3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.


3. $k_{\min }=6$.

Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.

Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be written in the form $\frac{m}{10^{T}-1}(m \in \mathbf{Z})$, we set
$$
a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}.
$$

By $a-b=\frac{m-n}{10^{30}-1}, a+k b=\frac{m+k n}{10^{30}-1}$, both are repeating decimals with a period of 15, subtracting them gives
$$
(k+1) b=\frac{(k+1) n}{10^{30}-1}
$$

is a repeating decimal with a period of 15, hence
$$
\left(10^{15}+1\right) \mid(k+1) n.
$$

Since $n$ is not a multiple of $10^{15}+1$ (otherwise, the period length of $b$ would be 15), $k+1$ must be a multiple of some prime factor of $10^{15}+1$.
Also, $10^{15}+1$ is not a multiple of $2,3,5$, thus,
$k+1 \geqslant 7 \Rightarrow k \geqslant 6$.
Next, provide an example for $k=6$.
Take $a=\frac{8}{7\left(10^{15}-1\right)}, b=\frac{1}{7\left(10^{15}-1\right)}$.
Notice that, $10^{3}+1=7 \times 143$, so the smallest period of $a, b$ is 30, $a-b=\frac{1}{10^{15}-1}, a+6 b=\frac{2}{10^{15}-1}$ has the smallest period of 15.