Daily calcudoku

The puzzle

Fill every row and every column with the digits 1 to 6, each exactly once, so that the digits in every outlined cage combine to its target with its operation. This one has 7 addition cages, 1 subtraction cage and 4 multiplication cages.

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Where to start

5 cells are printed for you: a 3 at row 2, column 4; a 1 at row 2, column 6; a 4 at row 4, column 4; a 4 at row 5, column 2; and a 1 at row 5, column 3. Fill those in first, then let each one rule its digit out of its row and column.

4 cages allow a single combination of digits — the safest openings this grid offers: the 15+ cage at row 1, column 4 (4 + 5 + 6); the 40× cage at row 3, column 6 (2 × 4 × 5); the 18× cage at row 5, column 5 (3 × 6); the 3× cage at row 6, column 5 (1 × 3).

Cage by cage

What each cage’s arithmetic allows on its own, before the rows and columns have a say — the most constrained first, which is the order worth looking at them in. Digits may repeat inside a cage only where the cells share no row or column.

• The 15+ cage at row 1, column 4 (3 cells) can only be 4 + 5 + 6.
• The 40× cage at row 3, column 6 (3 cells) can only be 2 × 4 × 5.
• The 18× cage at row 5, column 5 (2 cells) can only be 3 × 6.
• The 3× cage at row 6, column 5 (2 cells) can only be 1 × 3.
• The 6+ cage at row 1, column 1 (3 cells) could be 1 + 1 + 4 or 1 + 2 + 3 — the grid leaves only 1 + 2 + 3.
• The 13+ cage at row 1, column 3 (3 cells) could be 2 + 5 + 6 or 3 + 4 + 6 — the grid leaves only 2 + 5 + 6.
• The 8+ cage at row 6, column 1 (2 cells) could be 2 + 6 or 3 + 5 — the grid leaves only 2 + 6.
• The 12× cage at row 2, column 5 (3 cells) could be 1 × 2 × 6, 1 × 3 × 4 or 2 × 2 × 3 — the grid leaves only 1 × 2 × 6.
• The 7+ cage at row 2, column 1 (2 cells) could be 1 + 6, 2 + 5 or 3 + 4 — the grid leaves only 3 + 4.
• The 9+ cage at row 3, column 2 (3 cells) could be 1 + 2 + 6, 1 + 3 + 5, 1 + 4 + 4, 2 + 2 + 5 or 2 + 3 + 4 — the grid leaves only 1 + 3 + 5.
• The 1− cage at row 4, column 1 (2 cells) could be 2 − 1, 3 − 2, 4 − 3, 5 − 4 or 6 − 5 — the grid leaves only 6 − 5.
• The 11+ cage at row 5, column 4 (3 cells) could be 1 + 4 + 6, 1 + 5 + 5, 2 + 3 + 6, 2 + 4 + 5, 3 + 3 + 5 or 3 + 4 + 4 — the grid leaves only 2 + 4 + 5.

The moments that matter

Most of a solve is crossing candidates off a line until one is left. These are the steps where this grid actually turns:

1. Step 3 of 31. Work the 18× cage: with what the board already holds, only 3 can sit in row 5, column 5 and still hit the target. On its own that cage allows only 3 × 6.
2. Step 14 of 31. Work the 40× cage: with what the board already holds, only 2 can sit in row 4, column 6 and still hit the target. On its own that cage allows only 2 × 4 × 5.
3. Step 24 of 31. Work the 1− cage: with what the board already holds, only 5 can sit in row 5, column 1 and still hit the target. On its own that cage could be 2 − 1, 3 − 2, 4 − 3, 5 − 4 or 6 − 5; the digits already placed around it leave one.

How this grid gives way

Of the 31 cells left to fill after the 5 printed ones, 17 by crossing off what a cell’s row and column already hold until one digit was left; and 14 by a cage’s arithmetic leaving only one value that fits. No step needs a guess: every Cages daily is checked to have exactly one solution, reachable by reasoning alone.

The full solution