Gold bars: 1, 2, 4, 7, 13, 24

[sweetwater]

Label the seven vaults 0, 1, 2, 4, 7, 13, and 24. From each vault take bars in accordance to its label, i.e., from vault 0, no bars, vault 1, one bar, vault 2, two bars, and so on. There will be a total of 51 bars.

Weigh the 51 bars. Subtract the weight from 510 kilos and divide the difference by 10 kilos. The resultant value is the number of counterfeit bars. Refer to the table below to determine which vaults contain the counterfeit bars.

Value Vault Vault Vault
3 0 1 2
5 0 1 4
6 0 2 4
7 1 2 4
8 0 1 7
9 0 2 7
10 1 2 7
11 0 4 7
12 1 4 7
13 2 4 7
14 0 1 13
15 0 2 13
16 1 2 13
17 0 4 13
18 1 4 13
19 2 4 13
20 0 7 13
21 1 7 13
22 2 7 13
24 4 7 13
25 0 1 24
26 0 2 24
27 1 2 24
28 0 4 24
29 1 4 24
30 2 4 24
31 0 7 24
32 1 7 24
33 2 7 24
35 4 7 24
37 0 13 24
38 1 13 24
39 2 13 24
41 4 13 24
44 7 13 24

[AeroEng]

Sweetwater, Your answer is correct: You weigh 0, 1, 2, 4, 7, 13, 24 gold bars from the seven vaults. But you don't need a long table. First, you subtract the weight from 510. Then, you try to fit in the outcome the numbers of your vaults starting with the largest possible.
Example:
Weight = 479
510 - 479 = 31
The first vault is the largest vault number that can fit in 31. That is 24. Which leaves 7.
The second vault is the largest vault number that can fit in 7. That is 7. Which leaves 0.
The third vault is the largest vault number that can fit in 0. That is 0.
Answer: Vaults 0, 7, 24.