Seven gold links, one cut, seven days.
The gold bar puzzle: pay a worker one link per day for seven days, from a chain of seven links, making as few cuts as possible. Most people answer three cuts, some get two. The real answer is one, and it works because of how binary numbers cover a range.
You hire a worker for seven days. The agreed wage is one gold link per day, paid at the end of each day, never in advance and never in arrears. Your only gold is a chain of seven links, all identical, joined in a line.
Cutting a link is expensive, so you want to do it as rarely as possible. One cut opens a single link: the opened link becomes a piece on its own, and the chain separates into the segments on either side of it. Cutting the third link of seven, for example, leaves you three pieces: the opened link (size 1), a segment of 2, and a segment of 4.
One more rule, and it is the load-bearing one: the worker will make change. If handing over a larger piece overpays, they will return smaller pieces you gave them earlier.
What is the minimum number of cuts, and how does the week of payments actually go?
Three cuts (making seven loose links) obviously works. Before the hints, try to say precisely what a set of piece sizes must satisfy: after each day k, the pieces in the worker's hands must total exactly k. That turns a story about cutting into a question about which numbers a set of sizes can represent.
Stuck? Open hints one at a time
Each hint gives away one more layer. Stop the moment something clicks.
Hint 1 a nudge
Because the worker gives change, a day's payment is not "hand over one link". It is any exchange that leaves the worker holding exactly k links' worth after day k. So your pieces only need to make every total from 1 to 7 by combination.
Hint 2 the key move
Which set of piece sizes summing to 7 can express every number from 1 to 7 as a subset sum? Think of how binary notation writes 1 through 7 using only the place values 1, 2, and 4.
Hint 3 nearly the answer
You need pieces of sizes 1, 2, and 4. Can a single cut produce exactly those three pieces from a chain of seven? Count positions.
The solution
Reveal the full solution
One cut. Open the third link. The chain falls into pieces of size 1 (the opened link), 2, and 4.
The week then runs as a sequence of exchanges:
- Day 1: give the 1.
- Day 2: give the 2, take the 1 back.
- Day 3: give the 1 again.
- Day 4: give the 4, take back the 1 and the 2.
- Days 5, 6, 7: repeat days 1 to 3 on top of the 4.
After day k the worker holds exactly k links' worth, every day, which is the contract. The reason {1, 2, 4} works is that these are the binary place values: every integer from 1 to 7 has a unique three-bit representation, so every required daily total is some subset of the pieces. Handing a piece over sets its bit; taking it back as change clears it. The week of payments is literally counting from 001 to 111 in binary, one increment per day.
And one cut is clearly minimal, since zero cuts leave a single piece of 7 that cannot pay day 1. The puzzle is tight in a second way: three pieces can distinguish at most 2 to the power 3, minus 1, that is 7 non-empty subsets, so seven days is the absolute maximum three pieces can ever cover. The cut that produces {1, 2, 4} wastes nothing.
What this puzzle is really testing
Representation choice. The story pushes you toward thinking of payments as giving objects away; the solution reframes a payment as a reversible state change, at which point "make every total from 1 to 7" is the whole problem and binary is the standard answer. Doubling place values to cover a range with logarithmically few pieces is the same growth fact explored in the log n essay, and the "how many outcomes can my resources distinguish" count is the argument from the three-switches puzzle wearing gold.
Follow-ups to expect
A 63-link chain and a 63-day month: three cuts, opening links to leave singles of 1, 1, 1 and segments of 4, 8, 16, 32, and in general k cuts handle a chain of (k+1) times 2 to the power (k+1), minus 1 links. What if the worker refuses to make change? Then subset sums are useless, you need a piece for every prefix, and only all-singles works. Constraints on the exchange model, not the cutting, are what make the puzzle solvable; noticing which constraint carries the weight is the transferable skill, and the binary search on the answer deep-dive runs the same doubling logic inside real code.