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pick any 6 balls divide them into 3 each.
if they are same check the rest two and you will find the heavier.
if they are not same then choose 3 from heavier bowl and pick any two and weigh them, if they are same third one is heavier otherwise its clear.

so you have to weigh only two times.

Split balls into three sets: 2x sets of 3 balls each, and 1x set of 2 balls

First scale step: compare both sets of 3 balls. If one side is lighter, that 3-balls set goes to second step. If both sides are same, the set of 2-balls goes to second step.

Second scale step: If chosen set has two balls, just weigh and pick the lighter. If chosen set has 3 balls, compare any two and pick the lighter one, if they are same, then the third ball in the set.

plain common sense.. take 6 balls apart, split into 3 and 3. if scales are equal, measure the other two remaining balls. if scales not equal, choose any two from the heavier set and measure with one on each side. if equal, third ball is different, else the one where scale moves.

Awesome nd_100 thanks

I feel this is even simpler: - 2 sets x 4 balls = less groups. 1) Place one set of 4 balls on scale - record weight
2) Place second set on scale- record weight. = The group with the smaller weight hence has the smaller ball.
3) Divide the heavier group by 4.
4) From the lighter group of 4- Place 2 random balls on scale. = If this number is half of the homogeneous set then disregard
5) Chose the the group of in which is not eliminated by common factor- Drop one ball on the scale and if this is a factor then it is the seventh identical ball.

4 physical steps. If you can precisely do it easier please post!

2 weighings needed
ceil(log3(8)))

It will be ceil( log3(n) ); n=number of objects
where ceil (1.1) = 2;
log3(n) => logrithm of n with base 3.

plain common sense.. take 6 balls apart, split into 3 and 3. if scales are equal, measure the other two remaining balls. if scales not equal, choose any two from the heavier set and measure with one on each side. if equal, third ball is different, else the one where scale moves.

How about if you have N balls , then how many steps does it take to find the heavier one , and how do you find it ?

2 weighings needed.

and the logic is ceil(cube root of n) where n is the number of balls. it will work for all the cases.

how to find 2 heavy balls from 8 balls...and weight of 2 heavy balls is same...??

number of steps tends to be cube root of the number of balls.
separate the initial set into three groups, one for each side of the scale and one to be kept aside.
if the scale shows equal, you know the lighter one is in the set set aside.
if the scale tips one way, you know the lighter is in the set that the scale tipped away from.
solve recursively.

yes very simple and to the point answer.. its definitely 2.. good job punit.

yaah its easy ans 2

yes very simple and to the point answer.. its definitely 2.. good job punit.