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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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Updated on: 18 Dec 2016, 22:51
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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
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Originally posted by Rocky1304 on 18 Dec 2016, 09:19.
Last edited by Bunuel on 18 Dec 2016, 22:51, 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
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03 Jan 2017, 03:46
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
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18 Dec 2016, 09:53
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
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02 Jan 2017, 17:57
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
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02 Jan 2017, 22:55
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/mathnumbertheory88376.htmlThanks.



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Re: If the greatest integer k for which 3^k is a factor of n! is 8, what
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26 Mar 2017, 06:13
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
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27 Mar 2017, 03:34
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
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06 May 2017, 15:53
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
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27 Jun 2017, 18:43
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
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27 Jun 2017, 20:16
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
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16 Jul 2018, 11:01
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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