5. The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ .
5.1458.

Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore, $n=p^{6} q$.
If $p \geqslant 5$, then the number of digits in $n$ will exceed 4, which is a contradiction.
If $p=3$, then $p^{6}=729$, and when $q=2$, $n=$ $p^{6} q=1458$, which satisfies the condition.

If $p=2$, then $p^{6}=64$, in this case, $q$ must be chosen from 3, 13, 23, 43, ... these prime numbers with a unit digit of 3.
Multiplying 64 by these numbers, the smallest four-digit number obtained is
$$
p^{6} q=64 \times 23=1472>1458 .
$$

Therefore, the smallest four-digit number that satisfies the condition is 1458.