Events & Promotions
|
|
GMAT 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.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Apr 2912:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
TheUltimateWinner
Joined: 23 Feb 2015
Last visit: 16 Apr 2026
Posts: 2,465
Given Kudos: 1,990
Concentration: Finance, Technology
Schools: Harvard '24 Judge '23 LBS '24 McCombs '24 Mendoza Tepper '24 Ross '24 Cox Marshall '24 Fisher Kelley '24 Jones McDonough '24 Owen '24 Terry BU Simon '24 Katz GWU Mason Babson Gabelli Foster '24 Goizueta '24 Scheller Stern '23 CBS '23 Booth '23 Haas '25 Wharton '25 Stanford '26 Johnson'23 Fuqua '23 Kellogg '23 Kenan-Flagler'23 INSEAD Aug 22 intake Olin '23 Said '23 Sloan '23 Carey Arizona "22 Yale '23 UC Davis Madison '23 Anderson '23 Carroll Carlson '23 Darden '23
Schools: Harvard '24 Judge '23 LBS '24 McCombs '24 Mendoza Tepper '24 Ross '24 Cox Marshall '24 Fisher Kelley '24 Jones McDonough '24 Owen '24 Terry BU Simon '24 Katz GWU Mason Babson Gabelli Foster '24 Goizueta '24 Scheller Stern '23 CBS '23 Booth '23 Haas '25 Wharton '25 Stanford '26 Johnson'23 Fuqua '23 Kellogg '23 Kenan-Flagler'23 INSEAD Aug 22 intake Olin '23 Said '23 Sloan '23 Carey Arizona "22 Yale '23 UC Davis Madison '23 Anderson '23 Carroll Carlson '23 Darden '23
Posts: 2,465
03 Jun 2019, 13:16
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:09) correct
69%
(02:12)
wrong
based on 71
sessions
History
Date
Time
Result
Not Attempted Yet
What is the largest number with exactly \(3\) factors that divides \(n!\) ?
1) largest prime that divides \(n!\) is \(47\)
2) \(n!\) ends in exactly \(12\) zeroes in decimal representation
1) largest prime that divides \(n!\) is \(47\)
2) \(n!\) ends in exactly \(12\) zeroes in decimal representation
03 Jun 2019, 14:01
Kudos
Bookmarks
Not sure how the tags are generated for these questions, but this certainly isn't a sub-600 level problem. It should be a 700+ level problem, and it's also too awkward (and time-consuming) for the GMAT.
If a number has exactly 3 divisors, it must be the square of a prime, p^2 (which has the three divisors 1, p and p^2). We might first look at some random factorial, like 50!, say. The largest square of a prime that divides 50! will be 23^2, because if we write out the product 50! in full, we'll find both 23 and 46 = 2(23) in there, so we'll have two 23's. We won't be able to divide 50! by any larger square of a prime, because, considering the next larger prime, we'll only have one 29 (since we'd need to have 58 in our product to get another 29), and the same will be true for larger primes. So n! will only be divisible by p^2, where p is prime, if p < n/2.
If 47 divides n!, then n must be at least 47. If 47 is the largest prime factor of n!, then n cannot be 53 or greater, because then n! would be divisible by 53 too. So Statement 1 tells us 46 < n < 53. In each case, the largest prime less than or equal to n/2 is 23, so Statement 1 is sufficient.
If n! ends in 12 zeros, then n! is divisible by 10^12 = (2^12)(5^12). For any value of n > 1, n! will be divisible by a lot more 2s than 5s, so if n! is divisible by 5^12, it will always be divisible by 10^12. If you consider 50!, we'll have ten multiples of 5 if we write out our product in full, so 50! is divisible by at least 5^10. But we also have a 25 and a 50 in that product, which give us two extra 5's. So 50! is divisible by 5^12, and so are the factorials from 51! through 54!. Once we get to 55!, we'll have a number divisible by 10^13 though, so Statement 2 tells us 49 < n < 55. Again, in each case the largest prime less than or equal to n/2 is 23, so Statement 2 is also sufficient.
If a number has exactly 3 divisors, it must be the square of a prime, p^2 (which has the three divisors 1, p and p^2). We might first look at some random factorial, like 50!, say. The largest square of a prime that divides 50! will be 23^2, because if we write out the product 50! in full, we'll find both 23 and 46 = 2(23) in there, so we'll have two 23's. We won't be able to divide 50! by any larger square of a prime, because, considering the next larger prime, we'll only have one 29 (since we'd need to have 58 in our product to get another 29), and the same will be true for larger primes. So n! will only be divisible by p^2, where p is prime, if p < n/2.
If 47 divides n!, then n must be at least 47. If 47 is the largest prime factor of n!, then n cannot be 53 or greater, because then n! would be divisible by 53 too. So Statement 1 tells us 46 < n < 53. In each case, the largest prime less than or equal to n/2 is 23, so Statement 1 is sufficient.
If n! ends in 12 zeros, then n! is divisible by 10^12 = (2^12)(5^12). For any value of n > 1, n! will be divisible by a lot more 2s than 5s, so if n! is divisible by 5^12, it will always be divisible by 10^12. If you consider 50!, we'll have ten multiples of 5 if we write out our product in full, so 50! is divisible by at least 5^10. But we also have a 25 and a 50 in that product, which give us two extra 5's. So 50! is divisible by 5^12, and so are the factorials from 51! through 54!. Once we get to 55!, we'll have a number divisible by 10^13 though, so Statement 2 tells us 49 < n < 55. Again, in each case the largest prime less than or equal to n/2 is 23, so Statement 2 is also sufficient.
03 Jun 2019, 14:34
Kudos
Bookmarks
it is a 700+. If you see it for the first time, you may need at least 2.5 - 3 min to get it correct.
Given: A number with only three factors is a square of a prime number.
if this number is 4 (\(2^2\)), it is largest number with exactly 3 factors that divides \(n!\) when n = 4 or 5
if this number is 9 (\(3^2\)), it is largest number with exactly 3 factors that divides \(n!\) when n = 6, 7, 8, 9
and so on ...
Statement 1 : : largest prime that divides \(n!\) is \(47\)
It means that n ranges between (47 and 52) . It can't be \(\geq{53}\) because 53 will be the largest prime that divide it in this case.
the nearest square of a prime number to this range is \(23^2\), which is largest number with exactly 3 factors that divides \(n!\) when n ranges from 46 (\(2*23\)) to 57 ( 2*29 -1 ) --> sufficient
(from 58!, the largest prime square that divides it is \(29^2\))
Statement 2: \(n!\) ends in exactly \(12\) zeroes in decimal representation
It means that n ranges between (50 and 54). Number of zeros in 49! is 10 & Number of zeros of 55! is 13
the nearest square of a prime number to this range is \(23^2\), which is largest number with exactly 3 factors that divides \(n!\) when n ranges from (46 to 57) --> sufficient
D
Given: A number with only three factors is a square of a prime number.
if this number is 4 (\(2^2\)), it is largest number with exactly 3 factors that divides \(n!\) when n = 4 or 5
if this number is 9 (\(3^2\)), it is largest number with exactly 3 factors that divides \(n!\) when n = 6, 7, 8, 9
and so on ...
Statement 1 : : largest prime that divides \(n!\) is \(47\)
It means that n ranges between (47 and 52) . It can't be \(\geq{53}\) because 53 will be the largest prime that divide it in this case.
the nearest square of a prime number to this range is \(23^2\), which is largest number with exactly 3 factors that divides \(n!\) when n ranges from 46 (\(2*23\)) to 57 ( 2*29 -1 ) --> sufficient
(from 58!, the largest prime square that divides it is \(29^2\))
Statement 2: \(n!\) ends in exactly \(12\) zeroes in decimal representation
It means that n ranges between (50 and 54). Number of zeros in 49! is 10 & Number of zeros of 55! is 13
the nearest square of a prime number to this range is \(23^2\), which is largest number with exactly 3 factors that divides \(n!\) when n ranges from (46 to 57) --> sufficient
D
