13.144. The sum of all even two-digit numbers is divisible by one of them without a remainder. The quotient obtained differs from the divisor only in the order of the digits, and the sum of its digits is 9. What two-digit number was the divisor?
Solution.

Let's represent the divisor in the form $10a + b$. Then, according to the condition, we have $\left\{\begin{array}{l}\frac{2430}{10a + b} = 10b + a, \\ a + b = 9,\end{array}\right.$ from which $a = 4, b = 5$. The required number is 54.

Answer: 54.