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

17 comments

  1. 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. 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. Jeetender Kumar

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

  3. 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

  4. 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.

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