# Classic Windemere Castle Problem-algebraic pattern equation 6th grade

• Sep 5th 2012, 05:22 PM
jfkirkland
URGENT! Classic Windemere Castle Problem-algebraic pattern equation 6th grade
I'm needing assistance with a 6th grade math problem (algebraic pattern equations). It's obviously popular as I've found the question on the web, but unfortunately, not the answer. Here goes:

Evelyn is reading about Windemere Castle in Scotland. Many years ago, when prisoners were held in various cells in the dungeon area, they began to dig passages connecting each cell to each of the other cells in the dungeon. If there were 20 cells in all, what is the fewest number of passages that had to be tunneled out over the years?

The only thing I know for sure is the answer is NOT 19.

Thanks so much!
• Sep 5th 2012, 08:47 PM
Soroban
Re: URGENT! Classic Windemere Castle Problem-algebraic pattern equation 6th grade
Hello, jfkirkland!

Quote:

Many years ago, when prisoners were held in various cells in the dungeon area,
they began to dig passages connecting each cell to each of the other cells in the dungeon.
If there were 20 cells in all, what is the fewest number of passages that had to be tunneled out over the years?

The only thing I know for sure is the answer is NOT 19.

Thanks so much!

Let's call the 20 cells: .$\displaystyle A,\,B,\,C,\,D\,\hdots\,T$

Consider cell $\displaystyle A.$
We can connect it to $\displaystyle B,\,C,\,D\,\hdots\,T.$
There will be 19 passages.

Repeat this process for each of the cells.
There will be: $\displaystyle 20\times 19 \,=\,380$ passages.

But our list will have many duplicates.

. . $\displaystyle \begin{array}{c}{\color{red}AB} \\ {\color{blue}AC} \\ AD \\ \vdots \\ AT \end{array} \quad \begin{array}{c}{\color{red}BA} \\ {\color{green}BC} \\ BD \\ \vdots \\ BT\end{array} \quad \begin{array}{c}{\color{blue}CA} \\ {\color{green}CB} \\ CD \\ \vdots \\ CT \end{array} \quad \cdots$

The passage from A to B is the same as the passage from B to A;
the passage from A to C is the same as the passage from C to A;
. . and so on.

In fact, our list has twice as many items than is necessary.

Therefore, there are: .$\displaystyle \frac{20\times 19}{2} \:=\:190$ passages.
• Sep 5th 2012, 08:55 PM
amillionwinters
Re: Classic Windemere Castle Problem-algebraic pattern equation 6th grade
In reality, I would say that the number could depend on the way the cells were arranged; however, the answer that is probably expected is 190 tunnels.

Let$\displaystyle c =$ the number of cells, then:

$\displaystyle \frac{c}{2}(c-1) \rightarrow \frac{20}{2}(20-1) \rightarrow 10(19)=190$

or:

$\displaystyle 19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1=19 0$
• Sep 6th 2012, 07:30 AM
Soroban
Re: Classic Windemere Castle Problem (Part 1)
Hello again, jfkirkland!

Since the problem mentioned "algebraic pattern equations",
. . we may be expected to solve it by looking for a pattern.

The following detailed explanation is too long for a single post.
I must submit it in two installments.

Consider the first few cases and determine the number of passages.
Let $\displaystyle n$ = number of cells.

$\displaystyle n = 2$
Connect the two cells.
There is 1 passage.

$\displaystyle n = 3$
Connect the three cells.
We will have a triangle with a cell at each vertex.
There are 3 passages.

$\displaystyle n = 4$
Connect the four cells.
We will have a quadrilateral with a cell at each vertex: 4 sides.
. . And there are 2 diagonals.
There are: 6 passages.

$\displaystyle n=5$
Connect the five cells.
We will have a pentagon: 5 sides.
. . And there are 5 diagonals (forming a pentagram).
There are: 10 passages.

$\displaystyle n=6$
Connect the six cells.
We will have a hexagon: 6 sides.
. . And there are 9 diagonals.
. . (a Star of David, plus 3 "diameters".)
There are: 15 passages.

We have this table:

.$\displaystyle \begin{array}{c|c} n & f(n) \\ \hline 2 & 1 \\ 3 & 3 \\ 4 & 6 \\ 5 & 10 \\ 6 & 15 \end{array}$

Take the difference of consecutive terms,

. . $\displaystyle \begin{array}{c|cccccccccccc} \text{Sequence} & 1 && 3 && 6 && 10 && 15 && \cdots \\ \hline \text{Difference} && 2 && 3 && 4 && 5 && \cdots\end{array}$

We see that the differences are increasing.
The next term is: $\displaystyle 15 + 6 \:=\:21$
And the next is: $\displaystyle 21 + 7 \:=\:28$

$\displaystyle \text{We want the formula for the sequence: }\:1,3,6,10,15\,\hdots$
These are called "triangular numbers".
The reason will be obvious in this diagram.

$\displaystyle \begin{array}{cccccccccc} n=2 && n=3 && n=4 && n=5 && n=6 \\ &&&&&&&& \circ \\ &&&&&& \circ && \circ\:\circ \\ &&&& \circ && \circ\:\circ && \circ\circ\circ \\ && \circ && \circ\:\circ && \circ\circ\circ && \circ\circ\circ\circ \\ \circ && \circ\:\circ && \circ\circ\circ && \circ\circ\circ\:\circ && \circ\circ\circ\circ\circ \\ 1 && 3 && 6 && 10 && 15 \end{array}$
• Sep 6th 2012, 07:37 AM
Soroban
Re: Classic Windemere Castle Problem (Part 1)
Hello again, fjkirkland!

Strange and annoying!
The system will NOT allow me to post the second half of my explanation.
I'll try again later . . .
• Sep 6th 2012, 01:52 PM
Soroban
Re: Classic Windemere Castle Problem (Part 2)
Hello again, fjkirkland!

How do we find the value of the $\displaystyle n^{th}$ triangular number?

Consider $\displaystyle n=5,\;f(n) = 10.$

We have: .$\displaystyle \begin{array}{c}\circ \\ \circ\:\circ \\ \circ\circ\circ \\ \circ\circ\circ\:\circ \end{array}$

Left-justify the array: .$\displaystyle \begin{array}{c} \circ\qquad\; \\ \circ\:\circ\;\;\quad \\ \circ\circ\circ\;\;\; \\ \circ\circ\circ\:\circ \end{array}$

Append an inverted copy of the triangle: . $\displaystyle \begin{array}{c}\circ\bullet\bullet\bullet\bullet \\ \circ\circ\bullet\bullet\bullet \\ \circ\circ\circ\bullet\bullet \\ \circ\circ\circ\circ\bullet \end{array}$

The rectangle has $\displaystyle 4\cdot 5 = 20$ objects.
The triangle has half that many: .$\displaystyle \frac{4\cdot5}{2} \:=\:10$

The general formula is: .$\displaystyle \frac{n(n-1)}{2}$