Daily calcudoku

The puzzle

Fill every row and every column with the digits 1 to 7, each exactly once, so that the digits in every outlined cage combine to its target with its operation. This one has 5 addition cages, 8 subtraction cages, 4 multiplication cages and 1 division cage.

Want to try it before reading on? Play today's grid in Cages — free in the browser, no account needed.

Where to start

7 cells are printed for you: a 2 at row 1, column 1; a 1 at row 1, column 6; a 5 at row 2, column 3; a 5 at row 3, column 4; a 2 at row 4, column 5; a 2 at row 7, column 3; and a 1 at row 7, column 4. 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 30× cage at row 5, column 6 (5 × 6); the 18× cage at row 7, column 1 (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 30× cage at row 5, column 6 (2 cells) can only be 5 × 6.
• The 18× cage at row 7, column 1 (2 cells) can only be 3 × 6.
• The 28× cage at row 3, column 1 (3 cells) could be 1 × 4 × 7 or 2 × 2 × 7 — the grid leaves only 1 × 4 × 7.
• The 16+ cage at row 6, column 1 (3 cells) could be 3 + 6 + 7 or 4 + 5 + 7 — the grid leaves only 4 + 5 + 7.
• The 5− cage at row 2, column 1 (2 cells) could be 6 − 1 or 7 − 2 — the grid leaves only 6 − 1.
• The 10+ cage at row 3, column 5 (2 cells) could be 3 + 7 or 4 + 6 — the grid leaves only 3 + 7.
• The 5− cage at row 5, column 4 (2 cells) could be 6 − 1 or 7 − 2 — the grid leaves only 7 − 2.
• The 3÷ cage at row 5, column 7 (2 cells) could be 3 ÷ 1 or 6 ÷ 2 — the grid leaves only 3 ÷ 1.
• The 36× cage at row 3, column 7 (3 cells) could be 1 × 6 × 6, 2 × 3 × 6 or 3 × 3 × 4 — the grid leaves only 2 × 3 × 6.
• The 4− cage at row 1, column 2 (2 cells) could be 5 − 1, 6 − 2 or 7 − 3 — the grid leaves only 7 − 3.
• The 3− cage at row 1, column 4 (2 cells) could be 4 − 1, 5 − 2, 6 − 3 or 7 − 4 — the grid leaves only 6 − 3.
• The 3− cage at row 7, column 6 (2 cells) could be 4 − 1, 5 − 2, 6 − 3 or 7 − 4 — the grid leaves only 7 − 4.
• The 12+ cage at row 5, column 5 (3 cells) could be 1 + 4 + 7, 1 + 5 + 6, 2 + 3 + 7, 2 + 4 + 6 or 3 + 4 + 5 — the grid leaves only 1 + 5 + 6.
• The 1− cage at row 1, column 7 (2 cells) could be 2 − 1, 3 − 2, 4 − 3, 5 − 4, 6 − 5 or 7 − 6 — the grid leaves only 5 − 4.
• The 1− cage at row 4, column 1 (2 cells) could be 2 − 1, 3 − 2, 4 − 3, 5 − 4, 6 − 5 or 7 − 6 — the grid leaves only 5 − 4.
• The 1− cage at row 5, column 2 (2 cells) could be 2 − 1, 3 − 2, 4 − 3, 5 − 4, 6 − 5 or 7 − 6 — the grid leaves only 3 − 2.
• The 13+ cage at row 1, column 5 (3 cells) allows 8 combinations, from 1 + 5 + 7 to 4 + 4 + 5 — too many to start from; the cells around it settle it, as 2 + 4 + 7.
• The 11+ cage at row 3, column 3 (3 cells) allows 8 combinations, from 1 + 3 + 7 to 3 + 4 + 4 — too many to start from; the cells around it settle it, as 1 + 4 + 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 42. Somewhere in this line the 5 must go — and row 1, column 7 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 4 of 42. Work the 3− cage: with what the board already holds, only 3 can sit in row 2, column 4 and still hit the target. On its own that cage could be 4 − 1, 5 − 2, 6 − 3 or 7 − 4; the digits already placed around it leave one.
3. Step 31 of 42. Work the 36× cage: with what the board already holds, only 6 can sit in row 4, column 7 and still hit the target. On its own that cage could be 1 × 6 × 6, 2 × 3 × 6 or 3 × 3 × 4; the digits already placed around it leave one.

How this grid gives way

Of the 42 cells left to fill after the 7 printed ones, 21 by crossing off what a cell’s row and column already hold until one digit was left; 15 by a cage’s arithmetic leaving only one value that fits; 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