Can you work this out?

Question

After looking at the standard of Games & Recreation I decided this was better in here! Puzzle: You are given 8 metal cubes with exactly the same dimensions but one is heavier than the other 7. You are given a simple beam balance with standard 2 pans. What is the MINIMUM number of weighings you would need before you can say "That is the heaviest cube"

heavenlyhaggis

Answer 1

This a simplified version of the 'twelve billiard ball problem' in which you have not 8 but 12 apparently identical objects. Furthermore in the 12 bb problem you only know that one ball is DIFFERENT in weight, but NOT whether lighter or heavier.

In your example it's 2 weighings.

Place 3 cubes in each pan (6 in total). If they balance the heavy cube is one of the remaining two. Place one of these in each pan, the side that goes down is the heavier = 2 weighings.

If in your first weighing one pan goes down take two of the 3 cubes from this heavy side and place one in each empty pan. The third goes on the table.

If one side goes down, there's your cube; if it balances, the heavy cube is the third one you left on the table = 2 weighings.

Now try the 12 billiard ball problem! You have to be able state after NO MORE THAN THREE weighings which is the odd ball AND whether it is light or heavy.

Answer 2

One, you jsut have to be lucky enough to weigh it against a normal one on your first try ^_^

lol

Really though, if you don't want smart *** answers like mine, you should just say one cube has a different weight than all the others (not specifying heavier or lighter) and ask how many weighings it takes to find the wierd cube.

Also, try adding that when you start this, you have no previous knowledge of the cubes and you can't tell the difference without the scale, this prevents some of the more stupid answers I have seen.

And finally, specify it is a 2 pan scale that merely weighs one side against the other, that should seal up all of the holes in your problem.

Answer 3

3

Answer 4

With 2 pans, 1 weighing.

Answer 5

2 at the least...3 at the most

12345678

123x456
3x3 - If they are equal then weigh one of the original 3 against 1 not yet weighed...if they are equal then do the same to the other not yet weighed

If the result of above is not equal then let say
123 goes down and 456 goes up
then 14x25
if equal then 7x3
if equal then 6 is lighter
if 14x25 not equal then
lets say 14 goes down and 25 goes up
then the odd cube is either 1 or 5
do 1x7
if equal then 5 is lighter
if 1 goes down then 1 is heavier

you should be able to extrapolate this logic to other scenarios.

Answer 6

minimum is 2
weigh 123 vs 456 if they are same then 7 vs 8
if they are different then say 123 is heavier weigh 1 vs 2 if not same then the heavier one is the heaviest else 3

Answer 7

1 actually. first pan, cube with same weight like the other 6. second pan is the heaviest cube.

Answer 8

exactly 2 weighings are needed to find the heavier cube. No more. No less.

See the answer to this related question:
http://answers.yahoo.com/question/index;_ylt=AsuPI0y.nlEEYU5BzYFnsL_sy6IX?qid=1006052228707

Same answer as for the 9 ball problem with one lighter.

Answer 9

3 weighings

Answer 10

3 weighings
(4/4) (2/2) (1/1)