GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If the greatest integer k for which 3^k is a factor of n! is 8,
[#permalink]
23 May 2021, 23:45
23 May 2021, 23:45
1
1
Expert Reply
4
Bookmarks
00:00
Question Stats:
57% (01:59) correct
42% (01:38) wrong based on 19 sessions
Hide Show 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
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If the greatest integer k for which 3^k is a factor of n! is 8,
[#permalink]
24 May 2021, 01:38
24 May 2021, 01:38
3
GeminiHeat 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
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
arjunbir Thank you for pointing out my mistake, here's the edited version.
If n! is divisible by \(3^8\), then n! can have a maximum of eight 3s
n! could be between 18! and 20!
Why between 18 and 20?
Because, 18! will have eight 3s and so does 19! and 20!
[NOTE: It cannot be 21!, as we will have nine 3s then!]
i.e. (1)(2)(3)(4)(5)(6)(7)(8)(9) .... (18)(19)(20)
one 3 in 3, 6, 12, and 15
two 3s in 9, and 18
Now, if 20! is divisible by \(5^P\), p could only be 4 [5, 10, 15, 20]
Hence, option C
Re: If the greatest integer k for which 3^k is a factor of n! is 8,
[#permalink]
20 Aug 2021, 17:43
20 Aug 2021, 17:43
n!/(3^k) has to be an integer, and k is 8. This means, n! must have at least 8 3's. Since 8 3's are relatively easier to get through brute force: 21! has 8 3's.
21! also has 4 5's : 21/5 = 4 steps of 5, and there are no squares of 5 between 1 and 21, which leaves us with a final answer of 4 (C).
21! also has 4 5's : 21/5 = 4 steps of 5, and there are no squares of 5 between 1 and 21, which leaves us with a final answer of 4 (C).
