# Hermitcraft – How much Mycelium is worth the Diamond Throne?!

## What is Hermitcraft?

Hermitcraft it’s a youtube gaming series based on the popular sandbox game Minecraft with notorious players from the platform usually called hermitcrafters or just hermits. There are diverse locations in the server where the hermits converge, but most of the action develop in the Shopping District, since that is the place where players set up stores to sell and buy all in-game items.

Anuncios

Elections were made to designate a mayor for the Shopping District, and since it was built on a mushroom biome, the new elected mayor renewed the appearance of the land by changing mycelium for grass but not everyone was happy about, thus the Mycelium Resistance was born.

While the Mycelium vs. Grass conflict started to heat up, the mayor started a «buy back!» program to motivate neutral hermits to eradicate mycelium from the district. The deal was: 2 stacks of 64 blocks of Mycelium for 5 diamonds.

## The Grian math problem

The Mycelium Resistance «mother spore», Grian, saw this exchange program as the perfect plant to buy the diamond throne in exchange for all the mycelium harvested by resistance, so in his episode Hermitcraft 7: Episode 48 – BUYING THE THRONE he bumped into a curious math problem.

Anuncios

We know roughly that there are seven a half stacks of diamonds (in the Diamond Throne)… So if we do some math very quickly, I need my calculator… This is like a math paper examination of school…

After saying that, Grian, was amazed that he needed actual math from high school to solve this problem and after thinking a bit, he stated an actual math problem like those in math books:

If you have nine diamonds per diamond block, and there are sixty four blocks in a stack, and there are seven and a half stacks, how many mycelium do you need to farm in order to take the diamond throne?

What we are going to do is to translate this question into math equations to solve the problem.

Anuncios

## The math translation

Considering the Mycelium Resistance already have 34 shulker boxes full of mycelium, let’s state what are the equalities we have:

• $9$ Diamonds is equal to $1$ Diamond Block.
• $64$ Diamond Blocks are equal to $1$ Diamond Blocks Stack.
• $2$ Mycelium Stack is equal to $5$ Diamonds.
• $1$ Shulker Box full of Mycelium is equal to $27$ Mycelium Stacks.
• $34$ Shulker Box full of Mycelium is equal to $918$ Mycelium Stacks.
• $7 \frac{1}{2}$ diamond block stacks is equal to how many mycelium blocks?

First thing we need to do, is check how many diamonds are in $7 \frac{1}{2}$ diamond block stacks. Since seven and a half is a mixed number represented by $7\frac{1}{2}$, we have to rewrite it as a fraction, that is

$7\frac{1}{2} = 7 + \frac{1}{2} = \frac{15}{2}$

Then, we have $\frac{15}{2} \cdot 64$ diamond blocks, that is $480$ diamond blocks, and that is $480 \cdot 9 = 4 320$ diamonds.

We know that $2$ Mycelium Stacks are equal to $5$ Diamonds, in this way, we can state that if $M$ denotes a Mycelium Stack and $d$ denotes a diamond, then $\frac{M}{2} = \frac{d}{5}$. So, if we solve this equation for $M$, we have

$M = \frac{5}{2} \cdot d$

Considering this last equality, we can now replace $d$ with $4 320$, because that is the total of diamonds that the Diamond Throne, then, $M = \frac{2}{5} \cdot 4 320$ and that is $1 728$.

Anuncios

So, the Diamond Throne is worth $1 728$ mycelium stacks. However, the question is how many mycelium do you need to farm in order to take the diamond throne? We have to remember that the Mycelium Resistance already have 34 shulker boxes full of mycelium, that is $918$ Mycelium Stacks.

Then, the Mycelium Resistance needs to farm needs to farm $810$ mycelium stacks in order to take the diamond throne, and that is $810 \div 27 = 30$ shulker boxes full of mycelium.

# Point of intersection of two lines

If two straight lines intersect, we have mentioned that they intersect at a single point, however no mention has been made about the nature of this point. Graphically, the point of intersection between these two lines is the point where the two are exactly the same. From this fact, we can calculate the value of the coordinates that define it, formally, if we consider two lines expressed as follows

$l_1 : y = m_1 x + b_1$
$l_2 : y = m_2 x + b_2$

The point $P_0 = (x_0,y_0)$ is the intersection point of $l_1$ and $l_2$, if the values of $x_0$ and $y_0$ satisfy both equations at the same time. This is known as a system of linear equations that consists of two equations and two unknowns, nevertheless, we will not elaborate on this topic since noticing that the lines are expressed in the form slope intercept, we will simply equalize the expressions that define them to later calculate the value of the unknowns.

Let’s see with some examples how to calculate the intersection point between two lines using this technique.

Anuncios

## Examples

### Example 1

Calculate the intersection point between the lines $l_1 : y = 3x-3$ and $l_2 : y = -x + 1$.

For this we define our system of linear equations

$l_1 : y = 3x-3$
$l_2 : y = -x + 1$

We equal the two expressions that define these two lines, then we clear the variable $x$

$3x-3 = -x + 1$
$\Rightarrow 3x + x = 1 + 3$
$\Rightarrow 4x = 4$
$\Rightarrow x = \frac{4}{4}$
$\Rightarrow x = 1$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $y$. Let’s substitute the value of $x=1$ in $l_1$:

$y = 3(1)-3 \Rightarrow y = 3-3 \Rightarrow y=0$

Note that if we substitute the value of $x=1$ in the $l_2$ line, we get the same value for $y$:

$y = -(1) + 1 \Rightarrow y = -1+1 \Rightarrow y=0$

Therefore, we conclude that the intersection point between the lines $l_1$ and $l_2$ is $P_0 = (1,0)$ and we can also, locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 2

Calculate the intersection point between the lines $l_1 : y = -4x-2$ and $l_2 : y = \frac{1}{4}x + 3$.

For this we define our system of linear equations

$l_1 : y = -4x-2$
$l_2 : y = \frac{1}{4}x + 3$

We equal the two expressions that define these two lines, then we clear the variable $x$

$-4x-2 = \frac{1}{4}x + 3$
$\Rightarrow -4x - \frac{1}{4}x = 3 + 2$
$\Rightarrow -\frac{17}{4}x = 5$
$\Rightarrow x = -\frac{20}{17}$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=-\frac{20}{17}$. Let us substitute this value in $l_1$:

$y = -4\left( -\frac{20}{17} \right)-2 \Rightarrow y = \frac{80}{17} -2 \Rightarrow y = \frac{46}{17}$

Therefore, we conclude that the intersection point between the lines $l_1$ and $l_2$ is $P_0 = \left( -\frac{20}{17} , \frac{46}{17} \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

Anuncios

### Example 3

Calculate the intersection point between the lines $l_1 : y = x+5$ and $l_2 : y = 2$.

In this case it is necessary to raise a system of equations, because as $l_2$ is a horizontal line, we simply substitute the value of $y$ that defines it in the line $l_1$ and from there, we calculate the value of $x$. Then, if $y=2$ we have to

$2 = x+5 \Rightarrow -x = 5-2 \Rightarrow -x = 3 \Rightarrow x = -3$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left( -3 , 2 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 4

Calculate the intersection point between the lines $l_1 : y = -\frac{1}{5}x+2$ and $l_2 : x = -1$.

In this case it is necessary to raise a system of equations, because as $l_2$ is a vertical line, we simply substitute the value of $x$ that defines it in the line $l_1$ and from there, we calculate the value of $y$. Then, if $x=-1$ we have to

$y = -\frac{1}{5}(-1)+2 \Rightarrow y = \frac{1}{5}+2 \Rightarrow y = \frac{11}{5}$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left( -1 , \frac{11}{5} \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

Anuncios

### Example 5

Calculate the intersection point between the lines $l_1 : y = -3$ and $l_2 : x = 4$.

In this case it is necessary to raise a system of equations, because being $l_1$ a horizontal line and $l_2$ a vertical line, we can immediately conclude that the point of intersection between them is $(4,-3)$ and we can also locate it in the Cartesian plane.

We have seen the cases of intersections where the lines are expressed in the slope-ordered form, we now see the case in which we have lines expressed in a general way. formally, if we consider two lines expressed in the following way

$l_1 : a_1 x + b_1 y + c_1 = 0$
$l_2 : a_2 x + b_2 y + c_2 = 0$

Again, the point $P_0 = (x_0,y_0)$ is the point of intersection of $l_1$ and $l_2$, if the values of $x_0$ and $y_0$ satisfy both equations at the same time. However, the way to deal with this type of case is slightly different from a pending-ordered case.

In these cases it does not make sense to equalize the two expressions that define the lines, so the technique to find the solution consists in making operations between both equations to cancel one of the two variables. Let’s see with some examples how to calculate the solution of this type of systems of equations.

## Examples

### Example 6

Calculate the intersection point between the lines $l_1 : 2 x + 2 y - 1 = 0$ and $l_2 : - 2 x + y + 4 = 0$.

For this we define our system of linear equations

$l_1 : 2 x + 2 y - 1 = 0$
$l_2 : - 2 x + y + 4 = 0$

In this particular case, we can notice that in one equation is the expression $2x$ and in the other, the expression $-2x$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $y$, and thus obtain the value $y_0$ of our point of intersection.

$0x + 3y + 3 = 0$
$\Rightarrow 3y + 3 = 0$
$\Rightarrow 3y = -3$
$\Rightarrow y = -\frac{3}{3}$
$\Rightarrow y = - 1$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y=-1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y=-1$ in $l_1$:

$2 x + 2 (-1) - 1 = 0$
$\Rightarrow 2x - 2 - 1 = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow 2x = 3$
$\Rightarrow x = \frac{3}{2}$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( \frac{3}{2}, -1 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

Anuncios

### Example 6

Calculate the intersection point between the lines $l_1 : 2 x + 2 y - 1 = 0$ and $l_2 : - 2 x + y + 4 = 0$.

For this we define our system of linear equations

$l_1 : 2 x + 2 y - 1 = 0$
$l_2 : - 2 x + y + 4 = 0$

In this particular case, we can notice that in one equation is the expression $2x$ and in the other, the expression $-2x$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $y$, and thus obtain the value $y_0$ of our point of intersection.

$0x + 3y + 3 = 0$
$\Rightarrow 3y + 3 = 0$
$\Rightarrow 3y = -3$
$\Rightarrow y = -\frac{3}{3}$
$\Rightarrow y = - 1$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y=-1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y=-1$ in $l_1$:

$2 x + 2 (-1) - 1 = 0$
$\Rightarrow 2x - 2 - 1 = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow 2x = 3$
$\Rightarrow x = \frac{3}{2}$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( \frac{3}{2}, -1 \right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

Anuncios

### Example 7

Calculate the intersection point between the lines $l_1 : 3 x - 5 y + 2 = 0$ and $l_2 : x + y - 2 = 0$.

For this we define our system of linear equations

$l_1 : 3 x - 5 y + 2 = 0$
$l_2 : x + y - 2 = 0$

In the previous case we could cancel with relative simplicity the variable $x$ but in this particular case, we can notice that if we multiply the second equation by $5$ we obtain

$l_1 : 3 x - 5 y + 2 = 0$
$l_2 : 5x + 5y - 10 = 0$

Now, we can notice that in one equation is the expression $-5y$ and in the other, the expression $5y$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $x$, and thus obtain the value $x_0$ of our point of intersection.

$8x + 0y - 8 = 0$
$\Rightarrow 8x - 8 = 0$
$\Rightarrow 8y = 8$
$\Rightarrow y = \frac{8}{8}$
$\Rightarrow y = 1$

In this way, we can conclude that the coordinate value in the X-axis of the intersection point is $x=1$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $y$. Let’s substitute the value of $x=1$ in $l_1$:

$3 (1) - 5 y + 2 = 0$
$\Rightarrow 3 - 5y + 2 = 0$
$\Rightarrow -5x + 5 = 0$
$\Rightarrow -5x = -5$
$\Rightarrow x = \frac{-5}{-5}$
$\Rightarrow x = 1$

Therefore, we conclude that the intersection point between the $l_1$ and $l_2$ lines is $P_0 = \left( 1, 1 \right)$ and we can also, locate it in the Cartesian plane. We graph both lines making a table of values considering only the cutting points with the axes.

### Example 8

Calculate the intersection point between the lines $l_1 : 6 x - 5 y + 4 = 0$ and $l_2 : 4 x + 3 y - 5 = 0$.

For this we define our system of linear equations

$l_1 : 6 x - 5 y + 4 = 0$
$l_2 : 4 x + 3 y - 5 = 0$

In this case we must notice that the variables are accompanied by different coefficients, so it is not enough to multiply only one equation to cancel terms. We must then, multiply both equations by numbers that help us to cancel summands. Let us multiply the first equation by $4$ and the second equation by $-6$.

$l_1 : 24 x - 20 y + 16 = 0$
$l_2 : -24 x - 18 y + 30 = 0$

Now, we can notice that in one equation is the expression $24x$ and in the other, the expression $-24y$, therefore, we can add both equations to obtain that

Considering the resulting equation, we can clear the variable $x$, and thus obtain the value $x_0$ of our point of intersection.

$0x - 38y + 36 = 0$
$\Rightarrow -38y + 36 = 0$
$\Rightarrow -38y = -36$
$\Rightarrow y = \frac{38}{36}$
$\Rightarrow y = \frac{19}{18}$

In this way, we can conclude that the value of the Y-axis coordinate of the intersection point is $y = \frac{19}{18}$ and taking into account that this value is common in both lines, we can substitute it in the lines of our preference to calculate the value of $x$. Let’s substitute the value of $y = \frac{19}{18}$ in $l_2$:

$4 x + 3 \left( \frac{19}{18} \right) - 5 = 0$
$\Rightarrow 4 x + \frac{19}{6} - 5 = 0$
$\Rightarrow 4 x - \frac{11}{6} = 0$
$\Rightarrow 4 x = \frac{11}{6}$
$\Rightarrow x = \frac{11}{24}$

Therefore, we conclude that the point of intersection between the lines $l_1$ and $l_2$ is $P_0 = \left(\frac{11}{24},\frac{19}{18}\right)$ and we can also locate it in the Cartesian plane. We graph both lines making a table of values considering only the cut-off points with the axes.

# The Conjugate of a Sum

Next we will define an expression that is closely related to the difference of squares, because when we find the addition (or subtraction as the case may be) of two real numbers, we can define an expression that will allow us to write that subtraction as a difference of squares.

Formally, if $a$ and $b$ are two real numbers, the conjugate of the sum $(a+b)$ is defined as $(a-b)$. Similarly, the conjugate of the subtraction $(a-b)$ is defined as $(a+b)$. That is, the sign between the two is changed. The importance of the conjugate lies in the fact that the product of an addition by its conjugate is equal to a difference of squares, that is,

This equality can be deduced by performing the distributive property of the real numbers, let’s see then,

This type of expressions is often found in the development of algebraic operations and is used mainly to simplify operations, let’s see in the following examples how to identify the conjugation of some expressions:

Anuncios

### Examples

#### Example 1

Identify the conjugate of $12 - 5$. It does not have much sense to identify the conjugate of this expression because we can simply make the subtraction and obtain 7 as a result.

#### Example 2

Identify the conjugate of $\sqrt{12} - 5$. Note that one of the involved sums is the square root of twelve, so it cannot be subtracted with five, so we conclude that its conjugate is $\sqrt{12} + 5$.

#### Example 3

Identify the conjugate of $3 + \sqrt{8}$. Note that one of the summands involved is the square root of eight, so it cannot be added with three, so we conclude that its conjugate is $3 - \sqrt{8}$.

#### Example 4

Identify the conjugate of $3x - 7$. Note that one of the sums involved is three multiplied by one unknown, so it cannot be subtracted with seven, then, we conclude that its conjugate is $3x + 7$.

#### Example 5

Identify the conjugate of $15 + 4x$. Let’s notice that one of the involved sums is four multiplied by one unknown, therefore it cannot be added with 15, then, we conclude that its conjugate is $15 - 4x$.

#### Example 6

Identify the conjugate of $6 + \sqrt{x+2}$. This subtraction cannot be done, so we conclude that your conjugate is $6 - \sqrt{x+2}$. Noting that the sign inside the root does not change.

# Difference of two squares

When carrying out mathematical operations it is common to find subtractions between two numbers, however, when finding the subtraction of the squares of two numbers we will say that this is a difference of squares and it is of our particular interest because through the distributive property, we can express it as the product of two factors.

Formally, if $a$ and $b$ are two real numbers, then the difference of their squares will be equal to the sum of the first plus the second, multiplied by the subtraction of the first by the second, that is,

This equality can be deduced by performing the distributive property of the real numbers, let’s see then,

This type of expression is often found in the development of algebraic operations and is used mainly for factoring operations, let’s see in the following examples how to apply this operation

Anuncios

## Examples

### Example 1

Factorize the expression $5^2 - 3^2$. Note that in this case, we can simply apply the power of each of the summands and perform the subtraction directly.

$5^2 - 3^2 = 25 - 9$

$= 16$

### Example 2

Factorize the expression $x^2 - 9$. We notice that in this case, one of the summands is an x squared and the other one is a nine, so we cannot make the subtraction between them so we apply the difference of squares noting that nine is equal to three squared.

$x^2 - 9 = x^2 - 3^2$

$= (x-3)(x+3)$

### Example 3

Factorize the expression $x^2 - 2$. We notice that in this case, one of the summands is an x-squared and the other is two, so we cannot perform the subtraction between them so we apply the difference of squares noting that two can be rewritten as $2 = \left( \sqrt{2} \right)^2$.

$x^2 - 2 = x^2 -\left( \sqrt{2} \right)^2$

$= \left(x-\sqrt{2} \right) \left(x+\sqrt{2} \right)$

In this way, we can notice that if the square root of a number is not exact, it can be rewritten to use the difference of squares.

Anuncios

### Example 4

Factorize the expression $8 - x^6$. We notice that in this case, one of the summands is 8 and the other one is x to six, so we cannot make the subtraction between them so we apply the difference of squares noting that eight can be rewritten as $8 = \left( \sqrt{8} \right)^2$ and x to six as $x^6 = \left( x^3 \right)^2$.

$8 - x^6 = \left( \sqrt{8} \right)^2 - \left(x^3 \right)^2$

$= \left(\sqrt{8}-x^3 \right) \left(\sqrt{8}+x^3 \right)$

### Example 5

Factorize the expression $36x^4 - 5x^8$. We notice that in this case, we cannot make the subtraction between them so we apply the difference of squares using the observations exposed in the previous examples.

$36x^4 - 5x^8 = \left( 6x^2 \right)^2 - \left( \sqrt{5}x^4 \right)^2$

$= \left(6x^2-\sqrt{5}x^4 \right) \left(6x^2+\sqrt{5}x^4 \right)$

# The Notable Product

The remarkable product is a particular case of the distributive property that gives us the perfect square trinomial as a result and establishes that, if $a$ and $b$ are two real numbers, the square of the sum of them is equal to the first squared plus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the sum of two numbers by that same sum, let’s see then,

Similarly, if $a$ and $b$ are two real numbers, the square of the subtraction between the two is equal to the first squared minus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the substraction of two numbers by that same substraction, let’s see then,

This type of expression is often found in the development of algebraic operations because we cannot always carry out the sum that is inside the parentheses, let’s see in the following examples how to apply this operation:

## Ejemplos

### Ejemplo 1

Apply the notable product to expand the expression $(3 + 2)^2$. We add the two elements within the parentheses and square as follows:

$(3 + 2)^2$
$= 5^2$
$= 25$

### Ejemplo 2

Apply the notable product to expand the expression $(3 + \sqrt{2})^2$. Note that one of the addends involved is the square root of two, therefore it cannot be added with three.

$(3 + sqrt{2})^2$
$= 3^2 + 2(3)(sqrt{2}) + (sqrt{2})^2$
$= 9 + 6sqrt{2} + 2$
$= 11+6sqrt{2}$

### Ejemplo 3

Apply the notable product to expand the expression $(\sqrt[3]{6} - 4)^2$. Note that one of the addends involved is the cube root of six, therefore it cannot be subtracted with four.

$(sqrt[3]{6} - 4)^2$
$= (sqrt[3]{6})^2 -2(sqrt[3]{6})(4) + 4^2$
$= (sqrt[3]{6})^2 -8sqrt[3]{6} +16$

### Ejemplo 4

Apply the notable product to expand the expression $(x + 7)^2$. Note that one of the addends involved is an unknown, therefore it cannot be added with seven.

$(x+7)^2$
$= x^2 + 2(x)(7) + 7^2$
$= x^2 +14x + 49$

### Ejemplo 5

Apply the notable product to expand the expression $(2x-8)^2$. Note that one of the addends involved is an unknown multiplied by two, therefore it cannot be subtracted with eight.

$(2x-8)^2$
$= (2x)^2 - 2(2x)(8) + 8^2$
$= 4x^2 - 32x + 64$

### Ejemplo 6

Apply the notable product to expand the expression $(x^2 + x^5)^2$. Note that one of the addends involved is x squared and the other is x raised to five, therefore they cannot be added.

$(x^2 + x^5)^2$
$= (x^2)^2 + 2(x^2)(x^5) + (x^5)^2$
$= x^4 + 2x^7 + x^{10}$