If n is a positive integer, then `ubrace(111...1)_("2n times")-ubrace(222....2)_("n times")` is :

Options

A. a perfect square
B. a perfect cube
C. prime number
D. none

Answer

To solve the problem `ubrace(111...1)_("2n times") - ubrace(222...2)_("n times")`, we need to first express these numbers in a more manageable form.

### Step 1: Express the numbers
The number `ubrace(111...1)_("2n times")` can be expressed as:
- This is a number consisting of `2n` digits of `1`, which can be represented as:
\[ \text{Number of } 1's = \frac{10^{2n} - 1}{9} \]

The number `ubrace(222...2)_("n times")` can be expressed as:
- This is a number consisting of `n` digits of `2`, which can be represented as:
\[ \text{Number of } 2's = 2 \times \frac{10^n - 1}{9} \]

### Step 2: Set up the equation
Now we can set up the equation:
\[ \frac{10^{2n} - 1}{9} - 2 \times \frac{10^n - 1}{9} \]

### Step 3: Simplify the expression
We can combine the two fractions:
\[ = \frac{10^{2n} - 1 - 2(10^n - 1)}{9} \]
\[ = \frac{10^{2n} - 1 - 2 \cdot 10^n + 2}{9} \]
\[ = \frac{10^{2n} - 2 \cdot 10^n + 1}{9} \]

### Step 4: Factor the numerator
Notice that the numerator can be factored:
\[ 10^{2n} - 2 \cdot 10^n + 1 = (10^n - 1)^2 \]
Thus, we have:
\[ = \frac{(10^n - 1)^2}{9} \]

### Step 5: Final expression
This means our final expression is:
\[ \frac{(10^n - 1)^2}{9} \]

### Step 6: Analyze the result
Since \((10^n - 1)^2\) is always a perfect square, and dividing a perfect square by 9 gives us another perfect square (since \(9 = 3^2\)), we conclude that:
\[ \frac{(10^n - 1)^2}{9} \text{ is a perfect square.} \]

### Conclusion
Therefore, the answer to the original question is that the expression is a perfect square.

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