Yousaf has several plastic coins that are white on one side and grey on the other. He places \(16\) of these coins on a table with their white sides facing up, and flips them according to the following rules:
Flipping a coin changes the colour of the side facing up from either white to grey or from grey to white.
Yousaf flips exactly \(3\) different coins at a time. Each time he does this it is called a step.
Determine the fewest steps required until all \(16\) coins have their grey sides facing up.
Extension:
Suppose Yousaf now starts with \(2026\)
coins on the table with their white sides all facing up. Determine the
fewest steps required until all \(2026\) coins have their grey sides facing
up.
To simplify the language, a grey coin is a coin with its grey side facing up, and a white coin is a coin with its white side facing up. We start with \(16\) white coins and after some number of steps, we want to have \(16\) grey coins. There are four possible options for each step, assuming there are enough white or grey coins on the table. These options are listed in the following table, along with how each step affects the total number of grey coins:
| Step Description | Change in Number of Grey Coins |
|---|---|
| flip \(3\) white coins to grey | increase by \(3\) |
| flip \(2\) white coins to grey and \(1\) grey coin to white | increase by \(1\) |
| flip \(1\) white coin to grey and \(2\) grey coins to white | decrease by \(1\) |
| flip \(3\) grey coins to white | decrease by \(3\) |
Since we want the fewest steps possible, we should aim to flip \(3\) white coins to grey in as many of the steps as we can, as that will increase the total number of grey coins the fastest. If the total number of coins on the table were a multiple of \(3\), then in every step we could flip \(3\) white coins to grey. However \(16\) is not a multiple of \(3\). Therefore, in at least one of our steps we must not flip \(3\) white coins to grey.
To start with, in each of the first \(3\) steps we will flip \(3\) white coins to grey. There will then be \(3 \times 3 = 9\) grey coins and \(16-9=7\) white coins.
From here, we will only consider the remaining \(7\) white coins.
In step \(4\), we flip \(3\) white coins to grey. This results in \(3\) grey coins and \(4\) white coins, as shown.
In step \(5\), we flip \(2\) white coins to grey and \(1\) grey coin to white. This results in \(4\) grey coins and \(3\) white coins, as shown.
Finally, in step \(6\), we flip \(3\) white coins to grey. This results in \(7\) grey coins, as shown.
This method took a total of \(6\) steps to flip all \(16\) coins from white to grey. Since only one of these steps did not flip \(3\) white coins to grey, and we determined that at least one of the steps must do this, we could not have achieved our goal in fewer steps.
Therefore, the fewest steps until all \(16\) coins have their grey sides facing up is \(6\).
Solution to
Extension:
We can use a similar approach for \(2026\) coins. We notice that \(2026=3 \times 673 + 7\). Thus, if the first
\(673\) steps are to flip \(3\) white coins to grey, then we will be
left with \(7\) white coins. We already
determined that the fewest steps required to flip \(7\) white coins to grey is \(3\). Thus, the fewest steps required until
all \(2026\) coins have their grey
sides facing up is \(673+3=676\).