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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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23 Dec 2018, 19:25
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35% (medium)

Question Stats:

62% (01:03) correct 38% (01:06) wrong based on 74 sessions

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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7495 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: What is the smallest integer n such that 2^8 is a factor of n!? [#permalink] ### Show Tags 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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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

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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08 Jan 2019, 04:08
is there a shortcut?
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Joined: 26 Apr 2018
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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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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.

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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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.

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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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