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

Question

If  J  equals the number of positive even integers less than or equal to  2^{10} , and  K  equals the number of positive even integers less than or equal to  2^{20} , then  \frac{K}{J}=

(A)  \frac{21}{11}

(B)  2

(C)  20

(D)  2^{9} 
 
(E)  2^{10}

jim441 · Answer

IMO E is the correct answer.

My approach:-

Generalize the formula

-- so for 2^1 = 2 has only 2 as positive even integer ie 2^0=1 number of positive even integers.
-- so for 2^2 = 4 has only 2,4 as positive even integer ie 2^1=2 number of positive even integers.
-- so for 2^3 = 8 has only 2,4,6,8 as positive even integer ie 2^2=4 number of positive even integers.
-- so for 2^4 = 16 has only 2,4,6,8,10,12,14,16 as positive even integer ie 2^3=8 number of positive even integers.
so we can say that 2^n will have 2^(n-1) number of positive even integers

that means,
K=2^20 will have 2^19 number of positive even integers
& J=2^10 will have 2^9 number of positive even integers

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

BrentGMATPrepNow · Answer

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

hasanijtaba · Answer

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???

BrentGMATPrepNow · Answer

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

monkinaferrari · Answer

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

Fdambro294 · Answer

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*

Posted from my mobile device

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

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