To determine the quantity of positive integer pairs \((a, b)\) where \(a \leq b\) and \(ab = 4^{2017}\), we first analyze the prime factorization of \(4^{2017}\):
\[4^{2017} = (2^2)^{2017} = 2^{4034}\]
For \((a, b)\) to be positive integer pairs such that \(ab = 2^{4034}\), let \(a = 2^x\) and \(b = 2^y\), with the condition \(x \leq y\). The product becomes:
\[a \times b = 2^x \times 2^y = 2^{4034}\]
This implies:
\[x + y = 4034\]
Given the constraint \(a \leq b\), we have:
\[2^x \leq 2^y \implies x \leq y\]
Substituting \(x + y = 4034\) into \(x \leq y\) leads to:
\[x \leq \frac{4034}{2} = 2017\]
Consequently, the possible integer values for \(x\) range from 0 to 2017. For each selected value of \(x\), \(y\) is uniquely determined by \(y = 4034 - x\). This formulation ensures that \(y \geq x\) is satisfied due to the upper bound on \(x\). Therefore:
The count of valid values for \(x\) is:
\[2017 - 0 + 1 = 2018\]
Thus, there are 2018 pairs \((a, b)\) satisfying \(a \leq b\) and \(ab = 4^{2017}\).
Correct Answer: 2018