# 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

1. 3

2. Tamanna Gogia

3

3. Dhruv agarwal

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.

1. But this case is necessarily true
Not possibly true

1. 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

2. @dhruv
Typing error!
Its not necessarily true
It’s possibly true

4. Jeetender Kumar

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

5. Amitayu Krishna

3 chances needed

6. 3

7. 3

8. 2

9. Sahil chauhan

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

10. Parth kejriwal

1

11. Manish Jindal

Min. 2

12. 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.

13. Minimum no. of weighing required will be 2

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