If J equals the number of positive even integers less than or equal to

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

k=2^20/2
j=2^10/2

since every half int will be even and other half will be odd .
so,

K/j = 2^20/2 *2/2^10 = 2^10

am i right???

Your reasoning is solid.
How did you calculate 2^20/2 *2/2^10 to get 2^10?

Let's say we have 10 integers positive integers
So, there are 5 odd and 5 even.

Extending the same to the given question

2^20/2 = Even =K

2^10/2 = Even = J


=2^(20-10) = 2^10 = E

positive integers start with 1 (an odd number)

Any power of 2 is of course even

From 1 thru (2)^n …. (Inclusive)

Where n = some positive integer, there will be a symmetric number of Odd and Even Numbers.

In other words, (1/2) of the consecutive integers in the range will be odd
1 —to—- (2)^n - (1)

And the other (1/2) of the consecutive integers in the range will be even
2 ——to— (2)^n

K: all the consecutive even integers from 1 to (2)^20 , inclusive, will be simply (1/2) the positive integers in the range

Or K = (1/2) * (2)^20 = (2)^19

J: same reasoning, we get
J = (2)^9

K/J = (2)^(19 - 9) = (2)^10

*E*

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