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What is the smallest positive integer n for which n!/9^9 is an integer

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What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 16 Feb 2017, 02:35
1
3
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

48% (01:53) correct 52% (01:45) wrong based on 177 sessions

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Re: What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 16 Feb 2017, 03:08
9^9 = 3^18
so we must find which of the answers contain 18 multiples of 3.
36 has 12 multiples of 3 and we must not forger factors from 9,27 and 36.in total 36 has 17 factors of 3
thus 39 will have 18 factors of 3

answer is B
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What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post Updated on: 16 Feb 2017, 04:06
2
Bunuel wrote:
What is the smallest positive integer n for which n!/9^9 is an integer?

A. 36
B. 39
C. 54
D. 78
E. 81


\(9^9 = 3^{18}\)

lets consider the smallest number from the given options i.e. 36
number of multiples of 3 in 36! = 36/3 + 36/9 + 36/27 = 12 + 4 + 1 =17
so 39! will have 18 multiples of 3 and \(36!/3^{18}\) will surely be an Integer.

Hence Option B is correct.
Hit Kudos if you liked it 8-)

Originally posted by 0akshay0 on 16 Feb 2017, 03:22.
Last edited by 0akshay0 on 16 Feb 2017, 04:06, edited 1 time in total.
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Re: What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 16 Feb 2017, 03:28
1
Ans is B
3^18
Number of 3 in 39!
39/3=13
12/3=4
4/3=1
Hence
13+4+1= 18


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Re: What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 19 Feb 2017, 10:51
reena.phogat wrote:
Ans is B
3^18
Number of 3 in 39!
39/3=13
12/3=4
4/3=1
Hence
13+4+1= 18


Sent from my XT1068 using GMAT Club Forum mobile app


Number of 3 in 39!
39/3=13
12/3=4
4/3=1
Hence
13+4+1= 18


can you please elaborate how you are calculating this
Thanks
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What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 21 Feb 2017, 05:54
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1
Bunuel wrote:
What is the smallest positive integer n for which n!/9^9 is an integer?

A. 36
B. 39
C. 54
D. 78
E. 81


Official solution from Veritas Prep.

This is a number theory question – note the term “positive integer” and the fact that the question is essentially asking about the divisibility of \(n!\). As such, we will begin by prime factoring \(9^9\) as \((3^2)^{9}=3^{18}\). The real question is how far we have to go to find eighteen copies of 3.

The cop-out answer would be to simply take \(n=18∗3=54\); \(54!\) definitely contains at least eighteen factors of 3, since it contains \(3∗1, 3∗2, 3∗3, …, 3∗18\). However, \(54!\) actually contains far more than just the eighteen 3s, since several factors contain “bonus” 3s. For instance, \(54!\) contains 9, which is \(3^2\), and 27, which is \(3^3\), and even 54 itself, which is \(2∗3^3\). The correct answer must be less than 54.

At this point we could calculate the exact number of 3s in \(54!\) (it has twenty-six of them, as it turns out) and adjust from there, but it might be simpler to just try answers A and B, since one or the other must be correct.

\(36!\) contains twelve multiples of 3 (\(3∗1\) through \(3∗12\)), but it also contains four multiples of 9 (\(9∗1\) through \(9∗4\)), each of which provides an additional factor of 3, since \(9=3^2\). And \(36!\) Even contains one multiple of \(3^3=27\), which provides an “additional” additional 3. Taken together, \(36!\) contains \(12+4+1=17\) factors of 3, just short of our goal.

We can see at this point that \(39!\) provides exactly one more 3 than does \(36!\), since \(39!=39∗38∗37∗36!\).

Therefore, \(39!\) contains exactly 18 factors of 3, and answer B is correct.
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Re: What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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New post 21 Feb 2017, 09:27
3
1
Bunuel wrote:
What is the smallest positive integer n for which n!/9^9 is an integer?

A. 36
B. 39
C. 54
D. 78
E. 81


We need to determine the largest value of n such that n!/(9^9) is an integer. Let’s start by simplifying 9^9.

9^9 = (3^2)^9 = 3^18.

Thus, we need to find the smallest positive integer n such that n! contains at least 18 factors of 3 in its prime factorization.

Let’s start with answer choice A: 36.

We can determine the number of 3s in 36! by using the following shortcut in which we divide 36 by 3 then divide the quotient of 36/3 by 3, and continue this process until we no longer get a nonzero quotient.

36/3 = 12

12/3 = 4

4/3 = 1 (we can ignore the remainder)

Since 1/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 36!.

Thus, there are 12 + 4 + 1 = 17 factors of 3 within 36!.

Looking at answer choice B, 39, we see that 39! factorial will have one more factor of 3 than 36! (because 39 has a prime factor of 3), and thus, 39! will have 18 factors of 3.

Thus, 39 is the minimum value of n.

Answer: B
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Re: What is the smallest positive integer n for which n!/9^9 is an integer  [#permalink]

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