31 July 2015

One more day. One more flight. One more puzzle.

Try this one. It is not as simple as it might sound but not as complicated either.

You have six identical balls except two are painted red, two are painted blue and two are painted white. You know one ball of each color weighs 10 pounds and one ball of each color weighs 11 pounds. (So each color has one lighter ball and one heavier ball).

You have a scale and pan balance but no weights. Only the balance and the six balls.

What is the minimum number of weighings you have to do to determine which are the heavier balls and which are the lighter balls?

Posted July 31, 2015 by Rajib Roy in category "Puzzles


  1. By Kenneth Serrao on

    2 weighings. Red blue versus red white in 1st weighting. if this weighing balances then you weigh the two red balls and you know all the heavier and lighter combinations. If the first weighing doesn’t balance, then just weigh the two non-red balls (blue and white) from the first weighing and you know all the heavier and lighter combinations.

  2. By Rajib Roy on

    Kenneth, you might have missed one case. Let’s say initially they did not balance. [EDIT: Added after initial post: Say the red plus blue comes out heavier than red plus white.] So the heavier side has the heavy red for sure. But note that the blue and white can be heavy, light OR heavy, heavy OR light, light. In the second weighing if it is again balanced you do not know whether those two non-reds were both heavy or both light.

  3. By Kenneth Serrao on

    Sorry Rajib Roy. Clearly you remain as difficult (painful?? 🙂 ) a questioner as when you first interviewed me 16 years ago.

  4. By Rajib Roy on

    Kenneth, note that I edited and added a sentence above. My initial explanation of the miss might have not been complete.

  5. By Rajib Roy on

    Conversely, Kenneth, you have not become any better. Which would imply You Got The Job!!! 😉 😉

  6. By Rajib Roy on

    I do not think so, Kenneth. Take the above example. Red-Blue came heavier than Red-White. We know Red in Red-Blue has to be heavier. Following your logic, let’s say we replace the heavy red with the other White not sitting on the pan. So, we have White-Blue and Red-White. Now if you are disbalanced again, this could be because your original weighing had Blue and White as Heavy, Light OR Light, Light. You certainly have nailed the White one also – ha to be light. But you are still not sure which Blue is which.

  7. By Chris Corner on

    My answer would be 2.5.
    Weigh 3 (RWB) balls in each pan. One side must be heavier as there is no combination that balances.
    So either there are 2 or 3 heavy balls on the heavier side (HHH or HHL).
    Take any 2 of the heavy side balls and weigh against each other (say R and B – musician talking).
    If you are lucky, they don’t balance and you had HHL combination on prior weighing. Assume as example R was heavier on this weighing, then B is light and white was heavy. Success in 2 weighings.
    If they balance then you may have had a HHH combination or picked HH out of a HHL combination. Then you have to reweigh (assume it was R and B again) one of them against White. If they balance you had HHH. If they don’t, you have HHL and can perform same deduction as before. In sum, takes 2.5 weighings – 2 if you are lucky and 3 if not.

    1. By Rajib Roy on

      No. There is one heave red, one light red, one heavy white, one light white, one heavy blue and one light blue.

  8. By Manivannan Sadasivam on

    If the balls are a1,a2,b1,b2,c1,c2, there are two possibilities- lighter or heavier- for each of a1, b1, c1, allowing a total of 2^3=8 possibilities. That translates to three weighings unless each weighing can yield 3 pieces of information, 3^2=9 > 8.

  9. By Rajib Roy on

    ANSWER: 2 weighings

    Weighing 1: One Red and one Blue on one side and other Red and one White on another side.

    Option A:
    If we observed that both sides is equal, that means both side has one heavy and one light. Then all we have to do is

    Weighing 2: Take the Blue from previous weighing and put on one side. Then put the other Blue from the floor to another side. From that we know which is the heavier Blue and which is the lighter blue. If the blue from first weighing turned out to be heavier than the Red in that side in original weighing was lighter and vice versa. From that we know which Red is which. And similarly, from that, we know whether the White on the other side in the original weighing was heavier or lighter.

    Option B:
    If we observed (in Weighing 1) that one side is heavier – for example, the side with Red and Blue turned out to be heavier, that side has to have the heavier Red. (Remember, the lightest the side with heavier Red can be is one heavy Red and one light ball and the heaviest the other side can be is the light Red ball and one heavy ball. As you see there is no other way possible than the heavier side to have the heavy Red).

    Continuing with the case that the Red and Blue turned out to be heavier, we have three possibilities:

    Side with heavy Red ::::: Side with light Red
    has has
    (1) heavy Blue – heavy White OR
    (2) heavy Blue – light White OR
    (3) light Blue – light White

    To ascertain which one of this is the case,

    Weighing 2: Take out the two Red balls from Weighing 1. Put the two other balls from Weighing 1 on one side. Take the other two balls from the floor and put them on the other side. Observe what happens.

    If the side with original Weighing 1 balls is heavier, that means it is case (1) – both the blue and the white on that pan are heavier and the blue and white on the other pan is lighter.

    If the side with the original Weighing 1 balls is lighter, the reverse of the above is true – in other words, it is case (3)

    If both sides are equal, that means it is case (2) – the original weighing blue was heavy and white was light (and the floor blue was light and floor white was heavy)

    Of course, similar logic holds if in Weighing 1, the side with Red and White turned out to be heavier.

  10. By Manivannan Sadasivam on

    Rajibda, good puzzle. Here’s a simpler one from a story related to the origins of probability: A and B contribute to a stake of $60 in a game of chance. Each play is for one point and the first player to reach 3 points wins. When A is leading by 2 points to 1, they decide to stop the game. How should the $60 be split?


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