Numbers that divide every cyclic permutation of their multiples

consider a number n

get all its 3 digit multiples, and cyclically permute them, so

123 -> 312 -> 231

test whether n divides all these

what n satisfies this condition ? answer given by following code

def cyc(zz):
    return zz[-1]+zz[:-1]

def cyc2(a):
    c=[a]
    b=a
    while a!=cyc(b):
        c.append(cyc(b))
        b=cyc(b)
    return c

def mul(bb,n):
    d=1
    c=bb
    p=[]
    while len(str(c))<n+1:
        c=bb*d
        if len(str(c))==n:
           p.append(c)
        d+=1
    return p

def cycdig(a,b):
    z=[]
    for ss in range(1,a):
        tot=0
        for y in mul(ss,b):
            cou=0
            for x in cyc2(str(y)):
                if int(x)%ss==0:
                    cou+=1
            if cou==len(cyc2(str(y))):
                tot+=1
        if tot==len(mul(ss,b)):
            z.append(ss)
    return z

print (cycdig(100,3))

the answers are 1,3,9,27,37