找进位制和其中特殊数字，使其具有10进制里面7的这个特性，

[verdelite]

10进制里面，7有个特殊的地方。1/7=0.142857循环。2/7=0.285714循环。3/7=0.428571循环。。。。6/7=0.857142循环。

特点：都是142857这几个数字进行loop。

那么，n进制里面k有这个特性，找到n和k？
如果有许多，那么有什么n和k有什么特点？

[YWY]

10进制里面，7有个特殊的地方。1/7=0.142857循环。2/7=0.285714循环。3/7=0.428571循环。。。。6/7=0.857142循环。

特点：都是142857这几个数字进行loop。

那么，n进制里面k有这个特性，找到n和k？
如果有许多，那么有什么n和k有什么特点？

my hunch: n > k, gcd(n, k) = 1, k is prime, and n is a primitive mod k (the order of n mod k is k-1).

[YWY]

my hunch: n > k, gcd(n, k) = 1, k is prime, and n is a primitive mod k (the order of n mod k is k-1).

I don't think we need the condition n > k. So my best guess (for a sufficient and necessary condition) is that k is a prime number and n is a primitive root mod k.

[YWY]

I don't think we need the condition n > k. So my best guess (for a sufficient and necessary condition) is that k is a prime number and n is a primitive root mod k.

这其实就是长除法的本质：在n进制下，当我们算1/k的小数表达时，第一步的余数是n^1 mod k，第二步的余数是n^2 mod k，第i步的余数是n^i mod k。

If k is a prime and n is a primitive root mod k, then it takes precisely k-1 steps for us to get 1 as the remainder. Moreover, within these k-1 steps, each number from 1 to k-1 must appear as the remainder at precisely one of these k-1 steps. So when we do i/k with 0 < i < k, the remainders follow the same sequence except that the starting values are different. Once we understand the above, we should see that the condition is also necessary.

当n < k时，上面的关于remainder的结论依然正确，但下一步做长除得到的商却不和remainder一一对应（不同的remainder可能得到同样的商）。我刚算了3进制下的1/7（用3进制写就是1/21），得到的商用3进制写是1/21=0.010212010212……，2/21=0.0212010212……，10/21=0.10212010212……。如果大家认为这种情况不符合条件，那就额外加上n > k的条件。