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 5 addition cages, 3 subtraction cages, 5 multiplication cages and 1 division cage.

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Where to start

3 cells are printed for you: a 4 at row 1, column 5; a 3 at row 4, column 1; and a 1 at row 6, column 2. 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 18× cage at row 1, column 1 (1 × 3 × 6); the 30× cage at row 4, column 2 (5 × 6); the 5× cage at row 4, column 6 (1 × 5); the 18× cage at row 5, column 4 (3 × 6).

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 18× cage at row 1, column 1 (3 cells) can only be 1 × 3 × 6.
• The 30× cage at row 4, column 2 (2 cells) can only be 5 × 6.
• The 5× cage at row 4, column 6 (2 cells) can only be 1 × 5.
• The 18× cage at row 5, column 4 (2 cells) can only be 3 × 6.
• The 20× cage at row 2, column 3 (3 cells) could be 1 × 4 × 5 or 2 × 2 × 5 — the grid leaves only 1 × 4 × 5.
• The 6+ cage at row 5, column 1 (2 cells) could be 1 + 5 or 2 + 4 — the grid leaves only 2 + 4.
• The 9+ cage at row 1, column 6 (3 cells) could be 1 + 2 + 6, 1 + 3 + 5 or 2 + 3 + 4 — the grid leaves only 2 + 3 + 4.
• The 3− cage at row 1, column 4 (2 cells) could be 4 − 1, 5 − 2 or 6 − 3 — the grid leaves only 5 − 2.
• The 7+ cage at row 2, column 1 (2 cells) could be 1 + 6, 2 + 5 or 3 + 4 — the grid leaves only 1 + 6.
• The 3− cage at row 2, column 5 (2 cells) could be 4 − 1, 5 − 2 or 6 − 3 — the grid leaves only 6 − 3.
• The 3− cage at row 3, column 1 (2 cells) could be 4 − 1, 5 − 2 or 6 − 3 — the grid leaves only 5 − 2.
• The 2÷ cage at row 4, column 4 (2 cells) could be 2 ÷ 1, 4 ÷ 2 or 6 ÷ 3 — the grid leaves only 4 ÷ 2.
• The 9+ cage at row 5, 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 2 + 3 + 4.
• The 12+ cage at row 5, column 5 (3 cells) could be 1 + 5 + 6, 2 + 4 + 6, 2 + 5 + 5, 3 + 3 + 6 or 3 + 4 + 5 — the grid leaves only 1 + 5 + 6.

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 1 of 33. Somewhere in this line the 4 must go — and row 4, column 4 is the only cell that can still take it. Scanning a line for where a digit can still go — rather than what a cell can still hold — is what finds this one.
2. Step 2 of 33. Work the 2÷ cage: with what the board already holds, only 2 can sit in row 4, column 5 and still hit the target. On its own that cage could be 2 ÷ 1, 4 ÷ 2 or 6 ÷ 3; the digits already placed around it leave one.
3. Step 23 of 33. Work the 3− cage: with what the board already holds, only 5 can sit in row 3, column 1 and still hit the target. On its own that cage could be 4 − 1, 5 − 2 or 6 − 3; the digits already placed around it leave one.

How this grid gives way

Of the 33 cells left to fill after the 3 printed ones, 15 by a cage’s arithmetic leaving only one value that fits; 12 by crossing off what a cell’s row and column already hold until one digit was left; and 6 by finding the one place left in a line where a digit could still go. No step needs a guess: every Cages daily is checked to have exactly one solution, reachable by reasoning alone.

The full solution