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Exponents of prime p in n!

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Exponents of prime p in n!

by kartikshah » Fri Jul 27, 2012 5:10 am
Question 01

The exponent of 7 in 100C50 is:
(a) 0
(b) 2
(c) 4
(d) 6
(e) None of these

Source: www.gmatscore.com
OA is A

My question -- if the question had been modified and exponents of 2 had to be found in 100C50 then would the answer be 3?
I worked it out as under:
2s in 100! = 97
2s in 50! = 47

So 97 - (2)(47) = 97-94 = 3

Question 02

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
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by eagleeye » Fri Jul 27, 2012 8:32 am
kartikshah wrote:Question 01

My question -- if the question had been modified and exponents of 2 had to be found in 100C50 then would the answer be 3?
I worked it out as under:
2s in 100! = 97
2s in 50! = 47

So 97 - (2)(47) = 97-94 = 3
Yes. the answer is 3. You have done it correctly. Good Job!

Question 02
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
.

OA as A is incorrect. The correct answer is infinite number.

Here's my explanation:

Just start from the smallest option and check.
for n=1,2,3,4,5
n! 2*n
1! < 2*1
2! < 2*2
3! = 3*2 (equal, 6 divides 6
4! > 4*2 (4 and 2 are factors of 4!, 8 divides 24)
5! > 5*2 (5 and 2 are factors of 5!, 10 divides 120)
etc.

We can see that at first n! is less than 2n, then for n=3, it becomes equal to 2n. And for higher values, it outstrips 2n by larger and larger margins, and 2n will always divide n! . We need not check any further. And we see that infinite values of n satisfy.

The question would have been better written if it said "The number of positive integers n such that n! divides 2n" in which case answer would be only one value, making B the correct choice.

Let me know if this helps :)
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by GMATGuruNY » Fri Jul 27, 2012 10:58 am
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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by kartikshah » Fri Jul 27, 2012 4:51 pm
True!
It would make more sense if the question said 2^n.
Thanks for clarifying!
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