let a1,a2,...........,an be n nubers such that each ai is either 1 or -1.if a1a2a3a4+a2a3a4a5+........+ana1a2a3=0,

prove that 4 divides n

pls give me a solution

Results 1 to 2 of 2

- Nov 19th 2009, 05:33 AM #1

- Joined
- Nov 2009
- Posts
- 1

- Nov 19th 2009, 06:37 AM #2

- Joined
- Apr 2009
- Posts
- 678
- Thanks
- 1

Any of the ai's come in exactly 4 terms out of (a1a2a3a4,a2a3a4a5,...,ana1a2a3).

Let the sum be S0. Now just change the sign of any one of the ai. Let the new sum be S1. It is easy to see that 4|(S1-S0)

Now initially if all ai's =1; sum = n

Final sum when some ai's are -1 = 0

thus 4|(n-0) => 4|n