To solve the problem, let’s define the two-digit number as 10a + b, where a is the tens digit and b is the units digit.
According to the first part of the problem, the sum of the digits is five. We can represent this with the equation:
- a + b = 5
Next, when the digits are reversed, the new number becomes 10b + a. The problem states that this new number is greater than the original number by nine, which gives us this equation:
- 10b + a = (10a + b) + 9
Now, we can simplify this equation. Starting with:
- 10b + a = 10a + b + 9
Rearranging gives:
- 10b – b + a – 10a = 9
Which simplifies to:
- 9b – 9a = 9
Dividing the entire equation by 9, we find:
- b – a = 1
Now, we have two equations:
- 1. a + b = 5
- 2. b – a = 1
We can solve these equations simultaneously. From the second equation, we can express b in terms of a:
- b = a + 1
Now, substituting this expression for b into the first equation gives:
- a + (a + 1) = 5
This simplifies to:
- 2a + 1 = 5
Subtracting 1 from both sides results in:
- 2a = 4
Dividing both sides by 2, we find:
- a = 2
Now that we have a, we can find b using the equation b = a + 1:
- b = 2 + 1 = 3
Thus, the digits are a = 2 and b = 3, making the original two-digit number:
- 10a + b = 10(2) + 3 = 23
We can verify this solution:
- The sum of the digits is 2 + 3 = 5, which is correct.
- Reversing the digits gives 32, and 32 – 23 = 9, confirming it is also correct.
Therefore, the two-digit number is 23.