A two digit number ab is added to another number ba, which is obtained by reversing the digits then we get a three digit number. Thus a + b equals to :

Options

A. at least 18
B. 2ab
C. `2(a + b)`
D. `(a + b) ge 10`

Answer

To solve the problem step by step, we will analyze the two-digit number \( ab \) and its reverse \( ba \) and find the relationship between \( a \) and \( b \).

### Step 1: Define the two-digit numbers
Let the two-digit number \( ab \) be represented as:
\[ ab = 10a + b \]
where \( a \) is the tens digit and \( b \) is the units digit.

Similarly, the reversed number \( ba \) can be represented as:
\[ ba = 10b + a \]

### Step 2: Set up the equation
According to the problem, when we add these two numbers, we get a three-digit number:
\[ ab + ba = (10a + b) + (10b + a) \]
This simplifies to:
\[ = 10a + b + 10b + a = 11a + 11b = 11(a + b) \]

### Step 3: Analyze the result
Since \( ab + ba = 11(a + b) \) is a three-digit number, we know that \( 11(a + b) \) must be at least 100 (the smallest three-digit number). Therefore, we can set up the inequality:
\[ 11(a + b) \geq 100 \]

### Step 4: Solve the inequality
To find \( a + b \), we divide both sides of the inequality by 11:
\[ a + b \geq \frac{100}{11} \approx 9.09 \]
Since \( a + b \) must be an integer, we round up to the next whole number:
\[ a + b \geq 10 \]

### Conclusion
Thus, the sum of the digits \( a + b \) must be greater than or equal to 10.

### Final Answer
Therefore, the answer is:
\[ a + b \geq 10 \]

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