Dynamite Deal
Logic; Difficulty: easy, medium
Are you a savvy enough salesperson to close this deal?
Puzzle worm Sandi Sandman sent me this interesting puzzle that took me multiple tries to get to the best solution.
You have come into possession of eleven plutonium orbs that recently went missing from a nuclear facility. These orbs all look the same but each has a different weight: 1 oz, 2 oz, 3 oz, and so on, all whole numbers through 11 oz. You have arranged the orbs by weight in a box (you know how much each orb weighs). You are at a despot’s palace, trying to get him to buy these orbs.
The despot is convinced that this is the genuine set of eleven plutonium orbs that was recently reported missing and that he dearly needs for his country’s nuclear program. He knows the weights of the orbs are 1,2,3…10 and 11 oz, but he doesn’t know which is which and he sure as hell doesn’t trust you. He is willing to buy an orb only if he can be convinced which weight it is.
Unfortunately, there’s no weighing instrument at the palace. The only device available is a pressure-triggered explosive enclosed in a transparent chamber. This device will detonate if more than 11 oz of weight is ever placed on it!! So, you can place any combination of orbs on it as long as their combined weight doesn’t exceed 11 oz.
Part A: You are in a hurry to get the deal done and scoot off. What’s the minimum number of times you need to use the weighing device to convince the despot about the weight of any one orb?—without vaporizing yourself of course. Note: Each time you place a set of one or more orbs on the weighing device, it counts as one use of the device.
Part B: If the despot was willing to buy all eleven orbs as long as he is convinced about each orb’s weight, what is the minimum number of times you need to use the weighing device in this scenario?
*spoiler alert* solution and answer follow in the expandable sections below.
It should be obvious that a single weighing cannot identify any orb and so more than one is needed. But can it be done in two weighings? Turns out you can! Here’s how:
Place the 1, 2, 3, 5 oz orbs on the scale first. Then keeping the 1 oz orb on the scale, replace the 2,3,5 with the 4,6 oz orbs.
Reason with the despot that the only way to place 4 orbs is having A: 1,2,3,4 or B: 1,2,3,5. If it had been A, the lowest weight for the second weighing would have been 1 + 5 + 6 = 12 i.e. kaboom! So it had to have been 1,2,3,5 with the retained weight being the 1oz. And sold!!
Now that we’ve identified the 1 oz orb, how quickly can we verify the rest for the despot! First, let’s determine a lower bound for the number of min weighings required.
- There is no point placing the 11 oz orb on the scale. It can only go on by itself and that doesn’t give you any information. This implies the 11 oz will have to be identified by elimination.
- Hence all the other orbs will have to go on the scale at least once.
- The 10, 9,8,7 and 6 can never be weighed together in any combination without an explosion. To get the other orbs on the scale at least once with these five in five turns implies the weighings would have to be the pairs (10–1), (9–2), (8–3), (7–5) and (6–5). This doesn’t uniquely identify any orb.
Hence, we clearly need more than five weighings. But can we do it in six? Turns out there is a way to do it as an astute Reddit solver shows here.
The weighings are (in any order):
1–4–6
1–3–7
2–4–5
1–10
3–8
2–9
The key insight is that every triple has to contain either a 1 or a 2 and can never contain a 9 or 10. The whole solution is explained well here.
An alternative sequential solution in 7 moves that is a tad easier to follow is shown below.
The Solution:
| Weighing # | Orbs on scale | Single Orbs Identified | Orb Groups Identified |
| 1 | 1–2–3–5 | ||
| 2 | 1–4–6 | 1 | (2,3,5) (4,6) (7,8,9,10,11) |
| 3 | 7–4 | 4,6,7 | (2,3,5) (8,9,10,11) |
| 4 | 8–2–1 | 2,8 | (3,5) (9,10,11) |
| 5 | 8–3 | 3,5 | (9,10,11) |
| 6 | 9–2 | 9 | (10,11) |
| 7 | 10–1 | 10, 11 |
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