Inspired by Brett Manley

Logic Level 3

Suppose you have 8 watermelons that are exactly same in appearance. They are equal in weight except for one. It is heavier than the others. You want to find the heaviest one by using a two-pan balance, where each pan can only hold one watermelon.

In the worst case scenario, at least how many times do you have to use the two-pan balance?

Note: The watermelons cannot be cut (since you do not know how the weight is distributed). You have to weigh the whole watermelon.

Inspiration .

7 5 3 8 4 2 6

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Calvin Lin by Calvin Lin

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1 solution

Anthony Muleta
Jan 10, 2016

Weigh two watermelons at a time, if any two watermelons do not balance the scale then the lower one is the heavier one. Worst case scenario - the last pair of watermelons weighed contains the heavier one, which would require 4 weighings.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Similar to this one poisonous measurements .

Sravanth C. - 5 years, 5 months ago

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Oh haha, didn't see that one. Nice!

Calvin Lin Staff - 5 years, 5 months ago

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Just one doubt. It is not mentioned whether I can cut the watermelon or not. In case I can, I will take 1/2 each of number 1-2 on one side and 1/2 each of number 3-4 on other. If these are equal, I will continue with 1/2 each of number 5-6 on one side & 1/2 each of number 7-8 on other..the heavier side will then be compared further to find the heavier watermelon. So this can be done in 3 weighings.

A note mentioning that watermelons can't be cut may give clarity to the question.

Rohit Sachdeva - 5 years, 5 months ago

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@Rohit Sachdeva – Thanks. That's a reasonable consideration. I've added that in.

Calvin Lin Staff - 5 years, 5 months ago

Could anyone describe which of these steps are unnecessary? Remember, worst case.

  1. Watermelon 1 vs 2. If equal...
  2. Watermelon 3 vs 4. If equal again...
  3. Watermelon 5 vs 6. If still equal...
  4. Watermelon 7 vs 8. The different watermelon should be here. Then?
  5. Watermelon 7 to any other 1/2/3/4/5/6: This can be equal as well. Then the different watermelon should be watermelon 8.

So, are 4 steps generally minimum for all cases? I answered 6 steps (which I know it will be wrong anyway). But then I thought about the "worst case" above. I think it still requires 5 steps at most.

Maybe I should put on more concentration on the words.

Henny Lim - 5 years, 5 months ago

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The question states "It is heavier than the others". So, step 5 isn't required.

Calvin Lin Staff - 5 years, 5 months ago

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Oh... I see. So we don't need to determine which watermelon. Just have to find it? Got it.

Henny Lim - 5 years, 5 months ago

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@Henny Lim – Not quite sure what you are asking. We are told that the heavier watermelon is the odd-one-out. We have determined / found it in 4 weighings.

I agree that if the problem did not provide any information about the weight of the odd-one-out, then we will need a 5th weighing like you did, comparing 7 against 1 to see if 7 is indeed the odd-one-out.

Calvin Lin Staff - 5 years, 5 months ago

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