Let's analyze the problem step-by-step.
Problem Restatement
You have a two-digit number. When you interchange the digits of this number, the difference between the original number and the new number is 3.
Step 1: Define the digits
Let the two-digit number be 10x+y10x+y10x+y, where:
- xxx is the tens digit (1 through 9),
- yyy is the units digit (0 through 9).
The number obtained by interchanging the digits is 10y+x10y+x10y+x.
Step 2: Write the equation for the difference
The difference between the original number and the interchanged number is 3. So,
∣(10x+y)−(10y+x)∣=3|(10x+y)-(10y+x)|=3∣(10x+y)−(10y+x)∣=3
Simplify inside the absolute value:
∣10x+y−10y−x∣=∣9x−9y∣=3|10x+y-10y-x|=|9x-9y|=3∣10x+y−10y−x∣=∣9x−9y∣=3
∣9(x−y)∣=3|9(x-y)|=3∣9(x−y)∣=3
Step 3: Solve for x−yx-yx−y
9∣x−y∣=3 ⟹ ∣x−y∣=39=139|x-y|=3\implies |x-y|=\frac{3}{9}=\frac{1}{3}9∣x−y∣=3⟹∣x−y∣=93=31
Step 4: Analyze the result
Since xxx and yyy are digits (integers), the difference x−yx-yx−y must be an integer. But 13\frac{1}{3}31 is not an integer.
Conclusion
There is no two-digit number whose digits, when interchanged, differ from the original number by exactly 3.
Additional note
If the problem intended the difference to be a multiple of 9, for example 27 or 18, then the difference would be divisible by 9, which fits the form 9∣x−y∣9|x-y|9∣x−y∣. But for difference = 3, no such two-digit number exists. If you want, I can help you check for differences that are multiples of 9 or clarify the problem further!
