Strategy: I created this question to show one of the approaches you have at your disposal when you don't know how to solve certain question.
The strategy is to examine easier versions of the question until you spot a pattern.

For example \(2^2 = 4\), and there are \(2\) (aka \(2^1\)) positive even integers less than or equal to \(4\). They are 2 and 4
Similarly, \(2^3 = 8\), and there are \(4\) (aka \(2^2\)) positive even integers less than or equal to \(8\). They are 2, 4, 6, and 8
Likewise, \(2^4 = 16\), and there are \(8\) (aka \(2^3\)) positive even integers less than or equal to \(16\). They are 2, 4, 6, 8, 10, 12, 14, and 16
And \(2^5 = 32\), and there are \(16\) (aka \(2^4\)) positive even integers less than or equal to \(32\).

At this point, we might recognize the pattern, which tells us that \(2^n\) will have \(2^{n-1}\) positive even integers less than or equal to \(2^n\).

So, if \(J =\) the number of positive even integers less than or equal to \(2^{10}\), then \( J = 2^{10-1} = 2^{9} \)
Similarly, if \(K =\) the number of positive even integers less than or equal to \(2^{20}\), then \( K = 2^{20-1} = 2^{19} \)

So, \(\frac{K}{J} = \frac{2^{19}}{2^{9}} = 2^{19-9} = 2^{10}\)

Answer: E