heureka

Determine a constant k such that the polynomial

\(\displaystyle P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2)\)

is divisible by x+y+z

\(\begin{array}{|clcll|} \hline \\ P(x, y, z)\text{ is divisible by } x+y+z \\ \\ \hline \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = (x+y+z) \cdot f(x,y,z) \\ \text{Set } x+y+z = 0 \\ \text{so:} \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \cdot f(x,y,z) \\ x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \\ \\ \text{Set arbitrary } x+y+z = 2-1-1= 0 \qquad \text{so } x=2, \quad y=-1, \text{ and } z=-1 \\\\ P(2, -1, -1) = 2^5 + (-1)^5 + (-1)^5 + k\left(2^3+(-1)^3+(-1)^3\right)\left(2^2+(-1)^2+(-1)^2\right) = 0 \\ 32-1-1 + k(8-1-1)(4+1+1) = 0 \\ 30 + k\cdot 6 \cdot 6 = 0 \\ 30 +36k = 0 \\ 36k = -30 \\ k = -\dfrac{30}{36} \\ \mathbf{ k = -\dfrac{5}{6} } \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline x&=& 1 \\ y&=& 2 \\ z &=& 3 \\ x+y+z &=& 6 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline P(1, 2, 3) &=& 1^5 + 2^5 + 3^5 - \dfrac{5}{6}\left(1^3+2^3+3^3\right)\left(1^2+2^2+3^2\right) \\ &=& 1 + 32+ 243 - \dfrac{5}{6}\cdot(1 +8+27)(1+4+9) \\ &=& 276 - \dfrac{5}{6}\cdot 36\cdot 14 \\ &=& 276 - 5\cdot 6\cdot 14 \\ &=& 276 - 420 \\ &=& -144 \\ &=& -24\cdot 6 \quad \text{This is divisible by } x+y+z = 6 \\ \hline \end{array}\)

When I run up a staircase, my stride can carry me up 1, 2, or 3 steps at a time. In how many ways can I run up a 9-step staircase (given that my last stride lands me on the 9th step)?

Let \(T_n\) be the number of ways to climb n stairs taking only 1 or 2 or 3 steps:

\(\begin{array}{|rcll|} \hline T_n &=& \mathcal{T}_{n+2} \qquad \mathcal{T} \text{ is the Tribonacci number } \\\\ && \text{case of n=9 steps}: \\\\ T_{9} &=& \mathcal{T}_{11} \qquad \\ && \mathcal{T}_0 = 0 \\ && \mathcal{T}_1 = 0 \\ && \mathcal{T}_2 = 1 \\ && \mathcal{T}_3 = 1 \\ && \mathcal{T}_4 = 2 \\ && \mathcal{T}_5 = 4 \\ && \mathcal{T}_6 = 7 \\ && \mathcal{T}_7 = 13\\ && \mathcal{T}_8 = 24\\ && \mathcal{T}_9 = 44\\ && \mathcal{T}_{10} = 81\\ && \mathcal{T}_{11} = 149\\ && \mathcal{T}_{12} = 274\\ && \ldots \\ T_{9} &=& 149 \\ \hline \end{array}\)

Four boxes with ball capacity of 3 , 5 , 7 and 8 are given .

In how many ways can 19 same balls be put together in these boxes .

\(\begin{array}{|r|cccc|} \hline &\text{Box with} &\text{Box with}& \text{Box with}& \text{Box with} \\ &\text{capacity of 8} & \text{capacity of 7}& \text{capacity of 5} & \text{capacity of 3} \\ \hline 1. & 4 & 7 & 5& 3 \\ 2. & 5 & 6 & 5& 3 \\ 3. & 5 & 7 & 4& 3 \\ 4. & 5 & 7 & 5& 2 \\ 5. & 6 & 5 & 5& 3 \\ 6. & 6 & 6 & 4& 3 \\ 7. & 6 & 6 & 5& 2 \\ 8. & 6 & 7 & 3& 3 \\ 9. & 6 & 7 & 4& 2 \\ 10. & 6& 7 & 5& 1 \\ 11. & 7& 4 & 5& 3 \\ 12. & 7& 5 & 4& 3 \\ 13. & 7& 5 & 5& 2 \\ 14. & 7& 6 & 3& 3 \\ 15. & 7& 6 & 4& 2 \\ 16. & 7& 6 & 5& 1 \\ 17. & 7& 7 & 2& 3 \\ 18. & 7& 7 & 3& 2 \\ 19. & 7& 7 & 4& 1 \\ 20. & 7& 7 & 5& 0 \\ 21. & 8& 3 & 5& 3 \\ 22. & 8& 4 & 4& 3 \\ 23. & 8& 4 & 5& 2 \\ 24. & 8& 5 & 3& 3 \\ 25. & 8& 5 & 4& 2 \\ 26. & 8& 5 & 5& 1 \\ 27. & 8& 6 & 2& 3 \\ 28. & 8& 6 & 3& 2 \\ 29. & 8& 6 & 4& 1 \\ 30. & 8& 6 & 5& 0 \\ 31. & 8& 7 & 1& 3 \\ 32. & 8& 7 & 2& 2 \\ 33. & 8& 7 & 3& 1 \\ 34. & 8& 7 & 4& 0 \\ \hline \end{array} \)

Logarithm Finding, solving for k.

\(\text{Let $\log_{4}3=x$. Then $\log_{2}27=kx$. Find $k$.}\)

\(\large{ \begin{array}{|llcll|} \hline (1) & \log_{4}3=x & \text{or} & 4^x = 3 \\ (2) & \log_{2}27=kx & \text{or} & 2^{(kx)} = 27 \\ \hline \end{array} } \)

\(\large{ \begin{array}{|rcll|} \hline 2^{(kx)} &=& 27 \\ 2^{(\frac22 kx)} &=& 27 \\ 2^{(2\frac{kx}{2})} &=& 27 \\ \left(2^2\right)^{ \frac{kx}{2} } &=& 27 \\ 4^{( \frac{kx}{2} )} &=& 27 \\ 4^{( x\frac{k}{2} )} &=& 27 \\ \left(4^x\right)^{(\frac{k}{2} )} &=& 27 \quad & | \quad 4^x = 3 \\ 3^{(\frac{k}{2} )} &=& 27 \quad & | \quad 27 = 3^3 \\ 3^{(\frac{k}{2} )} &=& 3^3 \\ \frac{k}{2} &=& 3 \\ \mathbf{k} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} } \)

Part a)
In this multi-part problem, we will consider this system of simultaneous equations:

\(\begin{array}{lcr} 3x + 5y - 6z &=& 2 \\ 5xy - 10yz - 6xz &=& -41 \\ xyz &=& 6 \\ \end{array}\)

\(\text{Let $a = 3x$, $b = 5y$, and $c = -6z.$}\)

Determine the monic cubic polynomial in terms of a variable \(t\) whose roots are \(a\), \(b\), and \(c\).
Make sure to enter your answer in terms of \(t\) and only \(t\), in expanded form.

\(\begin{array}{|rcll|} \hline (t-a)(t-b)(t-c) &=& 0 \quad | \quad a=3x, \quad b = 5y \quad c=-6z \\ \left(t^2-t(a+b)+ab\right)(t-c) &=& 0 \\ t^3-t^2c-t^2(a+b)+t(a+b)c+tab-abc &=& 0 \\ t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ \\ \hline a+b+c &=& 3x + 5y -6z \\ &=& 2 \\ \\ \hline ac+bc+ab &=& -18xz-30yz+15xy \\ &=& 3\cdot(5xy-10yz-6xz) \\ &=& 3\cdot(-41) \\ &=& -123 \\ \\ \hline abc &=& 3x5y(-6z) \\ &=& -90xyz \\ &=& -90\cdot 6 \\ &=& -540 \\ \\ \hline t^3-t^2(a+b+c)+t(ac+bc+ab)-abc &=& 0 \\ t^3-t^2\cdot 2+t(123)-(-540) &=& 0 \\ \mathbf{t^3-2t^2 -123t +540} & \mathbf{=} & \mathbf{0} \\ \text{The roots are } t_1 = -12, t_2 = 5, t_3 = 9 \\ \hline \end{array}\)

Part b)
Given that \((x,y,z)\) is a solution to the original system of equations,

determine all distinct possible values of \(x+y\).

Roots are \(a\), \(b\), and \(c\)

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

\(\begin{array}{|rcll|} \hline a = 3x_1 &=& t_1 \\ 3x_1 &=& -12 \\ x_1 &=& -4 \\ \\ a = 3x_2 &=& t_2 \\ 3x_2 &=& 5 \\ x_2 &=& \frac53 \\\\ a = 3x_3 &=& t_3 \\ 3x_3 &=& 9 \\ x_3 &=& 3 \\\\ \hline x &=& \{-4,\frac53,3 \} \\ \hline \end{array} \)

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

\(\begin{array}{|rcll|} \hline b = 5y_1 &=& t_1 \\ 5y_1 &=& -12 \\ y_1 &=& -\frac{12}{5} \\ \\ b = 5y_2 &=& t_2 \\ 5y_2 &=& 5 \\ y_2 &=& 1 \\\\ b = 5y_3 &=& t_3 \\ 5y_3 &=& 9 \\ y_3 &=& \frac95 \\\\ \hline y &=& \{-\frac{12}{5},1,\frac95 \} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline c = -6z &=& t_1 \\ -6z &=& -12 \\ z_1 &=& 2 \\ \\ c = -6z &=& t_2 \\ -6z &=& 5 \\ z_2 &=& -\frac{5}{6} \\\\ c = -6z &=& t_3 \\ -6z &=& 9 \\ z_3 &=& -\frac32 \\\\ \hline z &=& \{2,-\frac{5}{6},-\frac32 \} \\ \hline \end{array}\)

Solutions:

\(\begin{array}{|r|r|r|r|} \hline x & y & z & \text{all distinct possible values of } x+y \\ \hline -4 & 1& -\frac32 & -3 \\ \hline -4 & \frac95& -\frac56 & -2.2 \\ \hline \frac53&-\frac{12}{5}& -\frac32 & -\frac{11}{15} \\ \hline \frac53& \frac95& 2 & \frac{52}{15} \\ \hline 3&-\frac{12}{5}& -\frac56 & \frac35 \\ \hline 3& 1& 2 & 4 \\ \hline \end{array}\)

There are constants  \(\alpha\) and \(\beta\) such that \(\dfrac{x-\alpha}{x+\beta} = \dfrac{x^2-80x+1551}{x^2+57x-2970}\).
What is \(\alpha+\beta\) ?

1.

\(\begin{array}{|rcll|} \hline && x^2-80x+1551 \\ &=& (x-40)^2 - 40^2+1551 \\ &=& (x-40)^2 - 1600+1551 \\ &=& (x-40)^2 - 49 \\ &=& (x-40)^2 - 7^2 \\ &=& [(x-40)-7][(x-40)+7] \\ &=& (x-47)(x-33) \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline && x^2+57x-2970 \\ &=& \left( x+\frac{57}{2} \right)^2 - \left(\frac{57}{2}\right)^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 28.5^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 3782.25 \\ &=& \left( x+28.5 \right)^2 - 61.5^2 \\ &=& [(x+28.5)-61.5][(x+28.5)+61.5] \\ &=& (x-33)(x+90) \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && \dfrac{x^2-80x+1551}{x^2+57x-2970} \\\\ &=& \dfrac{(x-47)(x-33)}{(x-33)(x+90)} \\\\ &=& \dfrac{x-47}{x+90} \\ \hline \end{array} \)

4.

\(\begin{array}{|rcll|} \hline \dfrac{x-\alpha}{x+\beta} &=& \dfrac{x-47}{x+90} \\ \\ \hline x-\alpha &=& x-47 \\ -\alpha &=& -47 \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{47} \\ \\ \hline x+\beta &=& x+90 \\ \mathbf{\beta} & \mathbf{=} & \mathbf{90} \\ \\ \hline \alpha + \beta &=& 47+90 \\ &=& 137 \\ \hline \end{array}\)

\(\mathbf{\alpha+\beta =137}\)

There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there? 

I assume \( \dfrac{12\cdot 10}{2}=60 \text{ hand shakes}\)

Medians AD and BE of triangle ABC are perpendicular. If AD = 15 and BE = 20,

then what is the area of triangle ABC?

not true to scale

\(\begin{array}{|rcll|} \hline A_{\triangle ABC}=A &=& \dfrac{ab\sin{C}}{2} \\ \hline \\ A_{\triangle EDC} &=& \dfrac{\dfrac{ab\sin{C}}{4}}{2} \\\\ &=& \dfrac{\dfrac{ab\sin{C}}{2}}{4} \\\\ &=& \dfrac{A}{4} \\ \hline \\ A_{ABDE} &=& \dfrac{AD\cdot BE}{2} \\\\ &=& \dfrac{15\cdot 20}{2} \\\\ &=& 15\cdot 10 \\\\ &=& 150 \\ \hline \\ A &=& A_{\triangle EDC} + A_{ABDE} \\ A &=& \dfrac{A}{4} + 150 \quad & | \quad \cdot 4 \\ 4A &=& A + 600 \\ 3A &=& 600 \quad & | \quad :3 \\ \mathbf{A} &\mathbf{=}& \mathbf{200} \\ \hline \end{array}\)

The area of triangle ABC is 200

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 $(A,B)$.

\(\text{Let $x^2 - x - 2 = (x-2)(x+1)$}\)

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad & | \quad \cdot (x-2)(x+1) \\\\ x + 7 &=& \dfrac{A(x-2)(x+1)}{x - 2} + \dfrac{B(x-2)(x+1)}{x + 1} \quad & | \quad x\ne2,\ x\ne -1 \\\\ x + 7 &=& A (x+1) + B(x-2) \\ \\ \hline x=-1: \quad -1 + 7 &=& A (-1+1) + B(-1-2) \\ \quad 6 &=& 0 -3B \\ \quad 6 &=& -3B \\ \quad -3B &=& 6 \quad | \quad :(-3) \\ \quad B &=& \frac{6}{-3} \\ \quad \mathbf{B} & \mathbf{=} & \mathbf{-2} \\ \hline x=2: \quad 2 + 7 &=& A (2+1) + B(2-2) \\ \quad 9 &=& 3A +0 \\ \quad 9 &=& 3A \\ \quad 3A &=& 9 \quad | \quad :3 \\ \quad A &=& \frac{9}{3} \\ \quad \mathbf{A} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)

Answer \((3,-2)\)

A rectangle has a base of 6x+7 and a height of 7. The area of the rectangle is 133in^2.

Solve for the missing sides.

\(\begin{array}{|rcll|} \hline A &=& \text{base} \times \text{height} \\ A &=&(6x+7) \times 7 \quad & | \quad A =133\ \text{in}^2 \\ 133 &=&(6x+7) \times 7 \quad & | \quad :7 \\ 19 &=& 6x+7 \\ \mathbf{6x+7} & \mathbf{=} & \mathbf{19} \\ \hline \end{array}\)

The missing side is 19.