## Puzzle of the Day : 6

# Find the Oddball

If you have two identical balls , one heavier than the other , you can easily determine which is heavier by putting them on opposite pans of a balance scale . If there are four balls , all the same weight except for one heavier one , you can find the heavier one in two weighings.

Suppose you have nine identical balls, one of which is heavier than the eight others . What is the smallest number of weighings needed for positively identifying the odd ball ?

**See our previous ‘Puzzles of the Day’:**

Puzzle of the Day : 5

Puzzle of the Day : 4

Puzzle of the Day : 5

Puzzle of the Day : 4

3

3

Minimum no. of weighings should be 1.

If we put four balls each on both sides of the balance, then we may get a case that both sides might weigh equal and the remaining ball will be the odd ball.

But this case is necessarily true

Not possibly true

To get a necessarily true solution

Total no of weighings will be 4

1st- 3 balls vs 3 balls

We’ll get 3 balls of comparatively higher weight

Then

2nd- remaining 3 balls with higher of 1st one

We’ll get total 3 balls which have higher weights

Then

3rd- 1vs1

Then 4th- 1vs higher of 3rd

Then the higher of 4th will be the highest

@dhruv

Typing error!

Its not necessarily true

It’s possibly true

the smallest number of weighings needed for positively identifying the oddball in 2.

3 chances needed

3

3

2

2

1st – weighing 2 pairs of 3 balls

So now we have one pair which has heavier ball

2nd – weighing 2 balls out if 3

If they are equal than the third ball is heavy

Otherwise heavier side of pan has heavy ball

1

Min. 2

2.

Divide in 3 pairs of three balls each.

Weigh any two pair. Possible outcomes: one of the weighed pair is heavier or both are of equal weight.

In either of the case, we have one pair of 3 balls left which is heavier than the other two.

Now take any two balls of this pair and weight. Possible outcomes: One of the ball is heavier or both are of same weight. In any case, we have identified one ball that is heavier.

So min. number of weighings required = 2.

Minimum no. of weighing required will be 2