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Problem 6

屋漏痕 2009-07-27 12:16:59 ( reads)

Mathematical induction

When n=2, obvious.

Assume the statement is true for n.

For n+1, write M as {b1,b2,…b(n)}, where b1
Choose a1, a2, …, a(n) such that a1+a2+…a(n) is not equal to b(n-1). By induction, a solution exists for instance with n steps.

If b(n-1)> a1+a2+…a(n), adding b(n) and a(n+1) won’t cause any conflict.

If b(n-1)

跟帖(6)

dynamic

2009-07-27 18:18:33

回复:Problem 6

屋漏痕

2009-07-28 05:53:51

回复:回复:Problem 6

dynamic

2009-07-28 09:14:51

回复:回复:回复:Problem 6

屋漏痕

2009-07-28 10:07:10

回复:回复:回复:回复:Problem 6

屋漏痕

2009-07-28 10:58:07

still off a littlt bit.

屋漏痕

2009-07-28 12:39:19

This one might be right, but I might be wrong.