Sum of 2 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?

Answer: The numbers are $42$ and $24$ . There are $2$ such numbers.

Step-by-step solution

Step 1: Represent the numbers algebraically

Let the tens digit be $x$ and the units digit be $y$ . The original two-digit number is given by $10x + y$ . When the digits are reversed, the units digit becomes the tens digit and vice versa, giving $10y + x$ .

Step 2: Form and simplify the first equation

The sum of the original number and the reversed number is given as 66. Combining the like terms yields $11x + 11y = 66$ . Dividing both sides by 11 gives $x + y = 6$ , which we label as Equation (1).

Step 3: Form the equations based on the difference of digits

We are given that the digits differ by 2. Since it is not specified which digit is greater, two cases arise: either $x - y = 2$ or $y - x = 2$ .

Step 4: Solve Case 1 where $x - y = 2$

Adding Equation (1) and Equation (2) eliminates $y$ , resulting in $2x = 8$ , so $x = 4$ . Substituting $x = 4$ into Equation (1) gives $4 + y = 6$ , which yields $y = 2$ . The number in this case is $10(4) + 2 = 42$ .

Step 5: Solve Case 2 where $y - x = 2$

Adding Equation (1) and Equation (3) eliminates $x$ , giving $2y = 8$ , so $y = 4$ . Substituting $y = 4$ into Equation (1) gives $x = 2$ . The number in this case is $10(2) + 4 = 24$ .

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Sum of 2 digit number and the number obtained by reversing the digits is 666666. If the digits of the number differ by 222, find the number. How many such numbers are there?