The same forced-combination rule on a large cage — where it locks five cells at once.
Check each large cage against the four totals its size makes unique: the minimum, the minimum + 1, the maximum − 1 and the maximum. A 5-cell cage totalling 34 can only hold 4, 6, 7, 8 and 9, so every other digit leaves all five cells in one stroke.
The forced-combination rule does not stop at pairs. A cage of any size still has a lowest and a highest possible total, and the two sums at each end are still unique — so a big cage sitting on one of them pins every one of its cells in a single stroke. The catch is that the second-from-the-end sums are no longer the obvious set. Five different digits top out at 5+6+7+8+9 = 35; one less than that is NOT '5,6,7,8,9 minus something obvious' — you have to work out which digit gives way. That is what this lesson walks through, and it is where the rule stops being a memory trick and starts being arithmetic.
A cage total is bounded on both sides. With n distinct digits the smallest reachable total is the run 1 + 2 + … + n, and the largest is the top n digits added together. Each extreme is reachable in exactly one way, because there is nothing left to swap in or out. That argument never mentions how big the cage is, so a 5-cell cage totalling 15 or 35 is pinned as hard as a 2-cell cage totalling 3 or 17 — and it pins five cells rather than two.
One step in from each extreme is unique as well, for a narrower reason: there is exactly one legal way to shift the total by a single point without repeating a digit. Every cage size therefore carries four single-combination totals, drawn from between 15 and 21 achievable totals. The payoff climbs with size — a 2-cell unique cage strips at most 7 candidates from each of 2 cells, while the 5-cell cage below strips 4 from each of 5 and hands you a digit set for more than half a box.
The extremes need no thought — they are the low run and the high run. The totals next to them are where players guess wrongly, so derive them instead. From the minimum, the digits are 1, 2, …, n; adding exactly 1 means swapping one digit for one not already present, and only the largest, n, has a free neighbour above it. From the maximum, the digits run from 10 − n up to 9, and only the smallest member of that run has a free neighbour below it. The digit that gives way at the top end is the small one, never the 9.
Applied to 5 cells: the minimum is 1 + 2 + 3 + 4 + 5 = 15, so 16 must be 1 + 2 + 3 + 4 + 6. The maximum is 5 + 6 + 7 + 8 + 9 = 35, so 34 must be 4 + 6 + 7 + 8 + 9. Notice what 34 is not. It is not the top five digits with the 9 reduced, which would repeat the 8. The 5 slides down to 4, and the set acquires a hole in the middle — 4, then nothing at 5, then 6, 7, 8, 9.
Take a 5-cell cage covering R1C2, R1C3, R2C2, R3C1 and R3C2 with a total of 34, on a board where no digit has been placed. Every one of those cells starts with all nine candidates, so the cage total is the only information on offer.
Enumerate around the total to see how sharp 34 is. Five distinct digits make 32 in 3 ways (2+6+7+8+9, 3+5+7+8+9, 4+5+6+8+9) and 33 in 2 ways (3+6+7+8+9, 4+5+7+8+9). At 34 the count collapses to 1: 4 + 6 + 7 + 8 + 9. A set containing a 5 would need 29 from four other distinct digits, and the largest four available are 6 + 7 + 8 + 9 = 30 — one too many, with no legal way to shed a single point once the 5 is spoken for.
So all five cells are drawn from 4, 6, 7, 8 and 9, and the digits 1, 2, 3 and 5 cannot appear in any of them: 4 candidates removed from each of 5 cells, 20 eliminations from one reading of one total. Be clear about what has not happened. No cell is solved — five digits across five cells still leaves 5 × 4 × 3 × 2 × 1 = 120 arrangements, and untangling those needs the surrounding houses.
Those five cells all lie in the top-left box, spanning rows 1–3 and columns 1–3. A box holds each of 1–9 once, so if the cage takes 4, 6, 7, 8 and 9, the four remaining cells of that box — R1C1, R2C1, R2C3 and R3C3 — hold 1, 2, 3 and 5 between them, totalling 11. This complement reading costs nothing and is usually where a large unique cage pays for itself.
The chain continues. R1C1 and R2C1 form a 2-cell cage totalling 8; read alone it has three combinations (1+7, 2+6, 3+5), but restricted to 1, 2, 3 and 5 only 3 + 5 survives, so those cells are a locked 3/5 pair. That leaves 1 and 2 for R2C3 and R3C3, which sum to 3. Those two belong to a 3-cell cage totalling 8 whose third cell is R3C4, so R3C4 = 8 − 3 = 5. One cage total has removed 20 candidates, fixed a pair and forced a digit two cages away — which is why containment is the first thing to check after reading a large cage.
The uniqueness ends abruptly one step from each extreme. Two steps in is never unique for a cage of 2–7 cells: five distinct digits make 33 in 2 ways and 17 in 2 ways, and the count keeps climbing further in — 32 admits 3, and a middle total such as 25 admits 12. Treating 33 as though it behaved like 34 is the common error, and it removes digits that belong on the board.
The reading also assumes the cage is empty. Once a digit is placed inside, the original size and total no longer describe what is left, so the deduction must be redone on the remaining cells against the reduced total; the step engine on this site offers a single-combination cage only for a cage of 2–7 cells with nothing placed in it. Two further limits: a cage of 8 cells is always single-combination, its digits being 1–9 minus the one digit equal to 45 − total, and the complement bonus needs containment — a cage crossing two boxes still yields its in-cage eliminations, but leaves no house with a known set of leftovers.
| Cage size | Lowest total | Lowest + 1 | Highest − 1 | Highest total |
|---|---|---|---|---|
| 2 cells | 3 = 1+2 | 4 = 1+3 | 16 = 7+9 | 17 = 8+9 |
| 3 cells | 6 = 1+2+3 | 7 = 1+2+4 | 23 = 6+8+9 | 24 = 7+8+9 |
| 4 cells | 10 = 1+2+3+4 | 11 = 1+2+3+5 | 29 = 5+7+8+9 | 30 = 6+7+8+9 |
| 5 cells | 15 = 1+2+3+4+5 | 16 = 1+2+3+4+6 | 34 = 4+6+7+8+9 | 35 = 5+6+7+8+9 |
| 6 cells | 21 = 1+2+3+4+5+6 | 22 = 1+2+3+4+5+7 | 38 = 3+5+6+7+8+9 | 39 = 4+5+6+7+8+9 |
| 7 cells | 28 = 1+2+3+4+5+6+7 | 29 = 1+2+3+4+5+6+8 | 41 = 2+4+5+6+7+8+9 | 42 = 3+4+5+6+7+8+9 |
Killer Sudoku keeps every rule of Classic Sudoku — fill the 9×9 grid so each row, column, and 3×3 box holds the digits 1 to 9 exactly once — and adds cages: dashed groups of cells with a small target sum, where no digit may repeat inside a cage. Puzzles usually start with zero given digits, so every deduction begins from the cage sums. That single extra rule unlocks a whole family of arithmetic techniques.
Practise this in Killer mode on Sudoku Challenge, where the in-game Next Step button walks the exact deduction taught in this lesson.
4 + 6 + 7 + 8 + 9 is the only set of five distinct digits totalling 34. Five digits cannot exceed 5 + 6 + 7 + 8 + 9 = 35, and shedding exactly 1 from that maximum without repeating a digit means lowering the 5 to a 4. No other five-digit set reaches 34.
Lowering the 9 to an 8 would repeat the 8 already in the set, and a cage cannot repeat a digit. The only digit in the run 5, 6, 7, 8, 9 with a free neighbour below it is the 5, whose predecessor 4 is unused. That is why the small end of the run always gives way.
Not directly. Once a cell in the cage is solved, the original size and total no longer describe the empty cells, so subtract the placed digits and re-read the smaller cage against its reduced total. The step engine on this site reserves the single-combination step for cages that are still completely empty.
Exactly one step from an extreme, no further. Two steps in is never unique for cages of 2 to 7 cells: five distinct digits make 33 in 2 ways and 17 in 2 ways. The count climbs further inward, with 32 admitting 3 and a middle total of 25 admitting 12.
Explore every Classic and Killer technique in the complete Learning Hub, or read how to play Sudoku to get started.