Daily calcudoku

The puzzle

Fill every row and every column with the digits 1 to 5, each exactly once, so that the digits in every outlined cage combine to its target with its operation. This one has 5 addition cages, 2 subtraction cages, 2 multiplication cages and 1 division cage.

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Where to start

2 cells are printed for you: a 3 at row 5, column 4; and a 5 at row 5, column 5. 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 6× cage at row 2, column 3 (1 × 2 × 3); the 40× cage at row 3, column 1 (2 × 4 × 5); the 3÷ cage at row 1, column 1 (3 ÷ 1); the 3+ cage at row 4, column 4 (1 + 2).

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 6× cage at row 2, column 3 (3 cells) can only be 1 × 2 × 3.
• The 40× cage at row 3, column 1 (3 cells) can only be 2 × 4 × 5.
• The 3÷ cage at row 1, column 1 (2 cells) can only be 3 ÷ 1.
• The 3+ cage at row 4, column 4 (2 cells) can only be 1 + 2.
• The 3− cage at row 1, column 4 (2 cells) could be 4 − 1 or 5 − 2 — the grid leaves only 5 − 2.
• The 7+ cage at row 1, column 5 (2 cells) could be 2 + 5 or 3 + 4 — the grid leaves only 3 + 4.
• The 5+ cage at row 3, column 4 (2 cells) could be 1 + 4 or 2 + 3 — the grid leaves only 1 + 4.
• The 5+ cage at row 5, column 2 (2 cells) could be 1 + 4 or 2 + 3 — the grid leaves only 1 + 4.
• The 2− cage at row 3, column 2 (2 cells) could be 3 − 1, 4 − 2 or 5 − 3 — the grid leaves only 5 − 3.
• The 11+ cage at row 1, column 2 (3 cells) could be 1 + 5 + 5, 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 1 of 23. Somewhere in this line the 5 must go — and row 1, column 3 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 3 of 23. Work the 3− cage: with what the board already holds, only 2 can sit in row 1, column 4 and still hit the target. On its own that cage could be 4 − 1 or 5 − 2; the digits already placed around it leave one.
3. Step 18 of 23. Work the 40× cage: with what the board already holds, only 2 can sit in row 5, column 1 and still hit the target. On its own that cage allows only 2 × 4 × 5.

How this grid gives way

Of the 23 cells left to fill after the 2 printed ones, 12 by crossing off what a cell’s row and column already hold until one digit was left; 9 by a cage’s arithmetic leaving only one value that fits; and 2 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