[Daily puzzle] Light bulb problem \ stacker news ~science

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747 sats \ 15 comments \ @south_korea_ln 3 Nov science

Too lazy on a Sunday night to go for an actual math problem. I'm sure many of you know this problem already. But if you haven't, here it is:

You are in a room with three switches. Each switch corresponds to one of three light bulbs in the next room, but you don’t know which switch goes to which bulb. You can turn the switches on and off as you like, but you may only go into the other room once.

How can you determine which switch controls which light bulb?

Previous iteration: #750725 (answer in #750751 with intuitive argument in #750897)

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133 sats \ 0 replies \ @denlillaapan 3 Nov

So all I have is this:

If there were only two bulbs, it'd be easy: turn one on and keep one off, go into room and check.

Problem is when 3, because there are two either dark or shining and we can't figure out which is which ...except if the light bulb can get warm (i.e., we use a third instrument to match the number of variables).

So: turn two on, leave them for a minute. Turn one of them off, then go into the room. There will be one light on, which obvs corresponds to the switch that's on. Now we touch the two dark ones, and whichever is still a little bit hot corresponds to the switch we had on but turned off. (The cold dark one = the switch we never turned on)

Am I getting, um, warm???

276 sats \ 1 reply \ @Aardvark 3 Nov

My friend couldn't remember if he left the light in in his garage on or not. He continually had to keep checking. Eventually he put a light sensor on a raspberry pi, and had it send him push notifications if the garage light was left on.

So clearly, the best option is to get 3 raspberry pi's with light sensors, and put them in the rooms.

0 sats \ 0 replies \ @JesseJames 3 Nov

that is the best answer...!

142 sats \ 0 replies \ @CruncherDefi 3 Nov

Pick a switch A. Continually turn it off and on for a really long time (we want the light bulb to burn out). Turn on switch B. Switch C is turned off.

You go into other room.

Burned bulb was controlled by A, lit one by B, non-burned unlit by C.

149 sats \ 8 replies \ @bordalix 3 Nov

I have one for you:

A friend of mine has 2 daughters (not twins) under 10 years old and wants you to guess the sum of their ages, but the only clue he gives is that the last digit of the product of their ages is the number of their house's door.

300 sats \ 1 reply \ @bordalix 9 Nov

Solution:

At first glance this puzzle seems unsolvable. But it has something weird, and this is the fact of “the last digit of the product of their ages” being a clue. How can something so vague be a clue? Unless it’s not that vague. Let’s explore it a little bit.

We make a table with all possible ages for each daughter and calculate the last digit of their product - i.e. (a*b).mod(10)

We can get rid of half the table (since 6x4 = 4x6) and since the problem says they are not twins, we can also get rid of the main diagonal, the one with the square numbers.

Now let's evaluate each digit on the table, and see what we get.

Let’s start with digit 1: it appears once (on 7x3) and the sum of the factors is 10.

Digit 2 on the other hand, appears 6 times (on 2x1, 4x3, 6x2, 7x6, 8x4 and 9x8) and the sum of factors is 3, 7, 8, 13, 12 and 17

Here is the final table

So, in order for the final digit to be a clue, it must tell us something unique. Looking at this table, we realize only 2 digits fulfill this, the digits 1 and 9. So, if the last digit of the product of their ages is 1, that means they are 7 and 3 years old. And if the final digit is 9, that means they are 9 and 1 years old.

Now we have two possible answers, how to proceed? Well, the problem is not to guess their daughters age, but the sum of it. And since in both cases the answer is 10, that’s your answer.

4 sats \ 0 replies \ @south_korea_ln OP 9 Nov

Beautiful!

0 sats \ 5 replies \ @Scroogey 3 Nov

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22 sats \ 4 replies \ @south_korea_ln OP 3 Nov

Is that because 1 as last number is the only one that gives a unique solution to multiplying two non equal values below 10? E.g. 11, 31, 41,... can't be obtained in such a way?

0 sats \ 3 replies \ @Scroogey 3 Nov

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22 sats \ 0 replies \ @south_korea_ln OP 3 Nov

I like this problem. The answer only hit me after i read your suggestion though. I assumed trolling at first...

0 sats \ 1 reply \ @Scroogey 3 Nov

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1 sat \ 0 replies \ @bordalix 3 Nov

Correct. Remember, he didn't ask for her ages, he asked for the sum of it. In both possible solutions, the sum is the same, 10 ;)

0 sats \ 0 replies \ @WeAreAllSatoshi 3 Nov

Is there any assumed initial state of the lights i.e. are they all assumed to be off before we start fiddling?

0 sats \ 0 replies \ @OT 3 Nov

Leave the door open between the rooms.