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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})


√log p + √log q + log √p + log √q = 67. √log p, √log q, log √p and log √q are all positive integers. What is the maximum value of log10 pq?

Solve for the maximum value of log10(pq) given a logarithmic equation involving √log p and √log q. This problem explores key logarithmic identities and algebraic techniques. Suitable for students in Grades 9-12.

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