PM the answer to me.

This is the answer I was looking for (obviously, there are other possibilities):

Go Steelers said:Remove 1 nail from box #1, 2 nails from box #2, 3 nails from box #3 and 4 nails from box #4. Then weight the 10 nails you just gathered.

If the total weight of those 10 nails is 20.1 grams, then box #1 is defective. If the total weight is 20.2 grams, then box #2 is defective. If the total weight is 20.3 grams, then box #3 is defective and if the total weight is 20.4 grams, then box #4 is defective.

This answer is the most complicated one, but it seems right from what I can see. If anyone finds a flaw that I missed, let me know. In the meantime, SilverGoat04 gets a point.

This question is closed.SilverGoat04 said:Let's call the four boxes A, B, C, and D. The full weight of each individual non-defective box is 400 and the full weight of the defective box is 420.

Take one nail out of box A and put it into B. Now the possibilities for the weight of B are:

402 (if both A and B contain non-defective nails)

402.1 (if A contains the defective nails)

422 (if B contains the defective nails)

Now, I would take one nail out of box C. The possible weights of box C would then be:

398 (if it is a non-defective box)

417.9 (if it is the defective box)

If I put both box B and box C onto the scale, the combined weight possibilities are:

1) 800 (if A, B, and C were non-defective)

2) 800.1 (if B and C were non-defective, but A has the defective nails, one of which is now inside of B)

3) 820 (if B already contained the defective nails, then C could not be defective and C would thus have a weight of 398)

4) 819.9 (if A and B were non-defective, but C is the defective nail box)

For scenario 1, I would conclude that box D is the defective box.

For scenario 2, I would conclude that A is the defective box.

For scenario 3, B contains the defective nails.

For scenario 4, C contains the defective nails.