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There are 8 boxes each containing 8 marbles. There are two kinds of marbles – Brazilian kind and Indian kind. Each marble of the Brazilian kind weighs 12 units and each marble of the Indian kind weighs 11 units. Of the 8 boxes there is one box in which all of the marbles are of the Indian kind, all others contain marbles of the Brazilian kind only. You are given a weighing scale (assume it’s digital). How will you find out which box has the marbles of the Indian kind with only one weighing?

I have just started studying discrete math and I am still getting the hang of combinatorics. Any contribution is very much appreciated!

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    $\begingroup$ Possible duplicate: puzzling.stackexchange.com/questions/45998/… $\endgroup$
    – Goose
    Commented Jul 10, 2020 at 0:04
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    $\begingroup$ @QuantumTwinkie I wouldn’t say so. This has a very different solution based on the fact only one weighing is allowed $\endgroup$ Commented Jul 10, 2020 at 0:09

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I’ve seen something similar before. You are missing one very key part of information in this question, and that’s

that you can take marbles out of these boxes. (This isn’t ruled out in the question, just not mentioned)

Without this I don’t think this is solvable. Assuming this was meant to be included, the solution is as follows:

Label all boxes $1$ - $8$. Now take $1$ marble from box $1$, $2$ from box $2$, $3$ from box $3$ etc until you have $36$ marbles.

Now put all the marbles on the weighing scale. The weight shown will be $36 \times 12 - X = 432 - X$

If you solve for $X$, then $X$ is the number box with the Indian marbles in them.

Why this works:

If all marbles were Brazilian the weight would be $36 \times 12 = 432$. But as not all marbles are Brazilian, not all weigh $12$. As Indian marbles weigh one less than Brazilian marbles, the amount the total weight is less than $432$ will be how many Indian marbles there are. As the number of marbles represents the number box, we know what box has the Indian marbles.

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  • $\begingroup$ You could reduce the number of marbles that need to be weighed by: rot13(gnxvat mreb zneoyrf sebz gur svefg obk, bar zneoyr sebz gur frpbaq, naq fb ba, svanyyl gnxvat frira zneoyrf sebz gur rvtugu obk. Nf orsber, lbh jbhyq trg n qvssrerag jrvtug sbe rnpu cbffvovyvgl.) $\endgroup$ Commented Jul 10, 2020 at 1:15
  • $\begingroup$ The same solution will work for 9 boxes. $\endgroup$
    – Laska
    Commented Jul 10, 2020 at 3:25

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