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The vexatious vault value veracity validation problem

Posted by Chris on August 22, 2014 – 6:45 am

This problem was originally posted by Karl Sharman (with a less idiotic title).

Whilst I was away under the pretence of work, a nearby bank uncovered a plot to swap the gold in their vaults with counterfeits. It was determined that all the gold bars in three of the Bank’s seven vaults were replaced with counterfeits. The other four vaults were uncompromised. The plot was foiled through the poor math skills of the thieves: while the real gold bars weigh ten kilograms, the counterfeits all weighed nine kilograms.

I was asked to work out which was the real gold, and which was the fake. I, being really bad at maths, so Chris tells me ;-) decided to recruit your help.

Your mission, should you wish to accept it, is determining which vaults have real gold, and which are just gold-plated bars of platinum.

The Bank Director has made the following generous offer: If you can determine the counterfeits using just one weighing on a scale, you can keep one bar as a souvenir.

Here are the rules:
This is a scale, not a balance, but you can weigh as many bars together as you like.
Only one weighing!
The bars will be handled by professional guards, so you won’t have a chance to “feel” their weights.
Each vault contains several hundred bars.
The guards have requested that you try to keep the number of bars you need to a minimum.

How do you do it, and what is the minimum number of bars…?

Good luck!

This post is under “Logic, Mathemagic, MathsChallenge” and has 12 respond so far.
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12 Responds so far- Add one»

  1. 1. Wizard of Oz Said:

    A fairly common type of problem . . .

    Take one bar from vault A, two from vault B, four from vault C, and so on up to 64 from vault G. Weigh all 127 of these bars, divide the total by 10, and convert the difference between this answer and 127 to binary.

    Where the 1’s are in this result will then tell you which vaults have the counterfeit bars, e.g. a result of 1010100 means that vaults G, E and C have the fake bars.

  2. 2. Chris Said:

    Wiz, your strategy would be correct if we had no idea how many vaults had been compromised.

    The last time it was posted, nobody said that no weighings are needed as the three vaults had already been determined. That’d be no fun though. So you know that exactly three vaults have been compromised, but you don’t know which three. That’s only possible in puzzleland.

  3. 3. Curtis Said:

    Well…. obviously you need to choose a different number from each vault, but to get a single solution, no combination of two numbers may = any other combination of two numbers
    e.g. if you used 1,2,3,4,5,6,7 then (1 + 4) = (2 + 3)

    nor may any combination of three numbers = any other combination of three numbers
    e.g. (as above) (1 + 4 + 5) = (1 + 3 + 6)

    so to get a unique solution: 1 from 1st etc….26 from 7th

    1,2,3,5,8,15,26 total bars = 60
    600 – weighed value = no of counterfeit bars = 1 solution

    …. but I can’t see an arithmatic progression so it’s probably wrong?

  4. 4. Chris Said:

    Hi Curtis. You are making a lot of the right noises. Unfortunately, 1 + 2 = 3 causes a problem.

  5. 5. Wizard of Oz Said:

    Curtis’s solution works equally well if you subtract 1 from each element, i.e. the set now becomes {0 1 2 4 7 14 25}. Each 3-way sum is now 3 less than before, and just 53 bars are now required.

    I can’t see what 1 + 2 = 3 has to do with it.

    Since there are 35 ways of distributing 3 fake vaults among the 7 possibilities then the 35 unique 3-way sums must lie in the range from 3 (0+1+2) to 46 (7+14+25), i.e. only 9 vacant numbers in this range for the sums. That doesn’t leave much room to find a distribution involving even fewer bars than 53.

    But if anyone can do it, Chris can.

  6. 6. Chris Said:

    Hi Wiz. You’re right about 1 + 2 = 3 being irrelevant – I jumped to a false conclusion based on the final numbers.

    You also did good in knocking off the 1 from Curtis’s results. But you can do even better.

  7. 7. Wizard of Oz Said:

    {0 1 2 4 7 13 24} requiring 51 bars

  8. 8. Chris Said:

    That’s the solution I was looking for. I think Curtis and Wiz need to share the bonus bar.

    Note that 7 = 4+2+1, 13 = 7+4+3 and 24 = 13+7+4.

  9. 9. Wizard of Oz Said:

    Actually I think platinum is currently worth about 10% more per ounce than gold, so 9 kg of platinum would be worth about the same as 10 kg of gold.

    So in switching the contents of the vaults maybe the thieves were just as bad at precious metal valuation as at maths.

  10. 10. Chris Said:

    Hi Wiz. I’ve just glanced through the previous solutions (see “return to the gold standard”). I had jumped to the wrong conclusion that the problem was only the usual one. I’m embarrassed to note that I hadn’t acknowledged that you had mentioned the Fibonacci sequence. The usual Fibonacci sequence doesn’t work. However, the extended one that starts: 1,2,4 and then sums the previous three terms to get the next one does work.

    My brain has gone on strike, so I can’t give a theoretical explanation of why it’s the best solution. It seems obvious that if the were V vaults and C had been compromised, then you’d need the first V terms (including 0) of Fibonacci type C sequence. The usual Fibonacci sequence I’m designating as type 2. Type 1 is simply 0,0,0,0,… or should be 1,1,1,1,…? The last time it was up, I did it the hard way – only slightly better than trial and error.

    Hi Curtis. I’m sorry that I’m focussing on Wiz. It was really your post that was the breakthrough.

  11. 11. Curtis Said:

    That’s fine Chris, I’m relaxed….. it’s interesting to learn. Most of the higher maths is beyond me but I like to have a stab at the more logic based problems……. still,….. rookie mistake to miss the zero selection, lol

  12. 12. DP Said:

    I’m with you Curtis.

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