The square root of a perfect square of n digits will have `n/2` digits if n is even.

Answer

To solve the problem, we need to understand how the number of digits in a perfect square relates to the number of digits in its square root, particularly when the number of digits (n) is even.

### Step-by-Step Solution:

1. **Understanding Perfect Squares**:
A perfect square is a number that is the square of an integer. For example, \(1^2 = 1\), \(2^2 = 4\), \(3^2 = 9\), \(4^2 = 16\), and so on.

2. **Digits in a Number**:
The number of digits in a number can be determined using the formula:
\[ \text{Number of digits} = \lfloor \log_{10}(x) \rfloor + 1 \]
where \(x\) is the number.

3. **Perfect Square of n Digits**:
If a number has \(n\) digits, the smallest perfect square with \(n\) digits is \(10^{(n-1)}\) and the largest is \(10^n - 1\).

4. **Finding the Range of Square Roots**:
To find the square roots of these perfect squares:
- The square root of the smallest perfect square \(10^{(n-1)}\) is \(10^{(n-1)/2}\).
- The square root of the largest perfect square \(10^n - 1\) is approximately \(10^{n/2}\).

5. **Number of Digits in the Square Root**:
- For the smallest perfect square, \(10^{(n-1)/2}\) will have \(\lfloor (n-1)/2 \rfloor + 1\) digits.
- For the largest perfect square, \(10^{n/2}\) will have \(n/2\) digits.

6. **Conclusion for Even n**:
If \(n\) is even, both calculations show that the square root of a perfect square with \(n\) digits will have exactly \(n/2\) digits.

### Final Statement:
Thus, we conclude that the square root of a perfect square with \(n\) digits will indeed have \(n/2\) digits when \(n\) is even.