- #26

CRGreathouse

Science Advisor

Homework Helper

- 2,820

- 0

Funny, I was just about to post the same thing.Xevarion's technique seems to work with 683.

And yes, composites do seem to work better -- I'll check that one, Xev.

- Thread starter approx
- Start date

- #26

CRGreathouse

Science Advisor

Homework Helper

- 2,820

- 0

Funny, I was just about to post the same thing.Xevarion's technique seems to work with 683.

And yes, composites do seem to work better -- I'll check that one, Xev.

- #27

- 76

- 0

- #28

- 15,393

- 685

Ouch. By hand!?! You did fine this time.Anyone who can program (DH?) want to check my work for me? I did this all by hand...

I realized you could even look at multiples of 2 (or 3) as moduli (spelling?). In case of multiples of 2 (or 3), multiples of 2 (or 3) have to be removed from the complement of [itex](A - B) \cup (B-A) \cup (A + B)[/itex] because we don't care about these values. With that, the non-multiples of 2 or 3 less than 100 that are in the complement of the reachable numbers modulo some integer p<=999 (or p<=133, for that matter) are 53, 71, and 95.

Here's the part where I pop in with some numerical evidence.

I have trouble with53, 71, and 95. If the conjecture holds for these numbers, then the constituent powers are at least 10^{100,000}. The other numbers = 1 and 5 mod 6 below 100 work.

Last edited:

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 890

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 15

- Views
- 2K

- Last Post

- Replies
- 28

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Replies
- 2

- Views
- 25K

- Last Post

- Replies
- 6

- Views
- 477

- Replies
- 6

- Views
- 6K

- Replies
- 1

- Views
- 4K