Easy Fools Gold Problem

You have 8 rocks. One contains real gold and is heavier than the other seven which all contain fools gold.

Using balancing scales, what is the least number of weighs you need to make to identify the real gold nugget?
Describe the method used.

3 -

1. Balance any four against the other four
2. Split the heaviest four into 2 pairs and balance against each other
3. Balance the heaviest pair against each other. The heaviest is the real gold.

Nope. This is the obvious answer, but it is not the most efficient way.

1. Weigh 3 against 3. If it balances, go to 2a. Otherwise 2b.
2a. Weigh the other two against each other - the heavier one is gold.
2b. Consider the heavier group of three from the first weighing, and weigh 2 of them against each other. If either is heavier, it's gold, otherwise the one you don't weigh is gold.

Originally posted by Acolyte
1. Weigh 3 against 3. If it balances, go to 2a. Otherwise 2b.
2a. Weigh the other two against each other - the heavier one is gold.
2b. Consider the heavier group of three from the first weighing, and weigh 2 of them against each other. If either is heavier, it's gold, otherwise the one you don't weigh is gold.

Or weigh 2 against 2 at the start, if one group of 2 is heavier do 2a on it, if the scales balance do 2b on the remaining 3.

Ok - how about if you don't know if the fools gold is lighter or heavier - but you know it's weight is different to real gold, what's the minimum then?

Ok - how about if you don't know if the fools gold is lighter or heavier - but you know it's weight is different to real gold, what's the minimum then?

Weigh 3 vs 3... if they balance, take any one of the six and weigh against the each of other two... whichever one DOESN'T balance is the one you're looking for...
(mimimum moves is 2... but maximum is 3)

Weigh 3 vs 3... if they DON'T balance, take one of the remaining two and weigh against each of the six... whichever one DOESN'T balance is the one you're looking for...
(minimum moves is again 2... but maximum is 7)

There's probably a way to streamline the second part....

Originally posted by iamatiger
Or weigh 2 against 2 at the start, if one group of 2 is heavier do 2a on it, if the scales balance do 2b on the remaining 3.

Ok - how about if you don't know if the fools gold is lighter or heavier - but you know it's weight is different to real gold, what's the minimum then?

Nope. There are 4 remaining, not 3. thus this won't work!

Originally posted by Acolyte
1. Weigh 3 against 3. If it balances, go to 2a. Otherwise 2b.
2a. Weigh the other two against each other - the heavier one is gold.
2b. Consider the heavier group of three from the first weighing, and weigh 2 of them against each other. If either is heavier, it's gold, otherwise the one you don't weigh is gold.

Correct: and beautifully explained too!

Originally posted by howardgee
Nope. There are 4 remaining, not 3. thus this won't work!

Yeah - realised that just after the option to edit the message expired 🙂 The second challenge where you don't know whether fools gold is light or heavy still stands though 🙂 The current best guess of 7 can be beaten significantly!

Originally posted by iamatiger
Yeah - realised that just after the option to edit the message expired 🙂 The second challenge where you don't know whether fools gold is light or heavy still stands though 🙂 The current best guess of 7 can be beaten significantly!

Let me try. Number the balls 1 to 8.
l: means the balance goes to the left
r: the balance goes to the right
=: the balance stays horizontal
1H means ball 1 is heavier
1L means ball 1 is lighter

Round 1: 1 2 3 vs 4 5 6

a) if l: then round 2: 1 4 vs 2 5

a1) if l: then round 3: 1 vs 7
if l: then 1H
if =: then 5L

a2) if =: then round 3: 3 vs 7
if l: then 3H
if =: then 6L

a3) if r: then round 3: 2 vs 7
if l: then 2H
if =: then 4L

b) if =: then round 2: 1 vs 7

b1) if l: then 7L
b2) if r: then 7H

b3) if = then round 3: 1 vs 8
if l: then 8L
if r: then 8H

c) if r: then round 2: 1 4 vs 2 5

c1) if l: then round 3: 4 vs 7
if l: then 4H
if =: then 2L

c2) if =: then round 3: 3 vs 7
if =: then 6H
if r: then 3L

c3) if r: then round 3: 5 vs 7
if l: then 5H
if =: then 1L

I hope I didn't mess up. Allways 3 actions, except in one case where 2 suffice.

Originally posted by Mephisto2
Let me try. Number the balls 1 to 8.
l: means the balance goes to the left
r: the balance goes to the right
=: the balance stays horizontal
1H means ball 1 is heavier
1L means ball 1 is lighter

Round 1: 1 2 3 vs 4 5 6

a) if l: then round 2: 1 4 vs 2 5

a1) if l: then round 3: 1 vs 7
if l: then 1H
if =: then 5L

a2) if =: then round 3: 3 v ...[text shortened]... f =: then 1L

I hope I didn't mess up. Allways 3 actions, except in one case where 2 suffice.

Looks comptetely correct! Congrats, nice one!

Originally posted by iamatiger
Looks comptetely correct! Congrats, nice one!

Here is another one.

You play a game with another person. In turn you have to call a number between 1 and 10 included. All numbers are added together. The player who has to reach or run over 100 as total loses. Your opponent starts with 7. What number do you call to make sure you will win?

Originally posted by iamatiger
Looks comptetely correct! Congrats, nice one!

No this is wrong! the right answer is above. 2 weighs is the minimum. See above! (Acolyte got it right)

Originally posted by howardgee
No this is wrong! the right answer is above. 2 weighs is the minimum. See above! (Acolyte got it right)

Sorry, I meant that his answer was correct for the problem I set. Acolytes was certainly correct for yours.

Originally posted by Mephisto2
Here is another one.

You play a game with another person. In turn you have to call a number between 1 and 10 included. All numbers are added together. The player who has to reach or run over 100 as total loses. Your opponent starts with 7. What number do you call to make sure you will win?

You should always call a number that makes 11 when added to his number (this means it's his go when it gets to 99), so you call 4 in response to his 7