13 | |||

10 | |||

4 | 16 | ||

7 |

The cells in this 4×4 grid puzzle contain the numbers 1-16 each once. There are two rules to the arrangement of the numbers:

- Any two consecutive numbers must share a row/column (1 and 16 should be considered consecutive).
- No row/column can contain three consecutive numbers.

Fill in the remaining numbers.

#### Solution

12 | 13 | 8 | 9 |

11 | 2 | 3 | 10 |

5 | 1 | 4 | 16 |

6 | 14 | 7 | 15 |

One super helpful insight for solving this puzzle is that numbers that are 2 apart cannot be in the same row/column. For example, 2 must be in the same row or column as both 1 and 3. If 1 and 3 were to be in the same row, 2 couldn’t share a column with both of them (can’t be in two place at once), and 2 can’t be in the same row as 1 and 3 (violates the 2nd rule of no three consecutive numbers), so this scenario can never occur.

With this implicit rule, one possible way to solve this is as follows:

- The column with 4 and 7 must contain (3 or 5) and (6 or 8).
- But since (5 and 7) and (4 and 6) cannot be in the same row/column, this column must contain 3 and 8.
- But since 8 cannot be in the same row/column as 10, we know exactly where 3 and 8 must go.

- 9 must be at the intersection of 8 and 10.
- 11 must share a row/column with 10, but can’t share a column with 9 or 13, so it must be in the first column.
- 2 must be the remaining number in the row with 11, 3, and 10.
- 12 must be at the intersection of 11 and 13.
- 1 must be at the intersection of 2 and 16.
- 5 must be the remaining number in the row with 1, 4, and 16.
- 6 must be the remaining number in the column with 12, 11, and 5.
- 14 must be the remaining number in the column with 13, 2, and 1.
- 15 must be the remaining number.