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.
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Originally posted by AcolyteOr 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.
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.
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 iamatigerNope. There are 4 remaining, not 3. thus this won't work!
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?
Originally posted by AcolyteCorrect: and beautifully explained too!
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.
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Originally posted by howardgeeYeah - 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!
Nope. There are 4 remaining, not 3. thus this won't work!
Originally posted by iamatigerLet me try. Number the balls 1 to 8.
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!
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.
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Originally posted by Mephisto2Looks comptetely correct! Congrats, nice one!
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.
Originally posted by iamatigerHere is another one.
Looks comptetely correct! Congrats, nice 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?
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Originally posted by Mephisto2You 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
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?