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标题: 2的幂与7的关系 [打印本页]

作者: 野 菜 花    时间: 2007-1-12 06:34
标题: 2的幂与7的关系

1. 找出所有的自然数 n 使得 2n -1 能被

7整除

2. 证明对任何自然数 n ,2n +1 都不能被

7整除
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作者: yinyin    时间: 2007-1-12 06:46
标题: 回复:2的幂与7的关系

1) n=23k-1, k=1, 2, ...  .


 

作者: yinyin    时间: 2007-1-12 07:05
标题: 回复:2的幂与7的关系

2)  Use the mathematical induction to show that 2n(mod 7) is in set {1,2,4}:www.ddhw.com
When n=1,2,or 3, 2n(mod 7) =2,4,1 respectively. For any given k>=3, assuming that 2k(model 7) is in set {1,2,4}, we know that 2(n+1)=2*2n(mod 7)is also in {1,2,4} since 2*1(mod 7)=2, 2*2(mod 7)=4, and 2*4(mod 7)=1.
Thus, (2n+1)(mod 7) is in set {2,3,5} and, therefore, 2n+1 cannot be exactly divided by 7 for any natural number n.
 
Note. For any natural numbers m and i, m(mod i) represents the remainder when m is divided by i.


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  本贴由[yinyin]最后编辑于:2007-1-11 23:22:26  


 

  本贴由[yinyin]最后编辑于:2007-1-12 0:45:59  


作者: 野 菜 花    时间: 2007-1-12 07:18
标题: 这么快!佩服![:-Q][@};-]

  这么快!佩服!





作者: 野 菜 花    时间: 2007-1-12 07:18
标题: PERFECT! [:-Q][@};-][@};-]

  PERFECT!





作者: 色盲    时间: 2007-1-14 22:32
标题: 回复:回复:2的幂与7的关系

貌似是n=3k, k=1,2,3...
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据说这世界是彩色的?

 






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