Boxes and coins problem

Given 10 boxes, each having 10 coins. The weights of the coins from one of the boxes are 10 grams , and the weights of the coins from the other 9 boxes are 9 grams.

Using exactly one weighting, how can you find which box contains the coins with 10 grams  weight?

Solution (drag over the blank space below to see it):

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Take 1 coin from first box, 2 coins from second box, … , 10 coins from the box number 10, and let S be the result of weighting.

Then S – (9 + 18 + … + 81 + 90 ) will be the number of the box that contains coins weighting 10 grams.

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3 Responses to Boxes and coins problem

  1. kookoo says:

    Nice…how about this one:

    12 coins problem
    You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weigh less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter?

  2. Anonymous says:

    Here is a solution, maybe not the nicest one. Let be the coins numbered from 1 to 12.
    Compare [1 2 3 4] with [5 6 7 8].
    1. They are equal. It means that the counterfeit is one of [9 10 11 12].
    Then compare [6 7 8] with [10 11 12], and we know that 6, 7, 8 are OK.

    1.A. If [6 7 8] = [10 11 12] then 9 is counterfeit, and by comparing it with any other coin we can determine if it is heavier or not.
    1.B. If [6 7 8] < [10 11 12] then we know that the counterfeit is one of 10, 11, 12 and that it heavier. Then if we compare 10 and 11 we have 3 situations: [10] = [11] means 12 is heavier and counterfeit, [10] < [11] means 11 is heavier and counterfeit and [10] > [11] means 10 is heavier and counterfeit.
    1.C. If [6 7 8] > [10 11 12] it means that the counterfeit is one of 10, 11, 12 and is lighter, and similar with B we can determine exacty which one.

    2. [1 2 3 4] < [5 6 7 8].It means that one of 1, 2, 3, 4 is lighter or one of 5,6,7,8 is heavier (***).
    Then comparing [1 2 5] with [4 6 9] leads to following situations:
    2.A. [1 2 5] = [4 6 9]. This reduces condition (***) to 3 is lighter or 7 is heavier or 8 is heavier. We can determine this by comparing 7 and 8, if they are equal then 3 is lighter.
    2.B. [1 2 5] < [4 6 9]. This reduces condition (***) to: 1 is lighter, 2 is lighter or 6 is heavier. The final result can be determined by comparing 1 and 2. If they are equal then 6 is counterfeit and heavier.
    2.C. [1 2 5] > [4 6 9]. Then (***) reduce to 5 heavier or 4 lighter. We can determine the final result by comparing 5 with 12 for example, if they are equal then 4 is lighter, otherwise 5 is heavier.

    3. [1 2 3 4] > [5 6 7 8]. Just replace 1 with 5, 2 with 6, 3 with 7 and 4 with 8 in the explanations from 2. and this case is also solved.

  3. Pingback: Bowls with candys | AlfaNotes

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