Riddles thread

[Ranzha]

Flip all the switches, then go through the door.

Then how do you know which switch turns the light on?

EDIT:

That wouldn't solve it, they need to find which one, not turn on the light.

Exactly. Now how do you do this?

 

[cmhardw]

Flip all the switches, then go through the door.

Then how do you know which switch turns the light on?

EDIT:

That wouldn't solve it, they need to find which one, not turn on the light.

Exactly. Now how do you do this?

Spoiler

Just because the door is solid doesn't mean it closes itself. I would open the door, leave the door open, and test which switch turns on the light in the, now visible, other room. Is this it?

Oooh someone posted a math riddle. When does 3 + 4 = 111?

 

[Sa967St]

Oooh someone posted a math riddle. When does 3 + 4 = 111?

Spoiler

when you write 3 and 4 in binary, 011 + 100= 111

 

[Hyprul 9-ty2]

A man married fifteen women but wasn't arrested for polygamy. How is this possible?

He's the priest-dude?

 

[Ranzha]

Just because the door is solid doesn't mean it closes itself. I would open the door, leave the door open, and test which switch turns on the light in the, now visible, other room. Is this it?

I'm sorry, I forgot to add that in the question.
I know I missed something!
I'll edit it again.

EDIT: I edited it.

 

[Sa967St]

A man married fifteen women but wasn't arrested for polygamy. How is this possible?

Spoiler

He's the priest-dude?

yes.

 

[nlCuber22]

A man married fifteen women but wasn't arrested for polygamy. How is this possible?

He's the priest-dude?

yes.

 

[edd5190]

A man married fifteen women but wasn't arrested for polygamy. How is this possible?

Spoiler

He's the priest-dude?

yes.

"They didn't catch him yet" was my second guess

 

[Forte]

A man married fifteen women but wasn't arrested for polygamy. How is this possible?

Spoiler

He's the priest-dude?

yes.

"They didn't catch him yet" was my second guess

It's both.

 

[cmhardw]

Oooh someone posted a math riddle. When does 3 + 4 = 111?

Spoiler

when you write 3 and 4 in binary, 011 + 100= 111

Nicely done!

How about the flaw in this "proof"?

assume x > 0 (no tricky divide by zero nonsense)

x^2 = x^2 ; identity, also it is true on the interval x > 0

x^2 = x*x ; (expand the exponent)

x^2 = x + x + x + ... + x ; (x instances of the variable x) ; expand the meaning of multiplication

d/dx (x^2) = d/dx (x + x + x + ... + x) ; (x instances of the variable x) ; differentiate both sides

2*x = 1 + 1 + 1 + ... + 1 ; (x instances of the number 1) ; evaluate the derivative, remembering that the derivative is a linear operator.

2*x = x ; (add together the x instances of the number 1)

2 = 1 ; divide by x, which is ok here because I differentiate on the interval of x > 0, and because of this restriction diving by x has a meaning and is not undefined

2 must really equal 1 :rolleyes:

Chris

 

[Forte]

I have a riddle.

You are in a room with three light switches. Beyond the only door in the room, there is another room with a light bulb fixed on the ceiling. One of the light switches turns the light bulb on. Your job is to find out which light switch turns on the light bulb in the other room. However, once you enter the room with the light bulb, there is no going back into the room with the light switches. You may not open the door to see if the light has turned on.
Keep in mind that the door is solid, and you cannot see traces of light coming from beyond it. Also, you may not trace wires.
How do you do it?

EDIT: Sorry for double post, but I think it was necessary.

EDIT 2: Added the statement "You may not open the door to see if the light has turned on."

[11:06:20 PM] Forte Shinko says: you just turn one switch on for a really long time
[11:07:29 PM] Forte Shinko says: then turn it off and turn another one on

go into the light room, examine the light:
[11:07:34 PM] Forte Shinko says: if it's hot, then it was #1
[11:07:41 PM] Forte Shinko says: if it's [lit up], then it's #2
[11:07:46 PM] Forte Shinko says: if neither, then #3

 

[Ranzha]

I have a riddle.

You are in a room with three light switches. Beyond the only door in the room, there is another room with a light bulb fixed on the ceiling. One of the light switches turns the light bulb on. Your job is to find out which light switch turns on the light bulb in the other room. However, once you enter the room with the light bulb, there is no going back into the room with the light switches. You may not open the door to see if the light has turned on.
Keep in mind that the door is solid, and you cannot see traces of light coming from beyond it. Also, you may not trace wires.
How do you do it?

EDIT: Sorry for double post, but I think it was necessary.

EDIT 2: Added the statement "You may not open the door to see if the light has turned on."

[11:06:20 PM] Forte Shinko says: you just turn one switch on for a really long time
[11:07:29 PM] Forte Shinko says: then turn it off and turn another one on

go into the light room, examine the light:
[11:07:34 PM] Forte Shinko says: if it's hot, then it was #1
[11:07:41 PM] Forte Shinko says: if it's [lit up], then it's #2
[11:07:46 PM] Forte Shinko says: if neither, then #3

Indeed!
Where did you find this knowledge?

 

[Sa967St]

Nicely done!

How about the flaw in this "proof"?

assume x > 0 (no tricky divide by zero nonsense)

x^2 = x^2 ; identity, also it is true on the interval x > 0

x^2 = x*x ; (expand the exponent)

x^2 = x + x + x + ... + x ; (x instances of the variable x) ; expand the meaning of multiplication

d/dx (x^2) = d/dx (x + x + x + ... + x) ; (x instances of the variable x) ; differentiate both sides

2*x = 1 + 1 + 1 + ... + 1 ; (x instances of the number 1) ; evaluate the derivative, remembering that the derivative is a linear operator.

2*x = x ; (add together the x instances of the number 1)

2 = 1 ; divide by x, which is ok here because I differentiate on the interval of x > 0, and because of this restriction diving by x has a meaning and is not undefined

2 must really equal 1 :rolleyes:

Chris

Spoiler

you didn't follow the Chain Rule when you took the derivative of (x + x + x + ... + x)

/not sure

 

[Forte]

Indeed!
Where did you find this knowledge?

Read it somewhere a long time ago. I used to be into riddles

 

[LewisJ]

It doesn't matter that he didn't follow the chain rule, all that matters is that d/dx(c) = 0 for all constants c. There is another variant of the 2 = 1 proof that does a bunch of factoring and then divides by 0 in the form of a binomial. Small easy to miss problems.

 

[Forte]

[20/12/09 8:35:58 PM] Forte Shinko: it's like
[20/12/09 8:36:06 PM] Forte Shinko: you can't differentiate x^x
[20/12/09 8:36:15 PM] Forte Shinko: the same way you differentiate x^2
[20/12/09 8:36:18 PM] Forte Shinko: or 2^x

I'm bad at explaining things

EDIT: So you can't take the derivative of (x+x+x . . .) (x times) like (x+x+x . . .) (12 times), because the times part has a variable too.

 

[(R)]

What am I
You usually have to pay to use me
I am often layed in the alley
once you stick your fingers in my holes, I'm ready to go

 

[Forte]

What am I
You usually have to pay to use me
I am often layed in the alley
once you stick your fingers in my holes, I'm ready to go

Bowling ball/prostitute/both in a world where females are bowling balls

 

[AndyRoo789]

What am I
You usually have to pay to use me
I am often layed in the alley
once you stick your fingers in my holes, I'm ready to go

Bowling ball/prostitute/both in a world where females are bowling balls

LOL!

 

[cmhardw]

Spoiler

you didn't follow the Chain Rule when you took the derivative of (x + x + x + ... + x)

/not sure

Spoiler

No, taking the derivative is sound. I took the derivative with respect to x of the function x^2 which really is just 2*x. The chain rule isn't necessary here, unless I defined x in terms of another variable.