Hopeful Solution (7 bar max)

[Ghost Post]

1. Take one from each vault(supposing there are 7 vaults and that all bars are real)the maximum weight should be 70 kilos(make sure you know which bar is from which vault)
2. place all on the scale and take one of at a time(by my standards you are still only weighing them once you are just subtracting from the weight)
3. record the weight after each bar is taken off to find out how each bar weighed
4. this should give you the answer
with only using 7 bars

[mwf]

If you read the long post. You would know that a weighing similar this was discused there. In that case, they were checking to bars as they were being placed on the scale.

If we were to use you method, I could do it with 6 bars. Put the 6 bars on the scale. 1 from each vult but last one and mark which vault they came from. If the weight is 58K then we know that the vault that we did not back one from is counterfit. If the weight is 57K then we know all 3 of the counterfit vaults must be from the 6 we took the gold from.



[This message has been edited by mwf (edited 03-15-2001).]

[QuikSand]

Do we really think that interpreting "one weighing" as really meaning "well, you can basically get 6 or 7 readings from the scale" leaves us with a very interesting puzzle?

I keep coming back to the fact that the relatively intuitive bare-bones interpretation of this puzzle:

7 vaults (or 6 if you prefer)
3 contain fakes
1 weighing, no trickery
use fewest bricks possible

...leaves us with a pretty nifty little puzzle (which it's clear from the numerous free-form posters has stumped a lot of people).

I again confess I'm pretty new to the GL site, but if the "official" answer involves any of the trickery-based answers (like the seven weighings as one concept) I'll be deeply disappointed.

- - -

That said, is it SOP here at the GL for puzzles to receive no "official" comment (or corrections) until an official answer is posted? It certainly seems that with the solid gold puzzle (and probably the secret chamber before it) a well-placed errata statement (at least clarifying the 3+3=7 problem) would have been helpful. Is that just out of the question?

[Ghost Post]

let the vaults' names be A,B,C,D,E,F and G.

take and weigh (together) 1, 2, 4, 8, 16, 32 and 64 bars from each of the vaults respectively. Call this weight W.

Then, 1270 - W = R.
And since you can only represent the number R by a unique sum combination of 3 numbers
from this list : 1, 2, 4, 8, 16, 32 and 64, then our problem is solved!

Of course, i'll ask for a platinum bar!!!

[mwf]

Edea, while you soulution will work. It will not use the least number of bricks.

The most of the discion on the this problem can be found in the thread "Solid gold - not adequate for real life situations"