Find the least number where moving the first digit to the end multiplies by 1.5.

Let's represent this number as  a*10^n+b,  where 1<=a<=9 and
b < 10^n.  Then the condition to be satisfied is:

3/2(a*10^n+b) = 10b+a

  3(a*10^n+b) = 20b+2a

   3a*10^n+3b = 20b+2a

  (3*10^n-2)a = 17b

            b = a*(3*10^n-2)/17

So we must have 3*10^n-2 = 0 (mod 17) (since a is less than 10, it
cannot contribute the needed prime 17 to the factorization of 17b).
(Also, assuming large n, we must have a at most 5 so that b < 10^n will
be satisfied, but note that we can choose a=1).  Now,

3*10^n-2 = 0 (mod 17)

3*10^n = 2 (mod 17)

10^n = 12 (mod 17)

A quick check shows that the smallest n which satisfies this is 15
(the fact that one exists was assured to us because 17 is prime).  So,
setting n=15 and a=1 (obviously) gives us b=176470588235294, so the
number we are looking for is

                        1176470588235294

and, by the way, we can set a=2 to give us the second smallest such
number,
                        2352941176470588

Other things we can infer about these numbers is that there are 5 of
them less than 10^16, 5 more less than 10^33, etc.