To solve this problem, we need to understand the expressions for \( p \) and \( q \) and how they relate to the given condition \( p + q = 66 \). The expressions for \( p \) and \( q \) are:

• \( p = \sum_{j=1}^n \log_{10} 2^j \)
• \( q = \sum_{j=1}^n \log_{10} 5^j \)

Let's break down these expressions:

Using logarithm properties, simplify each sum separately. We know:

• \(\sum_{j=1}^n \log_{10} 2^j = \log_{10} \left( 2^{1+2+3+\ldots+n} \right)\)
• \(1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}\)
• Thus, \(\sum_{j=1}^n \log_{10} 2^j = \log_{10} \left( 2^{\frac{n(n+1)}{2}} \right) = \frac{n(n+1)}{2} \log_{10} 2\)
• Similarly, \(\sum_{j=1}^n \log_{10} 5^j = \log_{10} \left( 5^{\frac{n(n+1)}{2}} \right) = \frac{n(n+1)}{2} \log_{10} 5\)

Accordingly, the sum \( p + q \) becomes:

• \(p + q = \frac{n(n+1)}{2} (\log_{10} 2 + \log_{10} 5)\)
• Since \(\log_{10} 10 = 1\), we have \(\log_{10} 2 + \log_{10} 5 = \log_{10} (2 \times 5) = \log_{10} 10 = 1\)
• Thus, \(p + q = \frac{n(n+1)}{2}\)

We know from the problem statement that \(p + q = 66\). Therefore:

• \(\frac{n(n+1)}{2} = 66\)
• Solving for \(n(n+1)\), we get:
• \(n(n+1) = 132\)
• The solutions to this quadratic equation \(n^2 + n - 132 = 0\) can be found by factorization or using the quadratic formula:
• Using trial and error or factoring directly, we find:
• \((n-11)(n+12) = 0\)

Therefore, the positive solution is \(n = 11\). This result satisfies the given choices:

• The correct range for \(n\) is \(9 < n < 12\).

Hence, the correct answer is \(9 < n < 12\).