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Originally posted by JO753:

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Originally posted by JEWELSTAITEFAN:
3. For practice, can you say how old is a boy born halfway through year 5 BC when his birthday comes in 5 AD?

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9

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Originally posted by JO753:
OK, Einstien! I got a good wun for you!

Youre in a starship thats capable uv reaching lite speed. You want to see the ej uv the universe (or the border to nothing, the next universe, the end uv the universe, the boundry beyond which there iz nothing - call it wut you like)

According to our current understanding uv fiziks, after you reach lite speed, how long duz it take you to get there by your calander?

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Originally posted by JO753:
Timez up.

The ansr iz that you instantly arrive at the end uv the universe frum your perspective.

Or you can think uv it az leaving the universe. Or reaching the end uv time.

But actually, it cant be dun according to our current understanding.

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Originally posted by JO753:
Timez up.

The ansr iz that you instantly arrive at the end uv the universe frum your perspective.

Or you can think uv it az leaving the universe. Or reaching the end uv time.

But actually, it cant be dun according to our current understanding.

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Originally posted by JO753:
I dont know. I'll haf to think about it.

By the lojik that leadz to the above conclusion, no. If you had a speedometer that sumhow detected your rate thru space, it woud get to 186,000 mps & then either show zero or error, bekuz there woudnt be any space outside the ship.

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Originally posted by JO753:
Its perspectiv. Frum inside the ship, time duznt seem to stop or chanje its rate.

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Originally posted by JO753:
Brevity iz the essens uv wit!

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Originally posted by JO753:
Its perspectiv. Frum inside the ship, time duznt seem to stop or chanje its rate.

----------------------------
DUZ XaT SEM RiT TQ YQ? - Jubal Early

http://www.nooalf.com

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Originally posted by JO753:
I think you missed the main point, JSF.

You arent in the universe anymore. Its gon! There iz no destination. Your starting point iz gon. If you programmed Whitefall into the navicomputer, you missed it.

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Originally posted by Magonsdaughter:
You really pronounce from as frum?

I say it the say as bob cop dock

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Originally posted by JEWELSTAITEFAN:
The time that it took you to get there still exists.

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Originally posted by JO753:

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Originally posted by Magonsdaughter:
You really pronounce from as frum?

I say it the say as bob cop dock

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My Webster'z showz it both wayz. I'll lissin for it on TV. I think it will be difficult to hear the differens.

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Originally posted by JO753:

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Originally posted by JEWELSTAITEFAN:
The time that it took you to get there still exists.

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Duz it?

Frum a perspectiv outside the ship, lets say 100 lite yirz away with a very good telescope, you woud see the ship flying really fast forever.

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Originally posted by 6IXSTRINGJACK:
This article has managed to put together thoughts I've had in places like while taking a hot shower.

This girl I know recently sent me a quiz on FB that said she got a 115 IQ. She's constantly telling me how smart I am and I have no doubt she wanted me to get like a 160 on it. I only had a 120.



Let me ask you someting J0.....

I just did the part where they asked the question:

Jack is looking at Anne, but Anne is looking at George. Jack is married, but George is not. Is a married person looking at an unmarried person?

A) Yes
B) No
C) Cannot be determined


I had to read the question once, look at the answers and read it again before answering the correct "A".

Am I dumber than the person who heard the question one single time and knew the answer?


I don't think I am. I may not ever be a salesman or a politician because of it, but that doesn't mean that I wouldn't make a great accountant or movie editor with my own strengths...

I don't think that the time it takes to get to an answer is a good indication of the quotient at all....



Do Right, Be Right. :)

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Originally posted by JEWELSTAITEFAN:
Heard the puzzle on Brooklyn 99 couple nights ago. They ended up claiming that there was no solution. The Capt had tried for 20 years to solve it.


If you are going to search for this answer on the net or web, don't bother posting here, please.

I admit I did not solve it by the end of the episode - that distraction named Melissa Fumero was the only reason to watch the show.

12 guys on an island.
1 guy is either heavier or lighter than the rest - 11 weigh the exact same.
There is a scale/balance - like a teeter totter. But it can only be used 3 times.

How do you determine which of the 12 is not the same weight?

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Originally posted by JEWELSTAITEFAN:
Heard the puzzle on Brooklyn 99 couple nights ago. They ended up claiming that there was no solution. The Capt had tried for 20 years to solve it.


If you are going to search for this answer on the net or web, don't bother posting here, please.

I admit I did not solve it by the end of the episode - that distraction named Melissa Fumero was the only reason to watch the show.

12 guys on an island.
1 guy is either heavier or lighter than the rest - 11 weigh the exact same.
There is a scale/balance - like a teeter totter. But it can only be used 3 times.

How do you determine which of the 12 is not the same weight?

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Originally posted by 1kiki:

Quote:

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Originally posted by JEWELSTAITEFAN:
Heard the puzzle on Brooklyn 99 couple nights ago. They ended up claiming that there was no solution. The Capt had tried for 20 years to solve it.

If you are going to search for this answer on the net or web, don't bother posting here, please.

I admit I did not solve it by the end of the episode - that distraction named Melissa Fumero was the only reason to watch the show.

12 guys on an island.
1 guy is either heavier or lighter than the rest - 11 weigh the exact same.
There is a scale/balance - like a teeter totter. But it can only be used 3 times.

How do you determine which of the 12 is not the same weight?

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easy

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Divide the group in half - six men and six men. The side with the heavier man goes down because he adds more weight. Repeat with those six as 3 men and 3 men.

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You now have 3 men, one of whom is heavier.

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Balance any two. If one is heavier the balance will go down. If it doesn't then the two are exactly the same and the one you haven't balanced is heavier. 3 seconds to come up with the answer, a couple of minutes to type.

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Originally posted by 1kiki:
ADHD struck - again! (skipped over the 'or lighter'). But it still works with a twist. 3/3 + 3/3 gives you which set is different. From there you do 1/1.

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Originally posted by 1kiki:
ADHD struck - again! (skipped over the 'or lighter'). But it still works with a twist. 3/3 + 3/3 gives you which set is different. From there you do 1/1.

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Originally posted by 1kiki:
ADHD struck - again! (skipped over the 'or lighter'). But it still works with a twist. 3/3 + 3/3 gives you which set is different. From there you do 1/1.

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Originally posted by JEWELSTAITEFAN:

Quote:

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Originally posted by JEWELSTAITEFAN:
Heard the puzzle on Brooklyn 99 couple nights ago. They ended up claiming that there was no solution. The Capt had tried for 20 years to solve it.


If you are going to search for this answer on the net or web, don't bother posting here, please.

I admit I did not solve it by the end of the episode - that distraction named Melissa Fumero was the only reason to watch the show.

12 guys on an island.
1 guy is either heavier or lighter than the rest - 11 weigh the exact same.
There is a scale/balance - like a teeter totter. But it can only be used 3 times.

How do you determine which of the 12 is not the same weight?

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Maybe this one is too hard.

Perhaps I should ask for the first step of the solution - which might be a clue for solution. I should admit that at this time I have not checked if there is more than one solution (method of finding the answer), so it is possible that more than one exists. I do recall that in the episode Santiago started saying that her answer started with 6 islanders on each side of the see-saw, and Capt holt immediately said no, that was wrong - however, Capt Holt also later said that there was no solution to the logic puzzle, and that statement is clearly false.

Which of the following is the first step of the solution?
1. Place 6 islanders on each side of the see-saw. (teeter-totter).
2. Place 5 islanders on each side of the see-saw.
3. Place 4 islanders on each side of the see-saw.
4. Place 3 islanders on each side of the see-saw.
5. Place 2 islanders on each side of the see-saw.
6. Place 1 islander on each side of the see-saw.

Does anybody have the answer among the above 6? Those who may have looked up an answer online need not respond here.

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Originally posted by 1kiki:
No - I worked it out on paper.

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None of which I'm going to post.

It looks like there's more than 1 right answer.

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Select to view spoiler:

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Originally posted by JEWELSTAITEFAN:

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Originally posted by JO753:

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Originally posted by JEWELSTAITEFAN:
3. For practice, can you say how old is a boy born halfway through year 5 BC when his birthday comes in 5 AD?

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9

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Correct. 5.5 BC plus 5.5 AD would be 11, but the correct answer is 9.

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