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What is the smallest integer n such that 2^8 is a factor of n!?

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Math Revolution GMAT Instructor
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What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 23 Dec 2018, 19:25
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[Math Revolution GMAT math practice question]

What is the smallest integer \(n\) such that \(2^8\) is a factor of \(n!\)?

\(A. 8\)
\(B. 10\)
\(C. 12\)
\(D. 14\)
\(E. 16\)

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Re: What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 23 Dec 2018, 19:27
=>
Since \(2 = 2^1, 4 = 2^2, 6 = 2^1*3, 8 = 2^3, 10=2^1*5\), the prime factorization of \(10! = 1*2*3*…*10\) has the form \(2^8*m\) for some integer \(m\), where \(m\) and \(2\) are relatively prime. Note that \(9! = 2^7*k\) for some integer \(k\), where \(k\) and \(2\) are relatively prime.

Therefore, the answer is B.
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Re: What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 23 Dec 2018, 19:33
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the smallest integer \(n\) such that \(2^8\) is a factor of \(n!\)?

\(A. 8\)
\(B. 10\)
\(C. 12\)
\(D. 14\)
\(E. 16\)



2^8= 16*16

so using given answer choices
10!/16*16 = integer value .. IMO B
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Re: What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 08 Jan 2019, 04:08
is there a shortcut?
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Re: What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 10 Jan 2019, 03:11
MathRevolution wrote:
=>
Since \(2 = 2^1, 4 = 2^2, 6 = 2^1*3, 8 = 2^3, 10=2^1*5\), the prime factorization of \(10! = 1*2*3*…*10\) has the form \(2^8*m\) for some integer \(m\), where \(m\) and \(2\) are relatively prime. Note that \(9! = 2^7*k\) for some integer \(k\), where \(k\) and \(2\) are relatively prime.

Therefore, the answer is B.


Could you explain in more detail?
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Re: What is the smallest integer n such that 2^8 is a factor of n!?  [#permalink]

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New post 15 Feb 2019, 18:36
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the smallest integer \(n\) such that \(2^8\) is a factor of \(n!\)?

\(A. 8\)
\(B. 10\)
\(C. 12\)
\(D. 14\)
\(E. 16\)


We see that 10! has 8 factors of 2. Let’s write out 10! as 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. Now let’s factor out each factor of 2 from 10!

2 → 1 factor of 2

4 = 2^2 → 2 factors of 2

6 = 2 x 3 → 1 factor of 2

8 = 2^3 → 3 factors of 2

10 = 2 x 5 → 1 factor of 2

However, 8! (or 9!) only has 7 factors of 2, so the smallest value of n is 10.

Answer: B
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Re: What is the smallest integer n such that 2^8 is a factor of n!?   [#permalink] 15 Feb 2019, 18:36
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