Problem - 4688
$\textbf{Lighting Bulb}$
There are $100$ bulbs, all are off, each of which is controlled by a switch. Joe was playing with them in the following way:
- In the first round, he toggled every switch. So, all the lights are on now.
- In the second round, he toggled switches $2$, $4$, $6$, $\cdots$, $100$. Now half are on and half are off.
- In the third round, he toggled switches $3$, $6$, $9$, $\cdots$, $99$,
- $\cdots$
- In the $10^{th}$ round, he toggled switch $10$, $20$, $\cdots$, $100$
- $\cdots$
- In the $100^{th}$ round, he toggled the switch $100$
Now, the question is, how many bulbs are on at the end?
$\textbf{Answer}$
There are totally $10$ bulbs are on: $1$, $4$, $9$, $\cdots$, $100$, i.e. whose sequences are square numbers.
$\textbf{Analysis}$
Let's start analysis one by one and try to find the patterns.
- The $1^{st}$ bulb will be on because its status will be only toggled in the $1^{st}$ round.
- The $2^{nd}$ bulb will be off because it will be turned on in the $1^{st}$ round and then turned off in the $2^{nd}$ round.
- The $3^{rd}$ bulb will be off because it will be turned on in the $1^{st}$ round, and then turned off in the $3^{rd}$ round
- The $4^{th}$ bulb will be on because it will be turned on in the $1^{st}$ round, off in the $2^{nd}$ round, and on in the $4^{th}$ round
- $\cdots$
Here are some observations:
- A bulb will only remain to be on if its switch is toggled for an odd number of times
- The switch $N$, $(N=1, 2, \cdots, 100)$, will be toggled in round $n$ if $n$ is a factor of $N$. For example, the $4^{th}$ switch will be toggled only in rounds $1$, $2$, and $4$ because these three numbers are all the factors of $4$.
Hence, we conclude that the $N^{th}$ bulb will remain on if and only if $N$ contains an odd number of factors. Only square numbers meet this requirement. There are $10$ square numbers between $1$ and $100$, hence the answer is $10$.