In [33]:

    

def one_per_x(n):
    r, f, m = [], [], 1
    while m not in f:
        f.append(m)
        m *= 10
        r.append(int(m/n))
        m %= n
    return ''.join(str(i) for i in r[f.index(m):])


    

In [34]:

    

one_per_x(7)


    

    
Out[34]:





'142857'

In [44]:

    

def max_range(f, t):
    from operator import itemgetter
    return max( ( (x, len(one_per_x(x))) for x in range(f, t) ), key=itemgetter(1))


    

In [45]:

    

max_range(2, 1000)


    

    
Out[45]:





(983, 982)

In [46]:

    

%timeit max_range(2, 1000)


    

    




1 loops, best of 3: 396 ms per loop

In [ ]: