Calcudoku for Monday, 31 August 2026

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 1 addition cage, 5 subtraction cages, 3 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

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

The first moves a careful solver finds:

Work the 4+ cage: with what the board already holds, only 2 can sit in row 1, column 4 and still hit the target.
Work the 4+ cage: with what the board already holds, only 1 can sit in row 1, column 5 and still hit the target.
Work the 4+ cage: with what the board already holds, only 1 can sit in row 2, column 4 and still hit the target.

How this grid gives way

Of the 22 cells left to fill after the 3 printed ones, 13 by crossing off what a cell’s row and column already hold until one digit was left; 8 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

Show all 22 steps

Work the 4+ cage: with what the board already holds, only 2 can sit in row 1, column 4 and still hit the target.
Work the 4+ cage: with what the board already holds, only 1 can sit in row 1, column 5 and still hit the target.
Work the 4+ cage: with what the board already holds, only 1 can sit in row 2, column 4 and still hit the target.
Cross off what the row and column already rule out, and row 4, column 1 has exactly one candidate left: 2.
Work the 10× cage: with what the board already holds, only 5 can sit in row 5, column 1 and still hit the target.
Cross off what the row and column already rule out, and row 3, column 1 has exactly one candidate left: 1.
Work the 4− cage: with what the board already holds, only 5 can sit in row 3, column 2 and still hit the target.
Cross off what the row and column already rule out, and row 5, column 4 has exactly one candidate left: 3.
Work the 2− cage: with what the board already holds, only 3 can sit in row 1, column 2 and still hit the target.
Cross off what the row and column already rule out, and row 1, column 1 has exactly one candidate left: 4.
Cross off what the row and column already rule out, and row 1, column 3 has exactly one candidate left: 5.
Cross off what the row and column already rule out, and row 2, column 1 has exactly one candidate left: 3.
Somewhere in this line the 2 must go — and row 2, column 2 is the only cell that can still take it.
Cross off what the row and column already rule out, and row 2, column 3 has exactly one candidate left: 4.
Cross off what the row and column already rule out, and row 3, column 3 has exactly one candidate left: 3.
Cross off what the row and column already rule out, and row 3, column 5 has exactly one candidate left: 2.
Cross off what the row and column already rule out, and row 4, column 2 has exactly one candidate left: 4.
Work the 4× cage: with what the board already holds, only 1 can sit in row 5, column 2 and still hit the target.
Cross off what the row and column already rule out, and row 4, column 3 has exactly one candidate left: 1.
Cross off what the row and column already rule out, and row 4, column 5 has exactly one candidate left: 3.
Work the 24× cage: with what the board already holds, only 4 can sit in row 5, column 5 and still hit the target.
Cross off what the row and column already rule out, and row 5, column 3 has exactly one candidate left: 2.

Show the finished grid