Results 16 to 17 of 17

- May 31st 2007, 07:12 AM #16

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Jun 4th 2007, 11:00 AM #17

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

The only answer is the trivial answer .

Follow, ecMathGeek's approach and arrive at:

Hence, is the only possible constant.

---

This problem was taken from my math book, which said it appeared as a problem in the 1990 Putnam. The problem I posed was a variation of that problem, not exactly the same.