kartikshah wrote:
The number of positive integers n such that
2n divides n! is
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
OA is
A BUT I did not understand the explantion provided by
www.gmatscore.com
I suspect that the portion in red should read as follows:
The number of positive integers n such that 2^n divides n! is
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
Let n=100.
Then n! = 100!
For the purposes of this problem, 2^n = the number of 2's that can be divided into 100!.
Count how many times EACH POWER OF 2 can be divided into 100!:
100/2¹ = 50.
The calculation above indicates that 100! includes 50 multiples of 2¹.
100/2² = 100/4 = 25.
The calculation above indicates that 100! includes 25 multiples of 2².
100/2³ = 100/8 = 12.
The calculation above indicates that 100! includes 12 multiples of 2³.
100/2� = 100/16 = 6.
The calculation above indicates that 100! includes 6 multiples of 2�.
100/2� = 100/32 = 3.
The calculation above indicates that 100! includes 3 multiples of 2�.
100/2� = 100/64 = 1.
The calculation above indicates that 100! includes 1 multiple of 2�.
Thus, 100! includes:
50 multiples of 2
25 multiples of 2²
12 multiples of 2³
6 multiples of 2�
3 multiples of 2�
1 multiple of 2�
Thus, the total number of 2's that can be divided into 100! = 50+25+12+6+3+1 = 97.
Since 2^100 is composed of 100 2's, 2^100 does not divide 100!.
The example above illustrates that there will always be more factors of 2 in 2^n than there are in n!.
Thus, there are no integers n such that 2^n divides n!.
The correct answer is
A.
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