5. Find all 4-digit numbers that are 7182 less than the number written with the same digits in reverse order.

ANSWER: 1909
Solution. Let's write the desired number as $\overline{a b c d}$

Then $\overline{a b c d}=\overline{d c b a}-7182$, from which $111(d-a)+10(c-b)=798$.

Obviously, $(d-a)$ can only be 7 or 8.

In the case $d-a=7$, we get $10(c-b)=21-$ which does not fit.

In the case $d-a=8$, we get $10(c-b)=-90$, hence $b-c=9$, from which $b=9, c=0$.

The equality $d$ - $a=8$ is possible when $d=8, a=0$ or $d=9, a=1$, but the first digit cannot be zero, therefore, $a=1, b=9, c=0, d=9$.