Optimizing Turns: Exploring Patterns in Turning Over Hexagonal Mats

[Natasha1]

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

 

[AKG]

I don't know why you did the stuff with turning 6 cards at a time. It asks you to investigate what happens when you have a different total number of mats to turn over, not when you have a different number of mats to turn over at a time. Figure out how many moves it takes to turn over 5, 6, 7, 8, 9, 10, 11, 12, or 13 mats, given that you must turn over 3 at a time. Once you figure out 5-10, then figuring out 11-13 should be a no-brainer, and you should have fully understood the pattern.

 

[Gokul43201]

This statement...

it is impossible to do if you have an odd number of cards and an even number of cards to turn

...nevertheless, is true.

Have you studied any modulo arithmetic yet ?