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

beiterspace, sorry there's just no hope for you!

Answer 1

I've got news for you: the standard isn't much better here. You need two weighing for certainty. On the first weighing, put three cubes on one side and three on the other. If the scale does not balance, take whichever group of cubes was heavier and place one of them on one side and one on the other - then the heavier cube is the odd one, or if the scale balances, the cube from the three that you didn't weigh on the first weighing. If the scale balances on the first weighing, then it must be one of the cubes you left out, so just weigh those two against each other.

Answer 2

Two. Put three cubes on each side to start with, and if that balances put the other two on separate sides by themselves. Whichever side goes down has the heaviest cube. Normally it would take three balances to get the heaviest cube (dividing the blocks evenly after each weighing), but this is the ideal case.

Now that I think of it, you could get it in one weighing if you happen to select the heaviest cube from all the others and one of the other seven. That is exactly as likely to happen as it is to happen in the two-weighing method I gave above, so it is superior.

Answer 3

Two. Put three cubes on each side to start with, and if that balances put the other two on separate sides by themselves. Whichever side goes down has the heaviest cube. Normally it would take three balances to get the heaviest cube (dividing the blocks evenly after each weighing), but this is the ideal case.

Now that I think of it, you could get it in one weighing if you happen to select the heaviest cube from all the others and one of the other seven. That is exactly as likely to happen as it is to happen in the two-weighing method I gave above, so it is superior.

Answer 4

7

Answer 5

5

Answer 6

Three. 4 on each side to start. Remove the lighter side's 4 cubes and move two from the heavier side to the other. Remove the cubes from the lighter side and move one cube from the heavier side to the other. The side that is heavier has the unique cube.

For the people that said two: this is only if you guess right. This won't work every time. You could do it in one attempt if you guess right.

Answer 7

Answer is two: one of the others has given a correct explanation that I shan't repeat. See the other string with the same question for my full answer.

Spaldingballa, luck has nothing to do with it only a logical method superior to the one you thought up!

You could get it in one if you started with one cube on each side but it wouldn't just be luck it would be DUMB LUCK since no logical or sensible method would begin by balancing one single cube against another!

Answer 8

2

Start with 4 on each side. weigh see which side is heavier.

then take 1 off on each side, if you've guessed right, the scale will be balanced and the cube that you took off on the heavy side is the one.

My way would need luck, but spaldingballa's way would get the answer for sure in exactly 3 weighs.

Answer 9

The minimum would be one. There is a probability that you could select the heaviest one in the first weighing. The most you would need is four if you select them in exclusive pairs.

Answer 10

3
1. 2 x 4 blocks
2. 2 x 2 blocks
3. 1 x 1 block