GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If 13!/2^x is an integer, which of the following represe
[#permalink]
23 May 2017, 08:54
23 May 2017, 08:54
Expert Reply
6
Bookmarks
00:00
Question Stats:
63% (01:29) correct
36% (01:16) wrong based on 231 sessions
Hide Show timer Statistics
If \(\frac{13!}{2^x}\) is an integer, which of the following represents all possible values of x?
A) 0 ≤ x ≤ 10
B) 0 < x < 9
C) 0 ≤ x < 10
D) 1 ≤ x ≤ 10
E) 1 < x < 10
Re: If 13!/2^x is an integer, which of the following represe
[#permalink]
22 Sep 2017, 08:01
22 Sep 2017, 08:01
6
Carcass wrote:
If \(\frac{13!}{2^x}\) is an integer, which of the following represents all possible values of x?
A) 0 ≤ x ≤ 10
B) 0 < x < 9
C) 0 ≤ x < 10
D) 1 ≤ x ≤ 10
E) 1 < x < 10
We can write 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1
or 13! = \(13 * (3*2^2) * 11 * (2*5) * 9 * (2^3) * 7 * (2*3) * 5 * (2^2) * 3 * (2) * 1\)
So maximum power of 2 = 2^10 ( adding all powers of 2 we get 2^10, so the value of x has to be ≤ 10 to make the fraction as integer)
since \(2^0\) =1 and it is divisible by any 13! so
we can consider 0 ≤ x ≤ 10
General Discussion
Re: If 13!/2^x is an integer, which of the following represe
[#permalink]
22 Sep 2017, 05:48
22 Sep 2017, 05:48
We should start by the left constraint. 2^0 is 1 and every number is divisible by 1, thus the lower bound is 0, included. Thus, we already remain with two options A and C. Then, we have to check if 10 is included or not. To do so, it is easy to remember that 2^10 = 1024. Thus, we have to check if 13! is divisible by 1024 or 2^10. My idea is to check if there is a way to compute 1024 using the prime factors of 13!.
13! = 1*2*3*4*5*6*7*8*9*10*11*12*13 Then, let's rewrite every number is its prime factors, i.e 12! = 1*2*3*(2*2)*5*(2*3)*7*(2*2*2)*(3*3)*(2*5)*11*(2*2*3)*13. Then it's easy to notice than this long multiplication can be rewritten as 12!=1*2^10*3^5*5^2*7*11*13.
Thus if we divide 13! by 2^10 we see that 2^10 compares in the number above thus it simplifies and what remains is an integer. Thus, 10 is the upper bound of our interval and answer is A!
13! = 1*2*3*4*5*6*7*8*9*10*11*12*13 Then, let's rewrite every number is its prime factors, i.e 12! = 1*2*3*(2*2)*5*(2*3)*7*(2*2*2)*(3*3)*(2*5)*11*(2*2*3)*13. Then it's easy to notice than this long multiplication can be rewritten as 12!=1*2^10*3^5*5^2*7*11*13.
Thus if we divide 13! by 2^10 we see that 2^10 compares in the number above thus it simplifies and what remains is an integer. Thus, 10 is the upper bound of our interval and answer is A!
