Divisibility (Posted on 2004-01-06)

	Submitted by Ravi Raja
Rating: 2.5000 (6 votes)
Solution:	(Hide)
The number x can take the values from 0 to 9 or any other value ending in these digits. Now, x ² will always end in: 0, 1, 4, 5, 6, or 9. So, (x ² + 2x) will end in: 0, 3, 4, 5,or 8. Therefore, (x ² + 2x + 3) will always end in: 1, 2, 3, 6, 7, or 8. Since in none of the cases the last digit is a 0 or a 5, hence it is not divisible by 5 and therefore never by 35 implying that there does not exist any natural number x such that (x ² + 2x + 3) will be divisible by 35 without leaving any remainder.

	Subject	Author	Date
	re: Simplest solution - no moduli!	Steve Herman	2016-06-11 08:15:09
	Simplest solution - no moduli!	JayDeeKay	2016-06-10 10:23:34
	Alternative Methodology	K Sengupta	2007-11-29 11:12:38
	umm	paul	2004-03-24 22:41:59
	re: Semantics	Ravi Raja	2004-01-16 02:45:36
	a bit clearer	zaphod	2004-01-07 03:19:36
	re(3): Solution	Gamer	2004-01-06 20:24:43
	re(2): Solution	Charlie	2004-01-06 13:18:15
	re: Solution	SilverKnight	2004-01-06 11:38:28
	Semantics	Dan Blume	2004-01-06 11:30:11
	Solution	Brian Smith	2004-01-06 10:56:19