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?
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.