A very tricky logic puzzle

[GlennB]

If you know this please stay schtum till tomorrow night at least - most people soon say it can't be done, so maybe we can watch some AN'ers squirming ....

 

You have a plain pair of balance-type scales, No weights, just a beam type balance like chemists used to use.

You have 12 identical-looking ball bearings.

One of the balls is odd, either heavier or lighter.

Your mission (should you accept it) is

1) to identify the odd ball

and

2) to tell whether it's heavier or lighter

 

you have to do this using nothing on the scales but the original 12 balls, in 3 weighing operations with a method that guarantees the result, rather just than getting lucky.

[Lid]

This may well be cheating, but you did not define or restrict what one weighing operation is.

 

WEIGHING OPERATION 1

Put one ball on each side of the scales. If they match put another ball on each side. Keep doing this until there is a mismatch. Note the pair which mismatched and which ball was heavier and which was lighter. I think this all counts as one weighing operation.

 

WEIGHING OPERATION 2

Compare the mismatched pair with two other balls (which will both be the standard weight). If they are heavier, then the ball identified earlier as heavier is the odd one. If they are lighter then it is the ball identified earlier as lighter.

Edited June 2, 2006 by Lid

[GlennB]

This may well be cheating, but you did not define or restrict what one weighing operation is.

 

WEIGHING OPERATION 1

Put one ball on each side of the scales. If they match put another ball on each side. Keep doing this until there is a mismatch. Note the pair which mismatched and which ball was heavier and which was lighter. I think this all counts as one weighing operation.

 

WEIGHING OPERATION 2

Compare the mismatched pair with two other balls (which will both be the standard weight). If they are heavier, then the ball identified earlier as heavier is the odd one. If they are lighter then it is the ball identified earlier as lighter.

 

Ah OK , that would be against the rules, which werem't defined very well An operation is putting some balls on the balance and seeing what happens. Adding or moving balls would be a new operation.

[Huge_Vitae]

Ah OK , that would be against the rules, which werem't defined very well An operation is putting some balls on the balance and seeing what happens. Adding or moving balls would be a new operation.

easy

[Dan]

I got it! But if you can do that one try this: (havent finished working it out yet!)

 

A camel works for a banana plantation that is 1,000 miles away from a market. The harvest contains 3,000 bananas that need to be moved to the market. The camel can carry 1,000 bananas at a time, however, the camel eats one banana for every mile it walks. What is the maximum number of bananas from the harvest that the camel deliver to the market?

 

Dan

[ayjay]

I got it! But if you can do that one try this: (havent finished working it out yet!)

 

A camel works for a banana plantation that is 1,000 miles away from a market. The harvest contains 3,000 bananas that need to be moved to the market. The camel can carry 1,000 bananas at a time, however, the camel eats one banana for every mile it walks. What is the maximum number of bananas from the harvest that the camel deliver to the market?

 

Dan

 

I'm guessing the answer is *one*, cos after the first trip to market there's one banana left but not enough bananas to get the camel back to the plantation for the rest.

[cidermonkey]

I've got him delivering 500 so far - just working on the permutations to see if I can get a few more.

 

I may be barking up the wrong tree and making an ar$e of myself but I think I've worked it out. Using whole bananas and whole miles I can get 602 bananas delivered to market.

Edited June 2, 2006 by cidermonkey

[GlennB]

easy

 

what would your first move be? i.e. how many each side?

[Huge_Vitae]

what would your first move be? i.e. how many each side?

4

[GlennB]

4

What if they don't balance - you've eliminated the remaining 4 standard balls, but don't know which side of the 8 the odd ball is.

What's the next move?