Let $$M$$ be the smallest positive integer with the following two properties:

(a) The leading digit of $$M$$ is equal to 3.
(b) If $$N$$ is the number obtained by moving this leading 3 to the units place, and shifting all the other digits one place to the left, then $$N = M/4$$. 

What is the sum of the digits of $$M$$?

If $$M$$ has $$k$$ digits then $$N = 10(M - 3 \times 10^{k-1}) + 3 = 10M - 3(10^k - 1)$$, and $$N = M/4$$ rearranges to $$M = \frac{4(10^k - 1)}{13}$$. Thus $$10^k \equiv 1 \pmod{13}$$, which first happens at $$k = 6$$, giving $$M = \frac{4 \times 999999}{13} = 307692$$, and indeed $$4 \times 76923 = 307692$$. The digit sum is $$3 + 0 + 7 + 6 + 9 + 2 = 27$$.