16.2.11 * A four-digit number satisfies the following conditions: (1) It is a multiple of 15 when 1 is added to it; (2) It is a multiple of 38 when 3 is subtracted from it; (3) The sum of this number and the number obtained by reversing its digits is a multiple of 10. Find this four-digit number.
Let the original four-digit number be $x=\overline{a b c d}$. From $15 \mid x+1$, we get $d=4$ or 9. Also, from $38 \mid x-3$, we know that $x$ is an odd number. Therefore, $d=9$.

Since $10 \mid \overline{a b c d}+\overline{d c b a}$, it follows that $a=1$. From condition (2), we have $\overline{1 b c 9}=38 l+3$, which means $\overline{1 b c 6}=38 l$. The unit digit of $38 l$ is 6, so the unit digit of $l$ is 7 or 2. Also, since $1006 \leqslant \overline{1 b c 6} \leqslant 1996$, we have $26<l<53$. Therefore, $l$ can only be $27, 37, 47, 32, 42, 52$.

From condition (1), we get $38 l+3=15 k-1$, which simplifies to $38 l+4=15 k$. Upon inspection, $l=37, 52$. This gives us the valid four-digit numbers 1409 and 1979. It is easy to verify that they both satisfy the conditions.