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If the greatest integer k for which 3^k is a factor of n! is 8, what

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If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post Updated on: 18 Dec 2016, 21:51
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If the greatest integer k for which 3^k is a factor of n! is 8, what is the largest possible value of p so that 5^p is a factor of n! ?

A) 2
B) 3
C) 4
D) 5
E) 6

Originally posted by Rocky1304 on 18 Dec 2016, 08:19.
Last edited by Bunuel on 18 Dec 2016, 21:51, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 03 Jan 2017, 02:46
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2
Rocky1304 wrote:
If the greatest integer k for which 3^k is a factor of n! is 8, what is the largest possible value of p so that 5^p is a factor of n! ?

A) 2
B) 3
C) 4
D) 5
E) 6


We know that every multiple of 3 will give us a 3 in n!
We also know that every third multiple of 3 will give us another 3.

Let's assume a value of n and see how many 3s it has. Say, n = 15

Number of 3s in 15! = 15/3 + 5/3 = 6

But we need 8 3s. So we must jump to the next multiple of 3 i.e. 18. It gives us 2 3s so we get a total of 8 3s. The next multiple of 3 is 21 which will give another 3 but we have only 8 3s. So the maximum value of n can be 20.

The largest value of p so that 5^p is a factor of 20! is 20/5 = 4

Answer (C)

For more on this concept, check:
https://www.veritasprep.com/blog/2011/0 ... actorials/
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 18 Dec 2016, 08:53
2
Rocky1304 wrote:
If the greatest integer k for which \(3^k\)is a factor of n! is 8, what is the largest possible value of p
so that \(5^p\) is a factor of n! ?


A) 2
B) 3
C) 4
D) 5
E) 6


\(\frac{n}{3} + \frac{n}{3^2} + \frac{n}{3^3} … = 8\)

\(n! = 18!, 19!, 20!\)

\([\frac{18}{3}] + [\frac{18}{3^2}] = [\frac{19}{3}] + [\frac{19}{3^2}] = [\frac{20}{3}] + [\frac{20}{3^2}] = 8\)

From \(21!\) we have additional factor of \(3\) hence \(3^9\).

\(18!\) has \(5^3\)

\(19!\) Still \(5^3\)

\(20!\) One more factor of 5 – \(5^4\)

Max power of \(5\) in \(n!\) is \(4\)

Answer C
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 02 Jan 2017, 16:57
Hello everybody,

Is someone out there who can explain me the first step that was made? Why can you divide n by 3 and 3^2 to figure out what the maximum value of n might be? In general, I really don't understand the applied approach.

Kind regards

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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 02 Jan 2017, 21:55
Bonachhilfe wrote:
Hello everybody,

Is someone out there who can explain me the first step that was made? Why can you divide n by 3 and 3^2 to figure out what the maximum value of n might be? In general, I really don't understand the applied approach.

Kind regards

Posted from my mobile device

Hi Bonachhilfe,

Please refer the Factorial section on the following page:
http://gmatclub.com/forum/math-number-theory-88376.html

Thanks.
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 26 Mar 2017, 05:13
I understand that n can be 18,19,20. However,the question never suggested that we have to consider the largest possible value of n. It says the largest possible value of k. So I substituted the value of n = 18 and not 20 while solving.
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 27 Mar 2017, 02:34
1
manishcmu wrote:
I understand that n can be 18,19,20. However,the question never suggested that we have to consider the largest possible value of n. It says the largest possible value of k. So I substituted the value of n = 18 and not 20 while solving.


This is what the question says: what is the largest possible value of p so that 5^p is a factor of n! ?

We need the largest possible value of p. We will get the largest possible value of p when n takes the largest possible value it can take. n can be 18, 19 or 20.
If n is 18, p is 3
If n is 19, p is 3.
If n is 20, then p is 4.
So 4 is the largest possible value of p.
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 06 May 2017, 14:53
1
C

18!, 19! and 20! has 8 x 3's
So then 18! & 19! has 3 x 5's, while 20! has 4 x 5's
So 4 x 5's
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 27 Jun 2017, 17:43
vitaliyGMAT wrote:
Rocky1304 wrote:
If the greatest integer k for which \(3^k\)is a factor of n! is 8, what is the largest possible value of p
so that \(5^p\) is a factor of n! ?


A) 2
B) 3
C) 4
D) 5
E) 6


\(\frac{n}{3} + \frac{n}{3^2} + \frac{n}{3^3} … = 8\)

\(n! = 18!, 19!, 20!\)

\([\frac{18}{3}] + [\frac{18}{3^2}] = [\frac{19}{3}] + [\frac{19}{3^2}] = [\frac{20}{3}] + [\frac{20}{3^2}] = 8\)

From \(21!\) we have additional factor of \(3\) hence \(3^9\).

\(18!\) has \(5^3\)

\(19!\) Still \(5^3\)

\(20!\) One more factor of 5 – \(5^4\)

Max power of \(5\) in \(n!\) is \(4\)

Answer C


Hi, could you please explain how you got 18!, 19!, and 20! from your first step?

Thank You!
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 27 Jun 2017, 19:16
Imo C
We are given 3^k is a factor of n! for k=8
That means we have to find out which factorial contains that much power of 3
18! has 8 powers of 3.
But here is the catch 19! and 20! also have 8 powers of 3 because 19 and 20 are not factors of 3.
If we take 21! it contains additional 3 i.e 20!*21=20!*3*7
So we have to stop at 20!
Now we have to find out maximum power of 5 power of 5 in 18!=3
power of 5 in 19!=3
power of 5 in 20!=4
So we have 4 as the maximum power.
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what  [#permalink]

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New post 16 Jul 2018, 10:01
Rocky1304 wrote:
If the greatest integer k for which 3^k is a factor of n! is 8, what is the largest possible value of p so that 5^p is a factor of n! ?

A) 2
B) 3
C) 4
D) 5
E) 6


Given: k = 8 is the greatest integer, such that n! is divisible by 3^k.

Hence we have 3*3*3....8 times in the expansion of n!

Hence we have n/3 + n/9 = 8, we get n = 18

However maximum value of n can be n = 20, as at n = 21, we will have 3 more than 8 times in n!

Now we need max power of 5 in 20!

Hence, 20/5 = 4


Answer C.



Thanks,
GyM
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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what &nbs [#permalink] 16 Jul 2018, 10:01
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