13.435 The desired number is greater than 400 and less than 500. Find it, if the sum of its digits is 9 and it is equal to 47/36 of the number represented by the same digits but written in reverse order.
Solution. The condition that the desired number is within the range from 400 to 500 makes the problem trivial. Therefore, let's solve this problem by replacing this redundant condition with the condition that the desired number is a three-digit number.

Let $\overline{x y z}$ be the desired number. According to the conditions, $\left\{\begin{array}{l}36(100 x+10 y+z)=47(100 z+10 y+x) \\ x+y+z=9\end{array} \Leftrightarrow\left\{\begin{array}{l}3553 x=110 y+4664 z \\ y=9-x-z\end{array} \Rightarrow\right.\right.$

$\Rightarrow 3553 x=110(9-x-z)+4664 z \Rightarrow 37 x=10+46 z \Leftrightarrow 37(x-z)=10+9 z$.

But $1 \leq z \leq 9$, from which $19 \leq 10+9 z \leq 91$. But within these bounds, there are only two numbers divisible by $37: 37,74$. Clearly, $74 \neq 10+9 z$. Hence $37=10+9 z, z=3 \Rightarrow x=4, y=2$.

Answer: 423.