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n is a product of 6 distinct prime numbers. m!/n is an integer.

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Math Revolution GMAT Instructor
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n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 11 Jan 2019, 04:49
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[Math Revolution GMAT math practice question]

\(n\) is a product of \(6\) distinct prime numbers. \(\frac{m!}{n}\) is an integer.
What is the smallest possible value of \(m\)?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)

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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 11 Jan 2019, 05:26
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(n\) is a product of \(6\) distinct prime numbers. \(\frac{m!}{n}\) is an integer.
What is the smallest possible value of \(m\)?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)



n = 2* 3*5*7*11*13
so m!/n can be an integer only when m = 13 !
IMO D
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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 11 Jan 2019, 05:26
n is a product of 6 distinct prime numbers. m!/n is an integer.
What is the smallest possible value of m?

Smallest 6 prime numbers = 2,3,5,7,11,,13
So m! should include all of them
implies smallest possible value of m = 13

Option D is correct
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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 11 Jan 2019, 11:30
m!/n is an integer.
For the value of n to be minimum, m!/n should be minimum.
But since n is a product of 6 distinct prime no.'s, the minimum value of n=2*3*5*7*11*13
So, the minimum value of m such that m!/n is an integer is 13
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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 13 Jan 2019, 18:04
=>

The smallest product of \(6\) distinct prime numbers is \(n = 2*3*5*7*11*13.\) \(13!\) Is the smallest factorial that is divisible by \(2*3*5*7*11*13\). Thus, the smallest possible integer value of \(m\) is \(13\).

Therefore, the answer is D.
Answer: D
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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.  [#permalink]

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New post 14 Jan 2019, 19:03
MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(n\) is a product of \(6\) distinct prime numbers. \(\frac{m!}{n}\) is an integer.
What is the smallest possible value of \(m\)?

\(A. 10\)
\(B. 11\)
\(C. 12\)
\(D. 13\)
\(E. 14\)


The smallest product of 6 distinct prime numbers is:

2 x 3 x 5 x 7 x 11 x 13, so we see that m! must have a prime of 13, so the least value of m is 13.

Answer: D
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Re: n is a product of 6 distinct prime numbers. m!/n is an integer.   [#permalink] 14 Jan 2019, 19:03
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