The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

### Solution

Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number can be written as 10x + y in the expanded form.

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.

According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66

x + y = 6 ... (1)

You are also given that the digits differ by 2. Therefore,

either x – y = 2 ... (2)

or y – x = 2 ... (3)

If x – y = 2, then solving (1) and (2) by elimination, you get x = 4 and y = 2. In this case, the number is 42.

If y – x = 2, then solving (1) and (3) by elimination, you get x = 2 and y = 4. In this case, the number is 24.

Thus, there are two such numbers 42 and 24.