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 number,


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.

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