The Ball Weight problem is as follows :
Suppose I give you 9 balls and a weigh balance. I tell you that one ball weighs lesser than the other 8( the other 8 balls being equal in weight obviously).
In how many tries of the weigh scale can you find out the lighter ball?
If your answer is 3, then you’re WRONG! 😛
The answer is 2
This is how you do it
Divide the 9 balls in groups of 3..
Lets the groups be A , B and C.
So group A will contain Ball 1 ,2 & 3.
and So group B will contain Ball 4 ,5 & 6.
and So group C will contain Ball 7 ,8 & 9.
Let us assume that Ball no.7 was the light one.
Now compare any two groups. If the weigh is equal, then its but obvious that the lighter ball is in the group not on the weigh scale.
If any one of them tilts, then also we can understand in which group of 3 , the lighter ball is.
This is Round 1. So, say we weigh group A & B, we find that the weigh is equal. Hence the lighter ball is in group C
Next, take those 3 balls in group C.
In a similar manner, choose any two balls at random and measure.
If the lighter ball happens to be within those two, it will show on the scale.
If the weight is equal, then you know for a fact that the lighter ball is the last remaining one(not on the weigh scale)
This is Round 2! Hence in just 2 tries of the weigh scale, we found out the lighter ball!
Confused ? 😛 Leave me a comment below if you have a doubt and Ill get back to you asap! Cheers! 😛