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Daily calcudoku

Monday, 10 August 2026 · 5×5 Medium

The puzzle

The 5×5 grid for 10 August 2026, unsolved 1− 3− 10+ 1− 3− 1− 3× 4× 3× 60× 5 2 2

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 and 4 multiplication cages.

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Where to start

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

The first moves a careful solver finds:

  1. Somewhere in this line the 1 must go — and row 1, column 3 is the only cell that can still take it.
  2. Somewhere in this line the 3 must go — and row 1, column 1 is the only cell that can still take it.
  3. Cross off what the row and column already rule out, and row 2, column 1 has exactly one candidate left: 4.

How this grid gives way

Of the 22 cells left to fill after the 3 printed ones, 11 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

Show all 22 steps
  1. Somewhere in this line the 1 must go — and row 1, column 3 is the only cell that can still take it.
  2. Somewhere in this line the 3 must go — and row 1, column 1 is the only cell that can still take it.
  3. Cross off what the row and column already rule out, and row 2, column 1 has exactly one candidate left: 4.
  4. Cross off what the row and column already rule out, and row 3, column 1 has exactly one candidate left: 1.
  5. Work the 3× cage: with what the board already holds, only 3 can sit in row 3, column 2 and still hit the target.
  6. Work the 1− cage: with what the board already holds, only 4 can sit in row 3, column 3 and still hit the target.
  7. Work the 1− cage: with what the board already holds, only 2 can sit in row 3, column 5 and still hit the target.
  8. Cross off what the row and column already rule out, and row 3, column 4 has exactly one candidate left: 5.
  9. Work the 3− cage: with what the board already holds, only 2 can sit in row 2, column 4 and still hit the target.
  10. Cross off what the row and column already rule out, and row 1, column 4 has exactly one candidate left: 4.
  11. Work the 10+ cage: with what the board already holds, only 5 can sit in row 1, column 5 and still hit the target.
  12. Cross off what the row and column already rule out, and row 1, column 2 has exactly one candidate left: 2.
  13. Work the 3− cage: with what the board already holds, only 5 can sit in row 2, column 2 and still hit the target.
  14. Cross off what the row and column already rule out, and row 2, column 3 has exactly one candidate left: 3.
  15. Cross off what the row and column already rule out, and row 2, column 5 has exactly one candidate left: 1.
  16. Cross off what the row and column already rule out, and row 4, column 4 has exactly one candidate left: 1.
  17. Work the 3× cage: with what the board already holds, only 3 can sit in row 4, column 5 and still hit the target.
  18. Cross off what the row and column already rule out, and row 4, column 2 has exactly one candidate left: 4.
  19. Work the 4× cage: with what the board already holds, only 1 can sit in row 5, column 2 and still hit the target.
  20. Cross off what the row and column already rule out, and row 5, column 3 has exactly one candidate left: 5.
  21. Cross off what the row and column already rule out, and row 5, column 4 has exactly one candidate left: 3.
  22. Work the 60× cage: with what the board already holds, only 4 can sit in row 5, column 5 and still hit the target.
Show the finished grid
The solved grid for 10 August 2026 1− 3− 10+ 1− 3− 1− 3× 4× 3× 60× 3 2 1 4 5 4 5 3 2 1 1 3 4 5 2 5 4 2 1 3 2 1 5 3 4