Hi,everyone!

How to get the bad ball between 12 balls in 3 times only?
I mean, there's a balance ,which has no measures.
What's more, we don't know whether the bad ball is too light or not.

The question confused me,if anyone has some ideas, just tell me, and I do hope we could discuss the problem... :)

Professor Layton and the Curious Village Walkthough: 131 Heavier or Lighter?

This should help

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Originally Posted by mathwiz3502

Professor Layton and the Curious Village Walkthough: 131 Heavier or Lighter?

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Hi,mathwiz3502.
I think there's something wrong with your link or I have no authority to get in.
Thanks!

Try again, and if that doesn't work Google Curious Village heavier or lighter, it is the answer to your question

Quote:

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Originally Posted by mathwiz3502

Try again, and if that doesn't work google Curious Village heavier or lighter, it is the answer to your question

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Would you mind just pating the answer for me, please?:)
It's greatly appreciated that you do it, and I think I would like to dicuss the problem here.

Thanks!

Ok, first of all lets think about 1/3 of the problem
Lets say you put 4 on both sides and they are equal
Then you have 4 unknown balls
Now, you weigh two unknown balls with two "100% equal" balls

If it shifts weigh the two unknown balls again 1 on each side and the one that shifts the same way is the one

If it doesn't shift weigh on of the balls you never weighed to any "100% equal ball" and if it shifts it is it if it doesn't the other ball is it
OK

On is one, sorry

By the way, I have an another question.

Could you tell me what is the difference between "He goes to the grocery for buying some goods" and "He goes to the grocery to buy some goods ", please?

Thanks!

I really don't understand how for is used.

Quote:

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Originally Posted by mathwiz3502

Ok, first of all lets think about 1/3 of the problem
lets say you put 4 on both sides and they are equal
then you have 4 unknown balls
Now, you weigh two unknown balls with two "100% equal" balls

if it shifts weigh the two unknown balls again 1 on each side and the one that shifts the same way is the one

if it doesnt shift weigh on of the balls you never weighed to any "100% equal ball" and if it shifts it is it if it doesnt the other ball is it
ok

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Thanks!

First "if" that you said was correct, but the second one wasn't. I mean, you don't know the ball is heavier or lighter...

Sorry, I just read you profile and I don't believe you will hear "he goes to the grocery (store) for buying some goods" from anyone

OK, if those two balls balance you know they are equal totaling 10 equal balls that you know.
Then if you weigh one unknown ball with a "100% equal" ball then if it balances you have process of elimination because you know 11 "equal" balls so the one you never weighed
If it shifts that unknown ball has to be it because it is different from an "equal" ball

Add to the third line "is the unbalanced one"

Quote:

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Originally Posted by mathwiz3502

I really don't understand how for is used.

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Thanks! I think for is not correct in that way...

I've done it this morning, and I think it's a quiet difficult problem, because it needs patience!
Now I'm going to post it, if one isn't a patient one just take it easy and you never understand what I say...
First, we have to divide the 12 balls into 3 groups, like A,B, and C.
Second, we weighe A and B. There're three possibilities:

1.A equals B, so we know the bad ball is in the group C, and it's a quiet easy question. For example, we mark the 4 balls in the group C as c1,c2,c3,c4. Then, weigh c1 and c2, there're 3 possibilities, if c1 equals c2, we just weigh c3 and another normal ball in group A, and we know which ball is the bad one. And if c1 lighter than c2, we just weigh c1 and another normal ball, then the question is solved.

2.A is lighter than B: now you have to take a paper or pen to draw what I say, or it's hard for you to understand!
We mark the 8 balls as a1,a2,a3,a4 and b1,b2,b3,before.
Take a4 and before off, then switch a1 and b1,I mean, we will weigh (b1,a2,a3) and (a1,b2,c3),there're 3 possibilities:
First, (b1,a2,a3) still is lighter than (a1,b2,c3), we could make sure the bad one is a2 or a3 or b2.
Then we weigh a2 and a3, if a2 is lighter than a3, we could know the bad one is lighter than other balls, and a2 is the bad one!
Because of the first weigh we've already known a2+a3 must be lighter than two normal balls!
Second, (b1,a2,a3) equals (a1,b2,c3), we could make sure the bad one is in (a4,before,b3), then we b3 and before, if b3 equals before we could know the bad one is a4. And if b3 is heavier than before ,we could make sure the bad one is b3or before, then we've known that (b3+b4) must be heavier than two normals balls, so the heavier one is the bad ball!

3.A is heavier than B, the situation is like the second one.
Good luck for reading this thread, and I think it's a challengeing problem.
It helps us to think in a multidimensional way.

That's it shihouzhuge! But quiet is a homophone and quite sounds the same way. Quite means what you used multiple times in your response and quiet means soft.
Good job

Quote:

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Originally Posted by mathwiz3502

That's it shihouzhuge! But quiet is a homophone and quite sounds the same way. Quite means what you used multiple times in your response and quiet means soft.
Good job

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:DThanks,mathwiz3502. But I don't know what you exactly mean in " But quiet is a homophone and quite sounds the same way. Quite means what you used multiple times in your response and quiet means soft. ":confused:

You used quiet in your response right, well quiet means soft, the word you are looking for is quite

Hi again, mathwiz3502. Now I get it.

It's a wrong word that I used in my thread, but I think it's too late for me to modify it...

You're a quite careful boy, Thanks!