# Math Help - Hexagonal-shaped mats puzzle

1. ## Hexagonal-shaped mats puzzle

Here my problem to solve: (Could anyone share some insight?)

You have seven hexagonal-shaped mats in a line. These mats all have to be turned over, but you can only turn over exactly three at a time.
You can choose the three from anywhere in the line.
A mat may be turned over on one move and turned back over again on another.

What is the smallest number of moves you can do this in?
Try with other numbers of mats.
Do you notice any patterns in your findings?
Can you explain why these patterns occur?

Here is my answer so far:

So the minimum number of turns for 7 cards, turning 3 at a time is 3 as follows:

+ 3 + ( 2 – 1) + 3 = 7

You always start and end with the number you are turning, so in the above you start with a +3 and end with a +3, so you just need to add another one (2-1) to make 7.

From here you can conclude that it is impossible to do if you have an odd number of cards and an even number of cards to turn e.g. if you are turning 6 cards at a time the only numbers you can make are even as follows:

+ 6

( 5 - 1) = 4

( 4 – 2) = 2

( 3 – 3) = 0 (pointless)

( 2 – 4) = 2

( 1 – 5) = 4

- 6

2. ## The Pattern

Although I can't answer all the questions, I can give you a nice pattern.

$\frac{m}{3}+r=t$

$m$=number of mats
$t$=the amount of turns, or flips, it takes
$\frac{m}{3}$is rounded down to the nearest whole number
$r$=the remainder of $\frac{m}{3}$

This equation works with all numbers except for 4, which requires 4 turns.

I hope that helps a bit.

3. Hello, Natasha!

I came up with a different general solution.

You have seven hexagonal-shaped mats in a line.
These mats all have to be turned over, but you can only turn over exactly three at a time.
You can choose the three from anywhere in the line.
A mat may be turned over on one move and turned back over again on another.

What is the smallest number of moves you can do this in?
Try with other numbers of mats.
Do you notice any patterns in your findings?
Can you explain why these patterns occur?

We have $N$ mats.

If $N$ is a multiple of 3, the mats can be inverted in $\frac{N}{3}$ moves.

If $N$ is not a multiple of 3, then $(1)\;N = 3k + 1$ or $N = 3k + 2$

(1) If $N = 3k + 1$, then after $k$ moves, there is one mat face up.

. . . Consider the face-up mat and any three face-down mats: $U\;D\;D\;D$

. . . Invert 1, 2, and 3: $D\;U\;U\;D$
. . . Invert 1, 2, and 4: $U\;D\;U\;U$
. . . Invert 1, 3, and 4: $D\;D\;D\;D$

. . . All the mats are face-down in $k + 3$ moves.

(2) If $N = 3k + 2$, then after $k$ moves, there are two mats face up.

. . . Consider the two face-up mats and any two face-down mats: $U\;U\;D\;D$

. . . Invert 2,3, and 4: $U\;D\;U\;U$
. . . Invert 1,3, and 4: $D\;D\;D\;D$

. . . All the mats are face-down in $k + 2$ moves.

Therefore, the number of moves is: $\begin{Bmatrix} k & \text{if }N = 3k \\ k + 3 & \text{if }N = 3k+1 \\ k+2 & \text{if }N = 3k + 2\end{Bmatrix}$

4. Impressive stuff