This article is from the Puzzles FAQ, by Chris Cole email@example.com and Matthew Daly firstname.lastname@example.org with numerous contributions by others.
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
and, by the way, we can set a=2 to give us the second smallest such
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.