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Question: I will give you 8 balls. In those eight balls, 7 balls have same weight and one ball has more weight than other balls.In how many minimum steps you can detect the ball with more weight using a common balance.
And also show the steps.


Answers: I will give you 8 balls. In those eight balls, 7 balls have same weight and one ball has more weight than other balls.In how many minimum steps you can detect the ball with more weight using a common balance.
And also show the steps.

Put two balls on one side, place to sets of three on the balance, if they balance then it is obviously one of the two remaining. If not, one of the three will go down. If that happens, two balls on the balance if they balance, the one left out is the heavy one. If the two sets of three balance, then it must be one of the two left, so put them on the balance, only two weightings!

Split and weigh them in two piles of four.

Take the heavier group and split and weigh it in two piles of two.

Take the heavier two and weigh them against each other to find the heavier one.

Put 4 in each balance the heavier split in to 2x2 then again the split will give you the heavier ball total 3 weighs.

4

Use a binary search.
1) Weigh 4 of the balls and note the weight.
2) Weigh the other 4. Discard the lighter set. Call the lighter of the two weights w.
3) Weigh two of the remainder and note the weight. If it's equal to half w discard these two otherwise discard the remaining two.
4) Weigh one of the two remaining balls. If it's greater than w/4 then this is your ball. Otherwise it's the other one.

Depends what you call a step though. Only 4 weighings in my version.

if you're lucky you could do it one - select two balls, one of which would ideally be heavier than the other on the scales
very interesting*

Weigh 4 against 4.
Then take the heavier 4 and weigh 2 against 2.
Then take the heavier 2 and weigh them. There you go.
3 steps.

throw them at something if one makes a mark then its the one lol idk im just guessing



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