heureka

by trial and error with function $\sigma_0(n)$ in wolfram alpha

For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors,
including $1$ and the number $110n^3$.
How many positive integer divisors does the number $81n^4$ have?

$\sigma_0(n)$ is the divisor function

$n=40: \begin{array}{|rcll|} \hline \sigma_0(110*40^3) &=& 110 \\ \sigma_0(81*40^4) &=& 325 \\ \hline \end{array}$

I have a bag with red and green marbles in it.
At the moment, the ratio of red marbles to green marbles is 3:2.
If I add 5 red marbles and 5 green marbles, the ratio will be 4:3.
How many red marbles will be in the bag after I add more?

$\begin{array}{|lrcll|} \hline (1)& \dfrac{r}{g} &=& \dfrac{3}{2} \\\\ & \mathbf{r} &=& \mathbf{\dfrac{3}{2}g} \\ \hline (2)& \dfrac{r+5}{g+5} &=& \dfrac{4}{3} \\\\ & r+5 &=& \dfrac{4}{3}(g+5) \\\\ & \dfrac{3}{2}g +5 &=& \dfrac{4}{3}(g+5) \\\\ & \dfrac{3}{2}g-\dfrac{4}{3}g &=& \dfrac{20}{3}-5 \\\\ & g\left(\dfrac{3}{2}-\dfrac{4}{3}\right) &=& \dfrac{5}{3} \\\\ & g\left(\dfrac{1}{6}\right) &=& \dfrac{5}{3} \\\\ & g &=& \dfrac{5}{3}* 6 \\\\ & \mathbf{g} &=& \mathbf{10} \\ \hline & \mathbf{r} &=& \mathbf{\dfrac{3}{2}g} \\ & r &=& \dfrac{3}{2}*10 \\ & \mathbf{r} &=& \mathbf{15} \\ & \mathbf{r+5} &=& \mathbf{20} \\ \hline \end{array}$

Solve $z^3 = 8*i$ in complex numbers.

$\begin{array}{|rcll|} \hline z^3 &=& 8*i \quad | \quad \text{cube root both sides} \\ z &=& \sqrt[3]{8i} \\ z &=& \sqrt[3]{8}\sqrt[3]{i} \\ \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline i &=& e^{i\frac{\pi}{2}} \\ \sqrt[3]{i}=i^{\frac{1}{3}} &=& \left(e^{i\left(\frac{\pi}{2}+ 2\pi*k \right)}\right)^{\frac13} \\ \mathbf{\sqrt[3]{i}} &=& \mathbf{e^{i\left(\frac{\pi}{6}+\frac{2\pi*k}{3}\right) } } \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline k=0: & \sqrt[3]{i} &=& e^{i \frac{\pi}{6} } = \cos(30^\circ)+i*\sin(30^\circ) \\ k=1: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{2\pi}{3}\right) } = \cos(150^\circ)+i*\sin(150^\circ) \\ k=2: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{4\pi}{3}\right) } = \cos(270^\circ)+i*\sin(270^\circ) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline z_1 &=& 2*\left( \cos(30^\circ)+i*\sin(30^\circ) \right) \\ z_1 &=& \sqrt{3} + i \\\\ z_2 &=& 2*\left( \cos(150^\circ)+i*\sin(150^\circ) \right) \\ z_2 &=& -\sqrt{3} + i \\\\ z_3 &=& 2*\left( \cos(270^\circ)+i*\sin(270^\circ) \right) \\ z_3 &=& -2i \\ \hline \end{array}$

In square ABCD, AD is 4 centimeters, and M is the midpoint of CD.
Let O be the intersection of AC and BM.
Let P be the intersection of DO and BC.
What is the ratio of BP to PC?

My attempt:

$\text{Let $BP = x,~ PC = 4-x,~D=(0,0)$}$

$\begin{array}{|rcll|} \hline \text{Line AC:} \quad y = 4-x \\ \text{Line BM:} \quad y = 2x-4 \\ \hline 2x-4 &=& 4-x \\ 3x &=& 8 \\ \mathbf{ x_{\text{O}} } &=& \mathbf{ \dfrac{8}{3} } \\ \hline y_{\text{O}} &=& 4-\dfrac{8}{3} \\ \mathbf{ y_{\text{O}} } &=& \mathbf{ \dfrac{4}{3} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dfrac{x_{\text{O}}}{y_{\text{O}}} &=& \dfrac{4}{4-x} \\\\ \dfrac{\dfrac{8}{3}}{\dfrac{4}{3}} &=& \dfrac{4}{4-x} \\\\ 2 &=& \dfrac{4}{4-x} \\\\ 2(4-x) &=& 4 \\ 4-x &=& 2 \\ x &=& 4-2 \\ \mathbf{x} &=& \mathbf{2} \\ \mathbf{4-x} &=& \mathbf{4-2=2} \\ \hline \end{array}$

$\dfrac{BP}{PC} = \dfrac{x}{4-x} = \dfrac{2}{2}$

Solve
$xy + x + y = 23 \\ yz + y + z = 31 \\ zx + z + x = 47$
in real numbers.

$\begin{array}{|rcll|} \hline xy + x + y &=& 23 \\ (x+1)(y+1) -1 &=& 23 \\ \mathbf{(x+1)(y+1)} &=& \mathbf{24} \qquad (1) \\ \hline yz + y + z &=& 31 \\ (y+1)(z+1) -1 &=& 31 \\ \mathbf{(y+1)(z+1)} &=& \mathbf{32} \qquad (2) \\ \hline zx + z + x &=& 47 \\ (z+1)(x+1) -1 &=& 47 \\ \mathbf{(z+1)(x+1)} &=& \mathbf{48} \qquad (3) \\ \hline \end{array}$

$\text{Let $a=x+1,~b=y+1,~c=z+1$}$

$\begin{array}{|rcll|} \hline ab &=& 24 \qquad (4) \\ bc &=& 32 \qquad (5) \\ ca &=& 48 \qquad (6) \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \dfrac{(4)*(6)}{(5)}: & \dfrac{ab*ca}{bc} &=& \dfrac{24*48}{32} \\ & a^2 &=& 36 \\ & \mathbf{a} &=& \pm \mathbf{6} \\ & x+1 &=& \pm 6 \\ & x_1 &=& 5 \\ & x_2 &=& -7 \\ \hline & \mathbf{(x+1)(y+1)} &=& \mathbf{24} \\ & (\pm6)(y+1) &=& 24 \\ & 6(y_1+1) &=& 24 \\ & y_1 +1 &=& 4 \\ & y_1 &=& 3 \\\\ & (-6)(y_2+1) &=& 24 \\ & y_2 +1 &=& -4 \\ & y_2 &=& -5 \\ \hline &\mathbf{(z+1)(x+1)} &=& \mathbf{48} \\ & (\pm6)(z+1) &=& 48 \\ & 6(z_1+1) &=& 48 \\ & z_1 +1 &=& 8 \\ & z_1 &=& 7 \\\\ & (-6)(z_2+1) &=& 48 \\ & z_2 +1 &=& -8 \\ & z_2 &=& -9 \\ \hline \end{array}$

Let $n$ be a positive integer greater than or equal to $3$.
Let $a,b$ be integers such that $ab$ is invertible modulo $n$ and
$(ab)^{-1} \equiv 2 \pmod {n}$.
Given $a+b$ is invertible, what is the remainder when
$(a+b)^{-1} \left(a^{-1}+b^{-1} \right)$ is divided by $n$?

$\begin{array}{|lrclcc|} \hline & && & \quad I. & \quad II. \\ & (ab)^{-1}=\dfrac{1}{ab} &\equiv& 2 \pmod{n} &| *a &| *b \\\\ \hline I. & \dfrac{a}{ab}&\equiv& 2a \pmod{n} \\ & \dfrac{1}{b} &\equiv& 2a \pmod{n} \\ & \mathbf{b^{-1}} &\equiv& \mathbf{2a \pmod{n}} \\\\ II. & \dfrac{b}{ab}&\equiv& 2b \pmod{n} \\ & \dfrac{1}{a} &\equiv& 2b \pmod{n} \\ & \mathbf{a^{-1}} &\equiv& \mathbf{2b \pmod{n}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\ &\equiv& 2(a+b)^{-1}(b+a) \\ &\equiv& 2*\left(\dfrac{a+b}{a+b}\right) \\ &\equiv& 2*1 \\ \mathbf{ (a+b)^{-1}(a^{-1}+b^{-1}) } &\equiv& \mathbf{ 2 \pmod{n} }\\ \hline \end{array}$

Let $n$ be a positive integer greater than or equal to $3$.

Let $a,b$ be integers such that $ab$ is invertible modulo $n$ and $(ab)^{-1} \equiv 2 \pmod{n}$.

Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by $n$?

The sum of the ages of the three Romano brothers is 63.
If their ages can be represented as consecutive integers,
what is the age of the middle brother?

$\begin{array}{|rcll|} \hline (r_2-1) + r_2 +(r_2+1) &=& 63 \\ 3r_2 &=& 63 \quad | \quad : 3 \\ \mathbf{r_2} &=& \mathbf{21} \\ \hline \end{array}$

The age of the middle brother is 21