Two rooms are connected by a hallway that has a bend in it so that it is impossible to see one room while standing in the other. One of the rooms has three lights switches. You are told that exactly one of the switches turns on a light in the other room, and the other two are not connected to any lights. What is the fewest numbers of times you would have to walk to the other room to figure out which switch turns on the light? And the follow up question is: Why is the answer to the preceding question “one”? (Look out, this question uses properties of real lights as well as logic.)
2007-09-09 01:04:32 · 2 answers · asked by Livelife 1 in Science & Mathematics ➔ Physics
You switch on the first switch, wait a minute or so, then switch it off again and switch the second switch on.
Then you go to the other room.
If the light is on, you know that the second switch controls the light.
If the light is off, you feel the light bulb. If it is warm, you know that the first switch controls the light.
If it is still cold, you know that it must be the third switch.
2007-09-09 01:50:48 · answer #1 · answered by lunchtime_browser 7 · 4⤊ 0⤋
Turn two of the switches on, then walk to the other room. If the light is not on, then it is the third switch that turned the light on.
Of course, if the light was on, then you would have to return, turn one switch off, and then go back to see if the light was on or off.
But the question asked "What is the fewest number of times..."; not "what is the least number of times you would have to return to identify which switch turned the light on".
The fewest number of times is 1; but only in the case when it was the third switch. You MAY have to return again, if you were unlucky.
The samee answer can be provided by turning one switch on, but your chance of being correct is then one in three, instead of one in two; but you MAY make a lucky guess.
Edit: The answer below is correct - very clever.
2007-09-09 01:37:50 · answer #2 · answered by AndrewG 7 · 0⤊ 0⤋