Jeff wrote:MrHamilton wrote:To be blue, 6 would have to travel with either 2 or 9, but it doesn't: therefore 6 is green and 5 is blue.

Hi Mr Hamilton, The above statement advised that 6 cannot travel with 2. But, I found that 6 can travel with 2 in the following cases. Could you explain what I have done wrong?

- Code: Select all
`*-------* *-------*`

| . 6 . | | . 6 . |

| . 2 . | | . 2 . |

| . . . | | . . . |

|-------| |-------|

| . . . | | 6 . . |

| 2 . . | | 2 . . |

| 6 . . | | . . . |

|-------| |-------|

| . . 6 | | . . 6 |

| . . . | | . . . |

| . . 2 | | . . 2 |

*-------* *-------*

Yes, thank you Jeff, you would have blue 2, 5, and 6 in C3R7-8-9 and one of them has to be green by the rule. Which one?