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 1 addition cage, 5 subtraction cages, 6 multiplication cages and 1 division cage.

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Where to start

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

2 cages allow a single combination of digits — the safest openings this grid offers: the 90× cage at row 4, column 4 (3 × 5 × 6); the 4× cage at row 4, column 5 (1 × 4).

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 90× cage at row 4, column 4 (3 cells) can only be 3 × 5 × 6.
• The 4× cage at row 4, column 5 (2 cells) can only be 1 × 4.
• The 20× cage at row 1, column 1 (3 cells) could be 1 × 4 × 5 or 2 × 2 × 5 — the grid leaves only 1 × 4 × 5.
• The 30× cage at row 4, column 1 (3 cells) could be 1 × 5 × 6 or 2 × 3 × 5 — the grid leaves only 2 × 3 × 5.
• The 15+ cage at row 5, column 6 (3 cells) could be 3 + 6 + 6 or 4 + 5 + 6 — the grid leaves only 4 + 5 + 6.
• The 3÷ cage at row 1, column 3 (2 cells) could be 3 ÷ 1 or 6 ÷ 2 — the grid leaves only 3 ÷ 1.
• The 12× cage at row 1, column 5 (2 cells) could be 2 × 6 or 3 × 4 — the grid leaves only 2 × 6.
• The 4− cage at row 2, column 3 (2 cells) could be 5 − 1 or 6 − 2 — the grid leaves only 5 − 1.
• The 36× cage at row 4, column 2 (3 cells) could be 1 × 6 × 6, 2 × 3 × 6 or 3 × 3 × 4 — the grid leaves only 2 × 3 × 6.
• The 3− cage at row 6, column 2 (2 cells) could be 4 − 1, 5 − 2 or 6 − 3 — the grid leaves only 4 − 1.
• The 2− cage at row 2, column 2 (2 cells) could be 3 − 1, 4 − 2, 5 − 3 or 6 − 4 — the grid leaves only 6 − 4.
• The 2− cage at row 2, column 4 (2 cells) could be 3 − 1, 4 − 2, 5 − 3 or 6 − 4 — the grid leaves only 4 − 2.
• The 2− cage at row 2, column 6 (2 cells) could be 3 − 1, 4 − 2, 5 − 3 or 6 − 4 — the grid leaves only 5 − 3.

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 2 of 31. Work the 4× cage: with what the board already holds, only 1 can sit in row 5, column 5 and still hit the target. On its own that cage allows only 1 × 4.
2. Step 3 of 31. Somewhere in this line the 1 must go — and row 1, 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.
3. Step 24 of 31. Work the 30× cage: with what the board already holds, only 2 can sit in row 6, column 1 and still hit the target. On its own that cage could be 1 × 5 × 6 or 2 × 3 × 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, 18 by crossing off what a cell’s row and column already hold until one digit was left; 12 by a cage’s arithmetic leaving only one value that fits; and 1 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