Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 Oct 2016, 17:30

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# n is a positive integer, and k is the product of all integer

Author Message
TAGS:

### Hide Tags

Director
Joined: 25 Oct 2008
Posts: 608
Location: Kolkata,India
Followers: 11

Kudos [?]: 722 [0], given: 100

n is a positive integer, and k is the product of all integer [#permalink]

### Show Tags

28 Sep 2009, 19:49
5
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

69% (02:25) correct 31% (01:45) wrong based on 147 sessions

### HideShow timer Statistics

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8
B. 12
C. 16
D. 18
E. 24
[Reveal] Spoiler: OA

_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
Followers: 8

Kudos [?]: 250 [0], given: 20

### Show Tags

28 Sep 2009, 20:52
tejal777 wrote:
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

Using mathematical brute force I got to 8 although I'm sure there is a much more elegant solution.
I figured 1440 is not too big a number for that type of formula so trying a few numbers

If n = 6: k = 1 x 2 x 3 x 4 x 5 x 6 = 720

K is a multiple of 1440 meaning k/1440 must not have remainder.

Using the above if n = 7 then k wont be divisible by 1440 evenly, since it's odd. (since 720 is 1440/2 and hence 7x1440/2 = 7x1440/2)
But if n = 8 then it will be divisible by 1440 since it's even.

That took me under 2 mins. It's probably not the best solution but made sense to me
Manager
Joined: 25 Mar 2009
Posts: 55
Followers: 1

Kudos [?]: 17 [0], given: 9

### Show Tags

28 Sep 2009, 21:25
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8
Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
Followers: 8

Kudos [?]: 250 [0], given: 20

### Show Tags

28 Sep 2009, 22:06
Mikko wrote:
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread:
what-is-the-least-value-for-n-82903.html
We have the same question but for 990. Factoring 990 we get = $$2*3^2*5*11$$
So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5.

Is that correct?
Manager
Joined: 25 Mar 2009
Posts: 55
Followers: 1

Kudos [?]: 17 [0], given: 9

### Show Tags

28 Sep 2009, 23:50
yangsta8 wrote:
Mikko wrote:
1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s
n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread:
what-is-the-least-value-for-n-82903.html
We have the same question but for 990. Factoring 990 we get = $$2*3^2*5*11$$
So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5. (I don't think that there is a correlation between Prime 5-the factor and the 5 in 2^5)

Is that correct?

It's coincident that you got 11 is the largest prime and n! has enough factors of 2, 3 or 5 in $$2*3^2*5*11$$ but in your topic, 5 is the largest Prime factor but not enough for the power 2^5 if stop at n=5 (only 2=2^1, 4=2^2=> max is 2^3).

U can approach anyway as long as n! is divisible by all the power-factors of the given number
Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 649
Followers: 41

Kudos [?]: 789 [0], given: 39

### Show Tags

24 Sep 2010, 05:06
Is not the question asked the smallest possible value of n? Is not 2?
_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 649
Followers: 41

Kudos [?]: 789 [0], given: 39

### Show Tags

24 Sep 2010, 08:45
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8
B. 12
C. 16
D. 18
E. 24

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2
_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Math Expert
Joined: 02 Sep 2009
Posts: 35275
Followers: 6636

Kudos [?]: 85571 [0], given: 10237

### Show Tags

24 Sep 2010, 08:57
Expert's post
5
This post was
BOOKMARKED
Baten80 wrote:
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

$$k$$ is the product of all integers from 1 to $$n$$ inclusive --> $$k=n!$$;

$$k$$ is a multiple of 1440 --> $$n!=1440*p=2^5*3^2*5*p$$, for some integer $$p$$ --> $$n!$$ must have at least 2 in power 5, 3 in power 2 and 5 as its factors. Now, $$7!=2*3*(2^2)*5*(2*3)*7=2^4*3^2*5*7$$ not enough power of 2 --> next #: $$8!=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5*7=1440*(2^2*7)$$ --> so lowest value of $$n$$ is 8.

_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 223

Kudos [?]: 1564 [0], given: 235

### Show Tags

24 Sep 2010, 09:01
Baten80 wrote:
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

$$1440=2^5*3^2*5$$ => n should have at least one 5, two 3's and 5 2's.

if you take n=6 => k = 6! = $$1*2*3*4*5*6 = 2^4 * 3^2 * 5$$ => does not satisfy we need one more 2.

so the next even number 8 is the answer.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 104

Kudos [?]: 899 [0], given: 25

### Show Tags

24 Sep 2010, 11:55
Baten80 wrote:
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8
B. 12
C. 16
D. 18
E. 24

To me Ans is 2.

LCM of 1440=2^5*3^2*5
K=n!
so, Ans. 2

k=n!
And 1440 divides k
1440 = 2^5 * 3^2 * 5

Counting the powers of 2 in n!
Upto 2 (1)
Upto 4 (3)
Upto 6 (4)
Upto 8 (7)
So n>=8

Counting powers of 3 in n!
Upto 3 (1)
Upto 6 (2)
So n>=6

Counting powers of 5 in n!
Upto 5 (1)
So n>=5

Combining, minimum n=8
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12212
Followers: 542

Kudos [?]: 151 [0], given: 0

Re: n is a positive integer, and k is the product of all integer [#permalink]

### Show Tags

26 May 2016, 23:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 22 Jun 2014
Posts: 456
Concentration: General Management, Technology
GMAT 1: 540 Q45 V20
GPA: 2.49
WE: Information Technology (Computer Software)
Followers: 12

Kudos [?]: 148 [0], given: 91

Re: n is a positive integer, and k is the product of all integer [#permalink]

### Show Tags

18 Jun 2016, 07:49
1440 = 2^5 * 3^2 * 5^1

We need five 2s, two 3s and one 5 in the n!. 8! contains all of these and it is the least of all. A is the answer.
_________________

---------------------------------------------------------------
Target - 720-740
helpful post means press '+1' for Kudos!
http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html
http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html

SVP
Joined: 17 Jul 2014
Posts: 1847
Location: United States
Schools: Stanford '19
GMAT 1: 550 Q39 V27
GMAT 2: 560 Q42 V26
GMAT 3: 560 Q43 V24
GMAT 4: 650 Q49 V30
GPA: 3.56
WE: General Management (Transportation)
Followers: 15

Kudos [?]: 217 [0], given: 114

Re: n is a positive integer, and k is the product of all integer [#permalink]

### Show Tags

23 Oct 2016, 11:21
tejal777 wrote:
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8
B. 12
C. 16
D. 18
E. 24

1440 = 144 * 10 = 12 * 12 * 2 * 5 = 2^5 * 3^2 * 5
we need at least 5 factors of 2, two factors of three, and 1 factor of five.
to have 2 factors of three, we need at least 6.
6! has 2^4, 3^2, and a 5.
but we need one more factor of two..therefore...8 would work!

A
Re: n is a positive integer, and k is the product of all integer   [#permalink] 23 Oct 2016, 11:21
Similar topics Replies Last post
Similar
Topics:
64 n is a positive integer, and k is the product of all integer 15 04 Nov 2010, 18:15
72 If the product of all the factors of a positive integer, N 27 25 Oct 2010, 08:24
85 If n is a positive integer and the product of all integers 47 28 Dec 2009, 11:47
If n is a positive integer and the product of all integers 4 07 Nov 2009, 11:02
If n is a positive integer and the product of all the intege 7 17 Oct 2007, 08:44
Display posts from previous: Sort by