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A group of 68 people attend a ball game. There were three times as many children as adults in the group.
Write a system of equations that you could use to set up this problem,
where a is the number of adults and c is the number of children in the group.
Solve the system of equations for c, the number of children in the group.

Let a = adults
Let c = children

\(\begin{array}{|lrcll|} \hline (1) & a + c &=& 68 \\ & a &=& 68-c \\\\ (2) & c &=& 3a \quad & | \quad a = 68-c \\ & c &=& 3(68-c) \\ & c &=& 204-3c \quad & | \quad + 3c \\ & c+3c &=& 204 \\ & 4c &=& 204 \quad & | \quad : 4 \\ & \mathbf{ c} & \mathbf{=}& \mathbf{51} \\\\ & a &=& 68-c \\ & a &=& 68-51 \\ & \mathbf{a} & \mathbf{=}& \mathbf{17} \\ \hline \end{array} \)

The number of children in the group is 51

The drama club sold 1500 tickets for the end-of-year performance.
Admission prices were $12 for adults and $6 for students.
The total amount collected at the box office was $15,480.
How many students attended the play?

Let a = adults

Let s = students

\(\begin{array}{|lrcll|} \hline (1) & a + s &=& 1500 \\ & a &=& 1500 -s \\\\ (2) & 12 a + 6 s &=& 15480 \quad & | \quad : 12 \\ & a + \dfrac{s}{2} &=& 1290 \\ & a &=& 1290 - \dfrac{s}{2} \\\\ \hline (1) = (2): & a = 1290 - \dfrac{s}{2} &=& 1500 -s \\ & 1290 - \dfrac{s}{2} &=& 1500 -s \quad & | \quad + s \\ & 1290 +s - \dfrac{s}{2} &=& 1500 \quad & | \quad -1290 \\ & s - \dfrac{s}{2} &=& 1500 -1290 \\ & \dfrac{s}{2} &=& 210 \quad & | \quad * 2 \\ & \mathbf{ s} & \mathbf{=}& \mathbf{420} \\\\ & a &=& 1500 -s \\ & a &=& 1500 - 420 \\ & \mathbf{ a} & \mathbf{=}& \mathbf{ 1080 } \\ \hline \end{array}\)

420 students attended the play.

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.
What is the 23rd term?

AP Formula:

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_5 &=& 9 \qquad & i = 5 \\ a_{32} &=& -84 \qquad & j = 32 \\ a_{23} &=& \ ? \qquad & k = 23 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ a_5(32-23) + a_{32}(23-5) +a_{23}(5-32) &=& 0 \\ 9\cdot 9 - 84 \cdot 18 - a_{23}\cdot 27 &=& 0 \\ a_{23}\cdot 27 &=& 9\cdot 9 - 84 \cdot 18 \\ a_{23}\cdot 27 &=& 81 - 1512 \\ a_{23}\cdot 27 &=& - 1431 \\\\ a_{23} &=& -\dfrac{ 1431 } {27} \\\\ \mathbf{ a_{23} } & \mathbf{=} & \mathbf{ -53 } \\ \hline \end{array}\)

The 23rd term is -53

For some integers that are not palindromes, like 91, a person can create a palindrome by repeatedly reversing
the number and adding the original number to its reverse. For example, .
Then , which is a palindrome, so 91 takes two steps to become a palindrome. Of all positive integers

between 10 and 100,
what is the sum of the non-palindrome integers that take exactly six steps to become palindromes?

Thank you Guest

How many different values can 

\(\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \) 
take for \(0 \leq x \leq 1 \)

lowest common multiple \(\mathbf{ \text{lcm}(1,2,3,4) = 12 }\)

\(\begin{array}{|c|c|c|c|} \hline x & \lfloor 1\cdot x \rfloor & \lfloor 2\cdot x \rfloor & \lfloor 3\cdot x \rfloor & \lfloor 4\cdot x \rfloor & \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \\ \hline \frac{0}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{1}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{2}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{3}{12} & 0 & 0 & 0 & 1 & 1 \\ \hline \frac{4}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{5}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{6}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{7}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{8}{12} & 0 & 1 & 2 & 2 & 5 \\ \hline \frac{9}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{10}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{11}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{12}{12} & 1 & 2 & 3 & 4 & 10 \\ \hline \end{array}\)

There are 7 different values: \(\mathbf{0, 1, 2, 4, 5, 6, 10}\)

2.

How many zeroes do we write when we write all the integers from 1 to243  in base 3?

\(\text{Let base $b = 3$} \\ \text{Let digit numbers $= n$ }\)

1. The one digit numbers \((n=1)\) don't have any zeroes. 

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 0\times \dfrac{2\cdot 3^0}{3} = 0 \text{ zeroes}\)

2. The two digit numbers \((n=2)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 1\times \dfrac{2\cdot 3}{3} = 2 \text{ zeroes}\)

3. The three digit numbers \((n=3) \) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 2\times \dfrac{2\cdot 3^2}{3} = 12 \text{ zeroes}\)

4. The four digit numbers \((n=4)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 3\times \dfrac{2\cdot 3^3}{3} = 54 \text{ zeroes}\)

5. The five digit numbers \((n=5)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 4\times \dfrac{2\cdot 3^4}{3} = 216 \text{ zeroes}\)

6. The six digit numbers \((n=6) \) have:

\(243 = 100000_3 = 5 \text{ zeroes}\)

\(\text{Ths sum $= 0 + 2 + 12 + 54 + 216 + 5 = 289$ zeroes } \)

We write 289 zeroes, when we write all the integers from 1 to 243  in base 3

SPQ is a straight line. We have sin RPQ = 7/25. What is sin RPS ?

\(\begin{array}{|rcll|} \hline RPS &=& 180^{\circ} - RPQ \\ \sin( RPS ) &=& \sin ( 180^{\circ} - RPQ ) \\ \sin( RPS ) &=& \sin ( RPQ ) \\ \sin( RPS ) &=& \sin ( RPQ ) \quad & | \quad \sin ( RPQ ) = \dfrac{7}{25} \\ \mathbf{\sin( RPS )} & \mathbf{=} & \mathbf{ \dfrac{7}{25} } \\ \hline \end{array}\)

In right triangle $ABC$,
we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$.
If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is an altitude of $\triangle ABC$,
then what is $\cos \angle ABM$?

\(\begin{array}{|rcll|} \hline \angle \text{ABM} &=& 90^{\circ}-\angle \text{CAB} \\ \cos(\angle\text{ABM}) &=& \cos(90^{\circ}-\angle \text{CAB}) \\ \cos(\angle \text{ABM}) &=& \sin(\angle \text{CAB}) \quad & | \quad \sin(\angle \text{CAB}) = \dfrac{BC}{\sqrt{AB^2+BC^2} } \\ \cos(\angle \text{ABM}) &=& \dfrac{BC}{\sqrt{AB^2+BC^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{10^2+24^2} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{\sqrt{676} } \\\\ \cos(\angle \text{ABM}) &=& \dfrac{24}{ 26 } \\\\ \mathbf{ \cos(\angle \text{ABM}) } & \mathbf{=} & \mathbf{ \dfrac{12}{ 13 } } \\ \hline \end{array} \)

What is the sum of all positive integers v for which lcm(v,20)=60.

\(\begin{array}{|r|r|r|r|c|} \hline v & \text{factor } v &\text{factor } 20 & \text{factor } 60 &\text{lcm}(v,20 ) \\ \hline 3 & {\color{red}3} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 6 & {\color{red}3} \times 2 & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 15 & {\color{red}3} \times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 30 & {\color{red}3} \times 2\times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 12 & {\color{red}3} \times {\color{blue}2^2} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline 60 & {\color{red}3} \times {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{green}5} & {\color{blue}2^2} \times {\color{red}3} \times {\color{green}5} & 60 \\ \hline \text{sum } = 126 \\ \hline \end{array} \)