I do this in nearly every forum I sign up for.

[tintytiny]

It's become a sort of tradition. Posting riddles.

 

I have a giant collection of riddles, and, well, I've gotta tell them somewhere, so I just victimize people wherever I go.

 

So here's the first one, to start us off. :) Don't know if there are any math geniuses around here, so I'll start with a lovely easy one.

 

Four people need to cross a bridge, in pitch black, to catch a train. (Don't ask me why, okay.) They've got seventeen minutes to get across the bridge before the train leaves.

 

They've only got one torch between them. Anyone crossing the bridge MUST carry the torch, which means each time people go across, someone has to come back to give the torch to the next people so they can see their way across. The bridge can only withstand the weight of two people.

 

The people all walk at different paces.

 

Person A - 1 minute to cross.

Person B - 2 minutes to cross.

Person C - 5 minutes to cross.

Person D - 10 minutes to cross.

 

If a pair goes across, they must travel at the rate of the slowest person.

 

In what order will they cross the bridge in order to ensure they get to the train in time?

 

Post if you know the answer. ^_^

[barkie]

I will need a better calculator than mine to work that out. :worried:

[Tynach]

That's easy. Person A is the one who goes back and forth on the bridge to keep holding the flashlight, and in that way it doesn't matter what order the rest of them go in. So, the pairs are like this:

 

AB

AC

AD

 

In no particular order.

 

I'm assuming that faster people can slow down to the rate of the slower people. Else, since no two people are the same rate, this entire system falls apart and becomes impossible.

 

Edit: Actually, if you count the time for the person going BACK across to give the flashlight to the next person... I don't see a way for it to be possible at all.

 

AB - 3 mins (2 across, 1 back)

AC - 6 mins (5 across, 1 back)

AD - 10 mins (don't have to go back)

 

Total: 19 minutes.

[tintytiny]

Barkie~ actually, for this problem, you don't need a calculator at all, unless you can't even add 1, 2, 5 and 10...

 

Tynach~ Yes, sorry, I edited it. :) And also, I do count the time going back. It's perfectly possible. ^_^

[shiranui_xiii]

Guys, we've found the real life Professor Layton! :laughingsmiley:

 

I've seen that riddle before, but the numbers were a bit different.

[Tynach]

I FOUND IT!!!

 

Notation: AB <A = "Person A and B go across, person A runs back."

 

Solution:

 

AB <B = 2+2 = 4

CD <A = 1+10 = 11

AB = 2

 

4+11+2 = 17

 

Edit: I was at first trying to keep every single move as the shortest possible, but then I tried doing it the opposite - get one or two moves to equal the largest allowable (I calculated 15), which would require the 'D' to move within the first couple moves, and I happened to guess the right one after a couple tries.

 

The first two moves add up to 15, and I realized the next move wouldn't be too large... And I guess I was right!

[Sweetdang]

Aha! Very clever! I love these. :)

 

More, more! (My turn to guess this time. :D )

[tintytiny]

Shira~ xDD Yeah, riddles just get told around until they come back to where they started.

 

Tynach~ Yes, you did. ^_^ Grats!

 

Sweetdang~ Right you are. xD Off we go.

 

You have seven coins, laid out like this:

 

    o

o o o o o

    o

 

As you can see, there's five in one row and three in the other. So, the question is: how can you move them so there's five in each row?

[Tynach]

You have seven coins, laid out like this:

 

    o

o o o o o

    o

 

As you can see, there's five in one row and three in the other. So, the question is: how can you move them so there's five in each row?

 

This is probably not right, but...

 

    o

ooooo

o

 

If you use a grid, and the only way to line them up is with a grid, then the three middle ones will be considered 'lined up' with the two on the top/bottom.

 

This is called aliasing when it comes to pixel graphics on a computer monitor. You can't draw a true diagonal line on a screen, so you draw a jagged one.

 

Edit: This is another lame 'answer' to the question:

 

ooooooo

 

Basically, it's two 'rows' overlapping each other.

₍oo⁽ooo₎oo⁾

[Sweetdang]

I am confused.

 

My brain cells are dying now, and I agree with Tynach on BOTH solutions. Also I liked the second one more. :P

[Bernard]

wait, just a quick question, can you stack a coin on top of another?

[tintytiny]

Tynach~ Nice try, but not exactly... er, what I was looking for. xD

 

Bernard~ Yeah, you can.

[Sweetdang]

If you can stack them, isn't this question magically answered? :D

 

  o

o o o

  o

 

And the one in the middle with 3 coins.

[Tynach]

If you can stack them, isn't this question magically answered? :D

 

  o

o o o

  o

 

And the one in the middle with 3 coins.

That's probably the correct answer, but myself being a 3D nerd, I disagree ;p

 

That would actually, from a 3D perspective, be 3 rows of 3... Not 2 rows of 5.

[tintytiny]

Yeah, that's the right answer. :) And Tynach, you're probably right, but for the purpose of this riddle, it's the only way. xD

 

Next.

 

You've got two hourglasses, one with 7 minutes' worth of sand and one with 11 minutes' worth of sand. Using only the two hourglasses, how do you accurately time 15 minutes?

[Sweetdang]

Turn both hourglasses over, so the sand starts. The moment the 7 mins one is over, you can start calculating the 15mins.

 

The remaining 4mins in the 11min one + when you finish you turn it over = 4+11 = 15mins! :D

[tintytiny]

Yep, you got it. xD Are these too easy? Do you want a harder one?

[Tynach]

Yep, you got it. xD Are these too easy? Do you want a harder one?

 

I think the winner of the previous one should post the next one, and so on. But that's just me ;p

[Bernard]

Well, i do think these are a bit too easy. But, i been late to solving each one because when i come on, the answer is already there. By the way, i agree that the winner should post up the next riddle. :P

[Sweetdang]

Butbut the winner doesn't knew any riddles! D:

 

Okayy hmm can it be a word one? :D No searching the answer online! >:[ (I don't even know if it's online... but everything is, honestly.)

 

I have a mouth

But I cannot eat.

I have a bed

But I cannot sleep.

I run all day and run all night -

What am I?

 

BEFORE you answer this I have to tell you that it's NOT Edward Cullen. Got that? Okay. :D

One more, cause it's so easy. :)

 

The more you take, the more you leave behind. What am I?

[tintytiny]

I know that one. It's a river. 0:-) And the second one's... a hole?

 

All right, a harder one then. :P

 

You have two eggs (let's pretend they're really, really hard to break, okay). There's a 100 storey building in front of you, and you know there's a certain floor that, if you drop an egg from it, it'll break, but if you go a floor lower and drop the egg, it won't. What's the least number of drops you have to do with the two eggs to be absolutely sure that you can find the designated floor? (You're allowed to break the eggs, as long as you can figure out the floor.)

[Tynach]

I know that one. It's a river. 0:-) And the second one's... a hole?

 

All right, a harder one then. :P

 

You have two eggs (let's pretend they're really, really hard to break, okay). There's a 100 storey building in front of you, and you know there's a certain floor that, if you drop an egg from it, it'll break, but if you go a floor lower and drop the egg, it won't. What's the least number of drops you have to do with the two eggs to be absolutely sure that you can find the designated floor? (You're allowed to break the eggs, as long as you can figure out the floor.)

 

Probably the wrong answer, but with the specific wording you chose, 2.

 

The least amount possible is 2, in case you happened to either:

A.) chose the exact floor first, or

B.) chose the floor above the exact floor first.

 

In the first case, the egg won't break, so you move up one level and you drop... And the egg will break. So you know know the correct floor.

 

In the second case, the egg will break, so you move down one level... And it doesn't break. So you know the correct floor.

 

But that's if you're EXTREMELY lucky.

 

The more you take, the more you leave behind. What am I?

I've heard this one before, but I don't remember what it is. 'A hole' doesn't sound right.

[Bernard]

The more you take, the more you leave behind. What am I?

 

I don't think it's the right answer because i think it's suppose to be an object, but my guess is time?

[Tynach]

I don't think it's the right answer because i think it's suppose to be an object, but my guess is time?

Sounds closer, but how do you 'leave behind' time?

[tintytiny]

Oh, I thought it might be a hole... like, you know, the more you take out of the hole, the more that's behind you. xD

 

Footsteps? Like, the more footsteps you take, the more you leave behind you?

 

This is what I said: 'What's the least number of drops you have to do with the two eggs to be absolutely sure that you can find the designated floor?'

 

If you're just guessing, well... you can't guarantee you'll guess correctly and have it done in two tries, right? ;) The minimum number you have to do to be absolutely certain you can find the floor within that number of tries, no matter what the floor happens to be.