)&\circ&(11,1,4)&\circ&(5,6,2,8,9)&\circ&(7,3)&\circ&(4,5,2,1) &\color{red} 1 & 4 &4 & 4 & 11 &11 & 11 \\ & & & & & & & & & & & 11 & 11 &11 & 11 & 1 &1 &3 \\ & & & & & & & & & & & 3 & 3 &7 & 7 & 7 &2 &2 \\ & & & & & & & & & & & 2 & 1 &1 & 1 & 4 &4 &4 \\ & & & & & & & & & & & 4 & 5 &5 & 6 & 6 &6 &6 \\ & & & & & & & & & & & 6 & 6 &6 & 2 & 2 &7 &7 \\ & & & & & & & & & & & 7 & 7 &3 & 3 & 3 &3 &5 \\ & & & & & & & & & & & 5 & 2 &2 & 8 & 8 &8 &8 \\ & & & & & & & & & & & 8 & 8 &8 & 9 & 9 &9 &9 \\ & & & & & & & & & & & 9 & 9 &9 & 5 & 5 &5 &\color{red}1 \\ \hline & & & & & & & & & & & \color{red}10 & 10 &10 & 10 & 10 &10 &\color{red}10 \\ \hline \end{array} } \)

Permutation cycles:
(1, 11, 3, 2, 4, 6, 7, 5, 8, 9) (10)

In the sequence

1, 2, 2, 4, 8, 32, 256,...,

each term starting from the third term) is the product of the two terms before it. For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32). 

This sequence can be continued forever, though the numbers very quickly grow enormous!

(For example, the 14th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the 35th term of the sequence?

NumTheSL5#5

I assume:

\(\begin{array}{|c|c|} \hline n & 12 \text{ positive factors} \\ \hline 60&12\\ \hline 72&12\\ \hline 84&12\\ \hline 90&12\\ \hline 96&12\\ \hline \text{sum }= \mathbf{402} \\ \hline \end{array}\)

Compute \(29^{13} - 5^{13}\) modulo 7

\(\begin{array}{|rcll|} \hline && \mathbf{29^{13} - 5^{13} \pmod{ 7}} \quad & | \quad \mathbf{29} \equiv {\color{red}1} \pmod{7} \\ &\equiv& {\color{red}1} ^{13} - 5^{13} \pmod{ 7} \\ &\equiv& 1 - 5^{13} \pmod{ 7} \quad & | \quad \mathbf{5} \equiv {\color{red}-2} \pmod{7} \\ &\equiv& 1 - \left({\color{red}-2}\right)^{13} \pmod{ 7} \quad & | \quad 2^{13} = 8192 \\ &\equiv& 1 - (-8192) \pmod{ 7} \\ &\equiv& 1 +8192 \pmod{ 7} \\ &\equiv& 8193 \pmod{ 7} \\ &\mathbf{\equiv}& \mathbf{3 \pmod{ 7}} \\ \hline \end{array}\)

1)   

What is the remainder when \(7^{2010}\) is divided by \(100\) ?

\(\begin{array}{|rcll|} \hline && \mathbf{7^{2010} \pmod{ 100}} \quad & | \quad \mathbf{7^4} \equiv {\color{red}1} \pmod{100} \\ &\equiv & 7^{4\cdot 502+2}\pmod{ 100} \\ &\equiv & \left(\mathbf{7^4}\right)^{502}7^2 \pmod{ 100} \\ &\equiv & \left({\color{red}1}\right)^{502}7^2 \pmod{ 100} \\ &\equiv & 7^2 \pmod{ 100} \\ &\mathbf{\equiv} & \mathbf{49 \pmod{ 100}} \\ \hline \end{array}\)

2) 

When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?

When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?

\(\begin{array}{|rcll|} \hline \mathbf{11011_b }&=& \mathbf{1}\cdot b^4 +\mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{1}\cdot b^1 +\mathbf{1}\cdot b^0 \\ \hline (11011_b)(b-1) &=&( \mathbf{1}\cdot b^4 +\mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{1}\cdot b^1 +\mathbf{1}\cdot b^0)(b-1) \\\\ &=& \mathbf{1}\cdot b^5+\mathbf{1}\cdot b^4 +\mathbf{0}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{1}\cdot b^1 \\ && \qquad\quad -\mathbf{1}\cdot b^4 -\mathbf{1}\cdot b^3 -\mathbf{0}\cdot b^2 -\mathbf{1}\cdot b^1-\mathbf{1}\cdot b^0 \\\\ &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 -\mathbf{1}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1-\mathbf{1}\cdot b^0 \\ \hline (11011_b)(b-1)+1001_b &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 -\mathbf{1}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1-\mathbf{1}\cdot b^0 \\ && \qquad \qquad\qquad + \mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{0}\cdot b^1+\mathbf{1}\cdot b^0 \\\\ &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 +\mathbf{0}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1+\mathbf{0}\cdot b^0 \\\\ &=& 100100_b \\ \hline \end{array} \)

for every base.

In triangle ABC, sin A : sin B : sin C = 5 : 5 : 6. Find cos C.

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{\dfrac{\sin(A)}{\sin(B)}} &=& \mathbf{\dfrac{5}{5}} \\\\ & \dfrac{\sin(A)}{\sin(B)} &=& 1 \\ \\ & \sin(A) &=& \sin(B) \\ & \mathbf{A} &=& \mathbf{B} \quad | \quad A=180^\circ-B \text{ is not possible}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(C) &=& \sin\Big(180^\circ-(A+B)\Big) \quad | \quad B=A \\ \sin(C) &=& \sin\Big(180^\circ-(A+A)\Big) \\ \sin(C) &=& \sin\Big(180^\circ-(2A)\Big) \\ \sin(C) &=& \sin( 2A ) \\ \mathbf{\sin(C)} &=& \mathbf{2\sin( A )\cos(A)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (2): & \mathbf{\dfrac{\sin(A)}{\sin(C)}} &=& \mathbf{\dfrac{5}{6}} \\\\ & \dfrac{\sin(A)}{2\sin( A )\cos(A)} &=& \dfrac{5}{6} \\\\ & \dfrac{1}{2 \cos(A)} &=& \dfrac{5}{6} \\\\ & 2 \cos(A) &=& \dfrac{6}{5} \\\\ & \mathbf{ \cos(A) } &=& \mathbf{ \dfrac{6}{10} } \\ \hline & \sin(A) &=& \sqrt{1-\cos^2(A) } \\ & \sin(A) &=& \sqrt{1-\dfrac{6^2}{10^2} } \\ & \sin(A) &=& \sqrt{\dfrac{10^2-6^2}{10^2} } \\ & \sin(A) &=& \sqrt{\dfrac{8^2}{10^2} } \\ & \mathbf{ \sin(A) } &=& \mathbf{ \dfrac{8}{10} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (2): & \mathbf{\dfrac{\sin(A)}{\sin(C)}} &=& \mathbf{\dfrac{5}{6}} \quad | \quad \mathbf{ \sin(A) = \dfrac{8}{10} } \\\\ & \dfrac{\dfrac{8}{10}}{\sin(C)} &=& \dfrac{5}{6} \\\\ & \sin(C) &=& \dfrac{8}{10}\cdot \dfrac{6}{5} \\\\ & \mathbf{ \sin(C) } &=& \mathbf{ \dfrac{24}{25} } \\ \hline & \cos(C) &=& \sqrt{1-\sin^2(C) } \\ & \cos(C) &=& \sqrt{1-\dfrac{24^2}{25^2} } \\ & \cos(C) &=& \sqrt{\dfrac{25^2-24^2}{25^2} } \\ & \cos(C) &=& \sqrt{\dfrac{7^2}{25^2} } \\ & \cos(C) &=& \dfrac{7 }{25 } \\ & \mathbf{\cos(C)} &=& \mathbf{0.28} \\ \hline \end{array}\)

Find all real numbers\(t\) such that \(\frac{2}{3} t - 1 < t + 7 \le -2t + 15\).

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} t - 1 &<& t + 7 \\ \dfrac{2}{3} t &<& t +8 \\ \dfrac{2}{3} t-t &<& 8 \\ -\dfrac{1}{3} t &<& 8 \\ \dfrac{1}{3} t &>& -8 \\ -8 &<&\dfrac{1}{3} t \\ -8\cdot 3 &<& t \\ \mathbf{ -24 } &\mathbf{<}& \mathbf{t} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t + 7 &\le& -2t + 15 \\ t &\le& -2t + 8 \\ 3t &\le& 8 \\ \mathbf{t } & \mathbf{\le} & \mathbf{\dfrac{ 8 } {3} } \\ \hline \end{array}\)

\(\mathbf{ -24 < t \le \dfrac{ 8 } {3} } \)

In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral.
Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.
Find the area of triangle APE in simplest radical form.

\(\begin{array}{|rcll|} \hline \text{area}_{[BCP]} &=& \dfrac{4x}{2}=2x \\\\ \text{area}_{[ABC]} &=& \dfrac{4^2}{2} = 8 \\\\ \text{area}_{[ABE]} &=& \dfrac{4\cdot 2\sqrt{3}}{2} = 4\sqrt{3} \quad | \quad h = 2\sqrt{3} \\ \hline \text{area}_{[APE]} &=& \text{area}_{[ABE]}+\text{area}_{[BCP]} -\text{area}_{[ABC]}\\\\ &=& 4\sqrt{3}+2x -8 \\ \hline \end{array}\)

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

\(\begin{array}{|lrcll|} \hline (1): & \tan(45^\circ) &=& \dfrac{x}{CQ} \quad | \quad \tan(45^\circ) = 1 \\ & 1 &=& \dfrac{x}{CQ} \\ & \mathbf{CQ} &=& \mathbf{x} \\ \hline (2): & \tan(30^\circ) &=& \dfrac{x}{4-CQ} \quad | \quad CQ=x \\\\ & \tan(30^\circ) &=& \dfrac{x}{4-x} \quad | \quad \tan(30^\circ) = \dfrac{\sqrt{3}}{3} \\\\ & \dfrac{\sqrt{3}}{3} &=& \dfrac{x}{4-x} \\\\ & (4-x)\sqrt{3} &=& 3x \\ & 4\sqrt{3}-x\sqrt{3} &=& 3x \\ & x\sqrt{3} + 3x&=& 4\sqrt{3} \\ & x(3+\sqrt{3})&=& 4\sqrt{3} \\\\ & x &=& \dfrac{4\sqrt{3}} {3+\sqrt{3}} \\\\ & x &=& \dfrac{4\sqrt{3}} {3+\sqrt{3}}\times \dfrac{(3-\sqrt{3})}{(3-\sqrt{3})} \\\\ & x &=& \dfrac{4\sqrt{3}(3-\sqrt{3})} {9-3} \\\\ & x &=& \dfrac{4\sqrt{3}(3-\sqrt{3})} {6} \\\\ & x &=& \dfrac{2\sqrt{3}(3-\sqrt{3})} {3} \\\\ & x &=& \dfrac{2\sqrt{3}\cdot 3 } {3} - \dfrac{2} {3} \cdot 3 \\\\ & \mathbf{x} &=& \mathbf{2\sqrt{3} - 2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{area}_{[APE]} &=& 4\sqrt{3}+2x -8 \quad | \quad x=2\sqrt{3} - 2 \\ &=& 4\sqrt{3}+2(2\sqrt{3} - 2) -8 \\ &=& 8\sqrt{3} -12 \\ \mathbf{\text{area}_{[APE]}}&=& \mathbf{4(2\sqrt{3} -3)} \\ \hline \end{array}\)

Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all $x$ such that $x\neq -1$ and $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

\(\begin{array}{|lrcll|} \hline & \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad x^2 - x - 2 = (x+1)(x-2) \\\\ & \dfrac{x + 7}{(x+1)(x-2)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \cdot (x+1)(x-2) \\\\ & x + 7 &=& \dfrac{A(x+1)(x-2)}{(x - 2)} + \dfrac{B(x+1)(x-2)}{(x + 1)} \\\\ & \mathbf{x + 7} &=& \mathbf{A(x+1) + B(x-2)} \\ \hline x=2 : & 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3 A \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& \mathbf{3} \\ \hline x=-1 : & -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& -3B \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& \mathbf{-2} \\ \hline \end{array}\)

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{3}{x - 2} - \dfrac{2}{x + 1}\)