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标题: 美国数学竟赛题 [打印本页]

作者: fzy    时间: 2005-4-28 19:52
标题: 美国数学竟赛题

Difficulty: +++
 
N is a positive integer satisfying the following condition:
 
For any two sets of N consecutive integers A, B, we can form N pairs (a, b) with a ∈ A, b ∈ B, each number is used exactly once, such that the sums of the pairs also form a set of N consecutive integers.
 
Find all such N. 
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作者: 野 菜 花    时间: 2005-4-28 23:39
标题: 回复:美国数学竟赛题

Did I misunderstand your question?
N=2k+1 (K=1,2,3,...)
A={a+1, a+2,..., a+N}
B={b+1,b+2,..., b+N}
 
Set of pairs: {(a+k,b+1), (a+k-1,b+3), (a+k-2, b+5),...,(a+1, b+N), (a+N, b+2),
(a+N-1, b+4),..., (a+k+1, b+N-1) }
==> a+b+k+1, a+b+k+2, a+b+k+3,..., a+b+N+1,a+b+N+2, a+b+N+3,...,a+b+N+k
N consecutive intergers
 
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作者: fzy    时间: 2005-4-28 23:42
标题: 回复:回复:美国数学竟赛题

You understood it corrctly. How about the other half (N = 2k)?
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作者: 野 菜 花    时间: 2005-4-28 23:53
标题: 回复:回复:回复:美国数学竟赛题

Suppose the consecutive interger series of sums of pairs starts with m+1
then
Na+(1+N)N/2 +Nb+(1+N)N/2
=N(a+b)+(m+1+m+N)N/2
Simplify it, we get
N=2m-1
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