# Number Puzzle #6 (I think)

I’m not keeping very good track of how many of these kind of puzzles I’m posting, so we’ll just say this is the 6th.

Problem: Place the digits 1 through 8 in the circles below such that no two adjacent circles contain consecutive digits.

Update: I was missing a couple of lines.  The picture is now correct.

## 10 thoughts on “Number Puzzle #6 (I think)”

1. And by adjacent I’m assuming connected by the lines?

2. I started with 1 and 8 in the center two circles because those circles are the “most connected” and those numbers have the least amount of consecutive digits.

From there added 7 and 2 to the top and bottom respectively since those were the only terms consecutive with 8 and 1 and needed to not be adjacent to them.

The rest was simply placing the numbers in…

Therefore:

```   (7)
(4)(1)(3)
(6)(8)(5)
(2)
```
3. I got an almost identical answer. I’m curious if there are other solutions, excluding rotations or reflections and excluding shifts (start with 2 instead of 1, etc.)

4. On my last comment, I take back what I said about the shifts. Starting with 2 instead of 1 won’t work. 1 and 8 are special numbers since they only have one adjacent value. Do they always end up in the middle?

5. Sean says:

Unless I’m misunderstanding what you mean by shift, I think they should work in just the same way one through eight does. Using only the numbers 2 through 9 would then leave 2 and 9 with only one adjacent number (at least in the context we’re working in)

6. I meant using 2, 3, …, 8, 1 as a shift, but still thinking of 1 and 2 as “consecutive”. That is, still using the digits 1-8. In a given solution, you can’t just replace 1’s with 2’s, 2’s with 3’s, …, 8’s with 1’s.

You are correct that the problem is solved similarly for any set of consecutive integers.