How to Play KenKen

KenKen is part Sudoku, part arithmetic. You fill an N×N grid with the digits 1 through N so that each row and column contains every digit exactly once (a Latin square). On top of that, the grid is divided into “cages” — outlined groups of cells — and each cage carries an operation and a target. The cage's cells, combined with the operation, must equal the target.

The Puzzle

Here's a 4×4 puzzle with 7 cages. Each cage's top-left corner shows its target and operation: 7+, 64×, 2÷, and so on. Cages with just one cell and a number alone are pre-filled values.

The Three Rules

Rule 1: Each row and column has every digit 1 to N exactly once

Standard Latin-square rule. On a 4×4 you use 1, 2, 3, 4.

Rule 2: Each cage's cells produce the cage target via the operation

• + (sum): the cage cells add up to the target.
• × (product): they multiply to the target.
• − (difference): 2-cell cage. Larger minus smaller equals the target.
• ÷ (division): 2-cell cage. Larger divided by smaller equals the target.
• = (single cell): the cell holds exactly that value.

Rule 3: Cages may contain repeats — but Latin still applies

A cage can hold duplicate digits, but only if those duplicates land in different rows and different columns (so the Latin rule is still satisfied). For instance, a 3-cell L-shape may legally contain two 3s if the two 3s are in different rows and different columns.

Walkthrough

Step 1 — Place the three single-cell cages

Three cages have just one cell each (operation =). They give us instant placements:

• 4 at row 1, column 2
• 1 at row 1, column 4
• 4 at row 4, column 1

Step 2 — The 7+ cage forces a {1, 3, 3} arrangement

Cage 7+ covers three cells: (1,1), (2,1), (2,2). Three cells from {1, 2, 3, 4} that sum to 7: either {1, 2, 4} or {1, 3, 3}.

Now apply Latin elimination. We just placed 4 at row 1 column 2 (blocks the entire row 1 and column 2 from being 4) and 4 at row 4 column 1 (blocks column 1). Each of the three cage cells is in either row 1, column 1, column 2, or all three:

• (1,1) — row 1 and column 1 both already have or will have 4 → can't be 4.
• (2,1) — column 1 already has 4 (at row 4) → can't be 4.
• (2,2) — column 2 already has 4 (at row 1) → can't be 4.

No cage cell can be 4. So {1, 2, 4} is impossible — the cage must be {1, 3, 3}.

Now place. The two 3s must sit in different rows AND different columns. (1,1) and (2,2) are in different rows and columns ✓. (1,1) and (2,1) are same column. (2,1) and (2,2) are same row. So the only valid placement is:

• 3 at (1,1) and (2,2)
• 1 at (2,1)

Step 3 — Latin cascades through row 1, the 64× cage, and column 4

Now Latin square fills in much of the rest:

• Row 1 has 3, 4, ?, 1 → the missing digit is 2 → (1,3) = 2.
• Row 2 already has 1 and 3, so its remaining cells (2,3) and (2,4) hold 2 and 4. Column 3 already has 2 — so (2,3) must be 4 and (2,4) must be 2.
• Cage 64× = (1,3) × (2,3) × (2,4) × (3,4) = 2 × 4 × 2 × (3,4) = 16 × (3,4) = 64 → (3,4) = 4.
• Column 4 now has 1, 2, 4 → missing 3 → (4,4) = 3.

Step 4 — Finish with cage 2÷ and the last few cells

Cage 2÷ covers (3,1) and (3,2). Column 1 has 3, 1, ?, 4 → so (3,1) = 2. Then 2÷(3,2) = 2 → (3,2) = 1; or (3,2)÷2 = 2 → (3,2) = 4. Column 2 already has 4 (at row 1) → 4 is blocked → (3,2) = 1.

Row 3 now has 2, 1, ?, 4 → (3,3) = 3.

Row 4: column 1 = 4, column 4 = 3; need 2 and 1 in cols 2 and 3. Column 2 already has 4, 3, 1 → (4,2) must be 2; therefore (4,3) = 1.

Tips for Beginners

• Single-cell cages are gifts. Place them first. They're Latin constraints in disguise.
• Enumerate cage candidate sets. For each cage, list all subsets of {1..N} (with possible repeats) that produce the target via the operation. Sum cages of 2 cells are quickest; product cages are surprisingly constrained too.
• Eliminate using Latin constraints. Every cell in a cage sits in some row and column. Existing placements eliminate digits — often crushing whole candidate sets like {1, 2, 4}.
• Watch repeats inside cages. A cage can repeat a digit, but only across different rows AND different columns. That constraint often pins down which arrangement of {a, a, b} works.
• Cascade aggressively. One placement narrows two rows' worth and two columns' worth of candidates, which often forces another placement immediately.

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