The problem appears to ask for the maximum value of $\log_{10} pq$ given the following equation and conditions:

$\sqrt{\log p} + \sqrt{\log q} + \log \sqrt{p} + \log \sqrt{q} = 67$ and that $\sqrt{\log p}, \sqrt{\log q}, \log \sqrt{p}, \log \sqrt{q}$ are all positive integers.

Step 1: Simplify the equation

We can begin by simplifying the given equation. First, recall that: $\log \sqrt{p} = \frac{1}{2} \log p \quad \text{and} \quad \log \sqrt{q} = \frac{1}{2} \log q$ Thus, the equation becomes: $\sqrt{\log p} + \sqrt{\log q} + \frac{1}{2} \log p + \frac{1}{2} \log q = 67$

Step 2: Assign variables to simplify further

Let $x = \sqrt{\log p}$ and $y = \sqrt{\log q}$. This implies: $\log p = x^2 \quad \text{and} \quad \log q = y^2$ Substituting these into the equation, we get: $x + y + \frac{1}{2} x^2 + \frac{1}{2} y^2 = 67$ Multiply the entire equation by 2 to eliminate the fractions: $2x + 2y + x^2 + y^2 = 134$ Rearrange this as: $x^2 + 2x + y^2 + 2y = 134$ This can be factored as: $(x+1)^2 + (y+1)^2 = 136$

Step 3: Solve for integer solutions

Now, we need to find integer values of $x$ and $y$ such that the sum of squares equals 136. Let's check pairs of integers whose squares sum to 136:

$136 - 1^2 = 135 \quad (\text{no solution for the remaining term})$ $136 - 2^2 = 132 \quad (\text{no solution for the remaining term})$ [ 136 - 7^2 = 87 \quad (\text{no solution for the remaining term})