5. Find all 4-digit numbers that are 8802 more than the number written with the same digits in reverse order.

ANSWER: 1099
Solution: Note that $x+8802<10000$, therefore, $x<1198$, so the first digit is 1, and the second digit is 0 or 1. The sum of x+8802 ends in 1, so the last digit is 9. We get a number of the form 10a9 or 11a9, we get an equation of the form $10 \mathrm{a} 9+8802=9 \mathrm{a} 01$ or $11 \mathrm{a} 9+8802=9 \mathrm{a} 11$. The first gives the solution $a=9$, the second has no solution.