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If 13!/2^x is an integer, which of the following represe
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23 May 2017, 08:54
23 May 2017, 08:54
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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!
