Suppose I have 2011 numbered lights in a row, all of which are initially off. Th

Question

Suppose I have 2011 numbered lights in a row, all of which are initially off. Then I toggle all of the light switches, so they are all on. So then I toggle all the even number light switches so the even numbered light are on and the odd number lights are still on. Then I toggle all the lights whose number is a multiple of 3. Then I toggle all the lights whose number is a multiple of 4. The I toggle all the lights whose multiples of 5, 6, 7 etc. I keep going until the last step, when I toggle the 2011th light by itself. At the end of this procedure, which lights are on? How many lights are on and how many are off?

Can you explain to me thorughly and clearly, i mean a step by step explaintion because I have read this so many times and I just can't figure it out.
Thank you very much for your help!!!

Explanation / Answer

Its a simple problem of calculating the number of divisors of a number. In your problem, you are just toggling the light switches, in multiples of 1, 2, 3, 4, 5.....so that when you come to a light switch, you make it off if it is on and vice versa. In the process, for a particular numbered light switch you are visiting it exactly the number of divisors of that number. For example, if you consider the light numbered 6, you will visit when you are at multiples of 1, 2, 3 and 6. Thus you are visiting it 4 times. So this light will finally be off, for you are starting from making all the lights on. That is, at 1 the light is on, at 2 it is off, at 3 it is on, and at 6 it is off.

Now its a simple job of taking a pen and paper and counting the number of divisors of numbers upto 2011, in order to determine whether that numbered light will be on or off finally. Just check it out, now its an easy job.

Hope that helps.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.