Doubt: LRDI All LRDI — The Marrowick Barrel Ledger

Think of this as: Larkin only has two lends in the whole season. Telling you which weeks they fall in removes all its freedom — because between those two lends, Larkin can only go up.

What the ledger actually gives you:

Thirty of the thirty-six working cells are blank. Nothing can be read off — everything has to be reconstructed.

The two counters that unlock everything. For each press write:

• $\ell(p)$ = the number of weeks that press was the lender
• $r(p)$ = the number of single barrels that press received over the five weeks

Every week has exactly one lender and exactly two receivers, so across the five weeks:

$\sum_p \ell(p) = 5 \qquad\text{and}\qquad \sum_p r(p) = 5 \times 2 = 10$

A press's holding only ever moves by −2 (a week it lends) or +1 (a week it receives), so

$\text{final}(p) = \text{start}(p) - 2\,\ell(p) + r(p)$

Two more facts worth holding: a lender hands its two barrels to two different presses, so nobody receives twice in one week; and a press cannot both lend and receive in the same week. Hence $\ell(p) + r(p) \le 5$.

Step 1 — Larkin's Start, from conservation alone.

The barrel stock never changes: 27 barrels, always. The five recorded starting holdings add to

$3 + 6 + 2 + 5 + 4 = 20$

so Larkin must have started with $27 - 20 = \mathbf{7}$.

Step 2 — Larkin lends exactly twice (this is the fact the what-if leans on).

Larkin's Least is 4, so it never sits below 4 and does stand on 4 at some point. With steps of only −2 and +1, the only way to land on 4 is to lend from 6 (arriving by +1 would mean standing on 3 first, below its Least). And reaching 6 from a start of 7 costs a lend whichever way you do it — $7 \to 5$ (lend) $\to 6$ (receive), or $7 \to 8$ (receive) $\to 6$ (lend). So $\ell(\text{Larkin}) \ge 2$. Meanwhile Cobbet (6 → 4), Fenwick (5 → 3) and Harrow (4 → 3) each must fall, so each lends at least once. Those bounds total $1 + 1 + 1 + 2 = 5$ — exactly the season's five lends. Every bound is tight: Larkin lends exactly twice.

Step 3 — Apply the condition.

Larkin lends in Week 1 and in Week 5. Since it lends exactly twice, those are its only two lends, so any barrels it receives fall in Weeks 2, 3 or 4.

1. Week 1: Larkin goes $7 \to 5$.
2. Weeks 2–4: with no lend available, Larkin can only stay put or rise. Call its holding entering Week 5 $v$, so $v \ge 5$.
3. Week 5: Larkin drops to $v - 2$ — its only remaining fall, and therefore its lowest holding of the season.
4. Larkin's Least is 4, and the Least column records a value actually reached. Every earlier holding is 5 or more, so the 4 must be the final one: $v - 2 = 4$, giving $v = 6$.
5. So Larkin received in exactly one of Weeks 2–4 ($5 + 1 = 6$), and finishes on 4.

Check every alternative — the number of receipts Larkin takes in Weeks 2–4 is the only free choice, and only one value survives:

💡 Teacher tip: A what-if is doing its job when it converts a count into a timetable. Knowing Larkin lends twice tells you little; knowing those two lends bookend the season means every receipt is trapped in the middle, and the last step becomes the only place its Least can live.

Answer: 4

The elegance is that the condition never mentions a single number, yet it pins one exactly. It works only because Larkin's two lends were already forced — with a third lend available, the final holding would still be up for grabs.