One Gold Bar, Two Cuts, Seven Evenings: 1, 2 and 4

A worker is hired for seven days and paid in gold. At the end of each day he must be holding exactly one more seventh of the bar than he was holding the evening before, so after day $d$ he holds $d/7$ of it and after day seven he has all of it. You are allowed to cut the bar twice.

Seven payments suggests seven pieces, and seven pieces need six cuts, because $c$ cuts of a single object always produce $c+1$ pieces. That schedule is not wrong. It pays the worker correctly on every one of the seven evenings, and if the contract had said nothing about cuts it would be the obvious way to run the job. It violates one constraint and nothing else.

The contract is on the holding, not on the handover

Read the requirement again. After evening $d$ the worker holds $d/7$. Nothing says a seventh of the bar has to change hands, and nothing says gold only travels one way. If the worker can hand pieces back as change, the daily transfer is a difference between two holdings rather than a payment in its own right.

Written that way it is a covering problem about subset sums, and the answer is the one that covers the most with the fewest parts. Measure everything in sevenths and the pieces are $1$, $2$ and $4$:

The ledger is a binary counter

Every whole number from one to seven is a sum of $1$, $2$ and $4$ in exactly one way, which is the statement that three binary digits count from $001$ to $111$. So the schedule writes itself: on evening $d$, the worker holds the pieces whose sizes are the set bits of $d$.

The carries are exactly the evenings where change comes back. On day two the small piece returns and the double goes out; on day four both small pieces return and the largest goes out. Every step hands over only pieces the employer is holding and takes back only pieces the worker is holding, which is the feasibility condition, and it holds because binary addition of one never needs a digit it does not have.

No evening moves more than three pieces, and the largest single motion is on day four. If you are asked to produce this at a whiteboard, the fast way to say it is that the worker's holding isthe day number in binary and the employer's is its complement.

Why one cut could never have worked

The problem hands you the number two, so the honest answer is a construction plus a reason nothing smaller works. The reason is a count.

With $m$ pieces there are $2^m$ subsets, one of them empty, so at most

distinct positive totals are available. Seven evenings need seven distinct totals, so $2^m - 1 \ge 7$ and $m \ge 3$. Three pieces cost two cuts. One cut would leave two pieces and at most three totals, which cannot cover seven evenings no matter where you place it.

Something stronger is true, and it is the part I did not expect. Among the four ways to split seven into three positive parts, only one works. Take $(1,1,5)$: its subset sums are $1, 2, 5, 6, 7$, missing three and four. Take $(1,3,3)$: sums $1, 3, 4, 6, 7$, missing two and five. Take $(2,2,3)$: sums $2, 3, 4, 5, 7$, missing one and six. Only $(1,2,4)$ gets all seven, and it gets each of them exactly once, with no slack anywhere. The bound in (2) is met with equality, so there is no room for a repeated total.

Longer contracts

Change seven days to $n$ and the same two arguments give a complete answer.

The lower bound is (2) again: $2^m - 1 \ge n$. For the construction, take the doubling pieces $1, 2, 4, \dots, 2^{\,m-2}$, which cover every total from $0$ to $2^{\,m-1} - 1$, and let the last piece be the remainder

Because $2^{\,m-1} \le n+1 \le 2^m$, this $L$ satisfies $1 \le L \le 2^{\,m-1}$. Subsets containing $L$ then cover $L$ through $n$, subsets without it cover $0$ through $2^{\,m-1}-1$, and since $L \le 2^{\,m-1}$ the two ranges leave no gap.

A hundred days therefore needs $\lceil \log_2 101 \rceil = 7$ pieces and six cuts, with sizes $1, 2, 4, 8, 16, 32$ and a last piece of $37$ hundredths. The count of cuts grows like the logarithm of the length of the contract, which is the same reason a set of weights or a set of coins is built out of doubling denominations.

What the puzzle needs you to grant

The worker has to accept change. This is not a technicality, it is the entire solution, and the problem is unsolvable in two cuts without it. A worker who keeps whatever he is handed can be paid in at most three distinct amounts, and by day four you are stuck.

The pieces have to be interchangeable by weight rather than individually earmarked. What makes the bar divisible into "a quantity" instead of seven distinct objects is that only the total matters, and the contract has to be about how much gold the worker holds and not how many lumps he holds. On day four he holds one piece, on day seven he holds three, and neither is a breach.

Finally, the pieces are not equal, so the phrase "paid a seventh a day" describes the arithmetic of the holdings rather than a physical transfer of equal shares. That is the part that feels like sleight of hand on first reading, and it is worth saying out loud rather than hoping nobody asks.

Sources and further reading

• The uniqueness that makes Fig. 2 work — Binary number and positional notation
• The covering problem in the definition above — Subset sum problem
• The same doubling logic applied to coinage — Change-making problem

Every claim above was checked before publication: the piece sizes by brute force over every way of cutting the bar into whole sevenths, which found $(1,2,4)$ to be the only three-piece solution and no two-piece solution at all, and the schedule by walking the seven evenings transaction by transaction with a ledger that refuses any step handing over a piece the employer does not hold.