Getting The Bases With 2007? (Posted on 2007-09-09)

	Submitted by K Sengupta
Rating: 3.0000 (1 votes)
Solution:	(Hide)
If possible, then 2*N^3 + 7 = y^3, when y is a positive integer......(*) Now N^3 (Mod 7) = 0, 1, 6, so that the lhs of (*) is congruent to 0, 2, 5 mod 7. Now, the rhs of (*) is y^3, which is congruent to 0, 1, 6 mod 7, giving: (N, y) (Mod 7) = (0, 0) Thus, (N, y) = (7M, 7z), for some integers M and z. Substituting this in (*), we obtain: 686*M^3 + 7 = 343*z^3 Or, 98*M^3 + 1 = 49*z^3 Thus, the lhs of the above equation is not divisible by 49, while the lhs is divisible by 49. This is a contradiction. Consequently, there do not exist any positive integer base N ≥ 8 such that 2007 is a perfect cube

	Subject	Author	Date
	re: another solution	Brian Smith	2007-10-19 10:59:09
	another solution	bubu	2007-10-18 13:11:30
	re: Hint	Praneeth	2007-09-11 01:02:36
	Hint	K Sengupta	2007-09-10 09:45:03