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1) n=23k-1, k=1, 2, ... . |
2) Use the mathematical induction to show that 2n(mod 7) is in set {1,2,4}: 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. 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 |
貌似是n=3k, k=1,2,3... |
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