January 15, 2019 January 15, 2019 10:00 PM PST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 09 Jan 2017
Posts: 5
Location: India
WE: Accounting (Accounting)

Largest possible value of exponent of prime number
[#permalink]
Show Tags
04 Aug 2017, 23:59
Question Stats:
54% (02:05) correct 46% (02:25) wrong based on 117 sessions
HideShow timer Statistics
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 The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too! Please give a kudos if you like the question!
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 07 Jun 2017
Posts: 102

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 00:07
waiting for the solution



Current Student
Joined: 06 Sep 2013
Posts: 33
Location: India
Concentration: Leadership, Strategy
GPA: 3.7
WE: Information Technology (Consulting)

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 01:24
DeathNoteFreak 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 The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too! Please give a kudos if you like the question! Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!. Since it is given that k is 8, this means that the following numbers are present in the expansion of n!: 3,6,9,12,15,18 Each of the above number gives the following number of 3's 1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2). In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7). Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20. Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5. Therefore, the maximum value of p would be 1+1+1+1=4 Hope this helps.



Intern
Joined: 09 Jan 2017
Posts: 5
Location: India
WE: Accounting (Accounting)

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 06:11
nitesh181989 wrote: DeathNoteFreak 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 The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too! Please give a kudos if you like the question! Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!. Since it is given that k is 8, this means that the following numbers are present in the expansion of n!: 3,6,9,12,15,18 Each of the above number gives the following number of 3's 1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2). In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7). Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20. Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5. Therefore, the maximum value of p would be 1+1+1+1=4 Hope this helps. If you still consider n=18, number of 5's would then be 3. What would you suggest to understand why to choose 20, instead of 18?? Thanks in advance!



Intern
Joined: 17 Feb 2016
Posts: 12

Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 09:05
DeathNoteFreak wrote: nitesh181989 wrote: DeathNoteFreak 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 The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too! Please give a kudos if you like the question! Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!. Since it is given that k is 8, this means that the following numbers are present in the expansion of n!: 3,6,9,12,15,18 Each of the above number gives the following number of 3's 1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2). In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7). Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20. Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5. Therefore, the maximum value of p would be 1+1+1+1=4 Hope this helps. If you still consider n=18, number of 5's would then be 3. What would you suggest to understand why to choose 20, instead of 18?? Thanks in advance! it says 'largest', thats why we'll have to take the maximum value of n for which 3 comes 8 times and in this case it happened to be 20 (just before 21) which also has one extra 5 (and question asks for maximum number of occurrence of 5)



Manager
Joined: 02 Nov 2015
Posts: 165

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 10:31
Stem says that 3^8 is the highest power that can divide n!. Basically we have to find the highest power of 5 less than or equal to 3^8. So the highest power of 5 will be 4. Because if it becomes 5 then 5^5 is quite later than 3^8. I don't know how much I could express myself but it's the gist of the problem. Answer is 4. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



Director
Joined: 04 Dec 2015
Posts: 738
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)

Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Aug 2017, 20:59
DeathNoteFreak 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 The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too! Please give a kudos if you like the question! I solved by following way. Hope it helps. \(3!\) has one \(3 = 3^1\) \(6!\) has two \(3 = 3^2\) \(9!\) has four \(3 = 3^4\) \(12!\) has five \(3 = 3^5\) \(15!\) has six \(3 = 3^6\) \(18!\) has eight \(3 = 3^8\) Also \(20!\) has eight \(3 = 3^8\) \(21!\) has nine \(3 = 3^9\) Given greatest integer \(k\) for which \(3^{k}\) is a factor of \(n!\) is \(8\). Therefore \(n!\) cannot be \(21!\). \(n!\) will be \(20!\). Therefor \(n!\) with largest possible value of \(5^p\) would be \(20!\). Lets check value of \(p\) in the above factorial to find the largest possible value of \(5^p\). \(20!\) has four \(5 = 5^4\) Therefore largest possible value of \(p\) so that \(5^{p}\) is a factor of \(n! = 4\) Answer (C)..._________________ Please Press "+1 Kudos" to appreciate.



Intern
Joined: 12 Oct 2017
Posts: 40

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
05 Nov 2017, 00:41
I solved this question in the following way: Step 1: find the largest possible value of n! by calculating how many "3" are presented in n! Because greatest integer k for which 3^k is a factor of n! is 8. => 3,6,9,12,15,18 are presented in n! Since we want to find the largest possible value of n! => we can increase the value of n as long as we dont add any more "3" =>n! = 20! Step 2: find the largest p by calculating how may "5" are presented in 20! We have 5,10,15,20 are presented in 20! and give us four "5" => p = 4 Hence, the answer is C.  Kindly press +1kudos if the explanation is clear! Thank you!



Manager
Joined: 24 Jun 2017
Posts: 122

Re: Largest possible value of exponent of prime number
[#permalink]
Show Tags
06 Nov 2017, 14:36
kumarparitosh123 wrote: Stem says that 3^8 is the highest power that can divide n!. Basically we have to find the highest power of 5 less than or equal to 3^8. So the highest power of 5 will be 4. Because if it becomes 5 then 5^5 is quite later than 3^8. I don't know how much I could express myself but it's the gist of the problem. Answer is 4. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app3^8 = 6561 5^4 = 625 and 5^5 is 3125 Thus why did you choose 4? can you please explain?




Re: Largest possible value of exponent of prime number &nbs
[#permalink]
06 Nov 2017, 14:36






