- #1

Olinguito

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In summary, the problem asks if it is possible for all $n$ switches to be on at the end, given that $n$ lights are arranged in a circle with each operated by exactly one of $n$ switches. It is possible if and only if $n\equiv0\text{ or }1\pmod4$. To prove this, we use induction with a step of 4. The base cases are $n=1$ and $n=4$, and the solution for $n=4$ is presented. Then, we show that if there is a solution for $n=k$, it can be extended to a solution for $n=k+4$. The solutions for $n=5$ and $

- #1

Olinguito

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- #2

castor28

Gold Member

MHB

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It is possible if and only if $n\equiv0\text{ or }1\pmod4$.

To prove that the condition is necessary, we note that each switch must be flicked an odd number of times. As the total number of flicks is $\dfrac{n(n+1)}{2}$, we must have:

$$\begin{align*}

n &\equiv\frac{n(n+1)}{2}\quad&\pmod2\\

2n &\equiv n(n+1)&\pmod4\\

n(n-1)&\equiv0 &\pmod4

\end{align*}$$

and, as $n$ and $n-1$ cannot both be even, the conclusion follows.

To prove that the condition is sufficient, we proceed by induction with a step of 4. The base cases are $n=1$ and $n=4$. The first case is obvious, and, for the second case, we have the solution:

$$

\begin{array}{c|c|c|c|c|}

&1&2&3&4\\

\hline

1&1&0&0&0\\

2&0&1&1&0\\

3&1&1&1&0\\

4&1&1&1&1\\

\hline

\text{total}&3&3&3&1

\end{array}

$$

where '1' stands for a flick, each column after the first corresponds to a switch and each row corresponds to a round.

We now assume that there is a solution for $n=k$ and show that we can extend it to a solution for $n=k+4$. We add 4 columns and 4 rows to the table.

The first $k$ rows are extended with 0. In the rows $k+1$ to $k+4$, we fill the first $k$ columns with 1, and the last 4 columns with a copy of the solution for $n=4$. This is indeed a solution, because:

- Row $i$ contains $i$ 1's.
- The parity of the first $k$ column totals has not changed; it is odd by the induction hypothesis.
- The parity of the last 4 column totals is odd as it is in the case $n=4$.

The solutions for $n=5$ and $n=8$ are shown below:

$$

\begin{array}[t]{c|c|c|c|c|c|}

&1&2&3&4&5\\

\hline

1&1&0&0&0&0\\

2&1&1&0&0&0\\

3&1&0&1&1&0\\

4&1&1&1&1&0\\

5&1&1&1&1&1\\

\hline

\text{total}&5&3&3&3&1

\end{array}

\qquad\begin{array}[t]{c|c|c|c|c|c|c|c|c|}

&1&2&3&4&5&6&7&8\\

\hline

1&1&0&0&0&0&0&0&0\\

2&0&1&1&0&0&0&0&0\\

3&1&1&1&0&0&0&0&0\\

4&1&1&1&1&0&0&0&0\\

5&1&1&1&1&1&0&0&0\\

6&1&1&1&1&0&1&1&0\\

7&1&1&1&1&1&1&1&0\\

8&1&1&1&1&1&1&1&1\\

\hline

\text{total}&7&7&7&5&3&3&3&1

\end{array}

$$

[/sp]

- #3

Olinguito

- 239

- 0

Splendid! That is precisely the solution I have in mind. (Happy)

Yes, it is possible for all n switches to be turned on at the end of the challenge problem. This depends on the specific arrangement and number of switches, as well as the actions taken to manipulate them.

The factors that determine if all n switches can be turned on include the number and arrangement of switches, the initial state of the switches, and the rules for manipulating them. Additionally, the solution may also depend on the specific problem constraints and limitations.

Yes, it is possible for all n switches to be turned on with a finite number of moves. However, the number of moves required will depend on the specific problem and the approach used to solve it.

Some strategies that can be used to solve this challenge problem include trial and error, logical reasoning, and mathematical techniques such as algebra and combinatorics. It may also be helpful to break the problem down into smaller, more manageable parts.

Yes, this challenge problem can have real-world applications in fields such as computer science, engineering, and game theory. It can also help develop critical thinking and problem-solving skills that can be applied to various situations.

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