NEW HERE? LEARN MORE
closeclose
Last visit was: 23 Apr 2024, 20:28 It is currently 23 Apr 2024, 20:28
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

n is a positive integer, and k is the product of all integers from 1

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
shrive555
Manager
Manager
Joined: 15 Sep 2010
Status:Do and Die!!
Posts: 207
Own Kudos [?]: 2131 [195]
Given Kudos: 193
 Q29  V6 GMAT 3: 430  Q31  V19
Send PM
n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   04 Nov 2010, 18:15
19
Kudos
176
Bookmarks
00:00
A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

70% (02:13) correct 30% (02:22) wrong based on 2180 sessions
Hide Show 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
ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92883
Own Kudos [?]: 618618 [62]
Given Kudos: 81563
Send PM
CAT Tests Forum Quiz Mode
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   04 Nov 2010, 18:28
24
Kudos
38
Bookmarks
Expert Reply
shrive555 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

8
12
16
18
24

please share your method. Thanks



\(k\) 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.

Answer: A (8).
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14816
Own Kudos [?]: 64884 [27]
Given Kudos: 426
Location: Pune, India
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   04 Nov 2010, 18:35
17
Kudos
10
Bookmarks
Expert Reply
shrive555 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

8
12
16
18
24

please share your method. Thanks


When we say k is the product of all integers from 1 to n, we mean k = n!
Now 1440 = 1.2.3.4.5.6.2
Now if k is a multiple of 1440, it has to be at least 8!. This is so because it cannot be 6! due to the extra 2 factor at the end; it cannot be 7! either, again because the extra 2 in the end will be unaccounted for.
If k = 8! = 1.2.3.4.5.6.7.8, then k = 1440*7*4 i.e. k will be a multiple of 1440.
Answer A
User avatar
MacFauz
VP
VP
Joined: 02 Jul 2012
Posts: 1011
Own Kudos [?]: 3117 [22]
Given Kudos: 116
Location: India
Concentration: Strategy
Schools: Ross '16 (D) McCombs '16 (D) Kenan-Flagler '16 (WA) Olin '16 (M$) Tepper '16 (D)
GMAT 1: 740 Q49 V42
GPA: 3.8
WE:Engineering (Energy and Utilities)
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   21 Oct 2012, 00:29
9
Kudos
13
Bookmarks
kapsycumm 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


Factorizing 1440, we get \(2^5 * 3^2 * 5^1\)

So, k should have at least five 2s, two 3s and one 5.

a) 8! = 2*3*4*5*6*7*8. Above condition is satisfied. Hence answer is A
General Discussion
User avatar
rockzom
Manager
Manager
Joined: 25 Oct 2010
Status:Planning to retake.
Affiliations: Alpha Psi Omega
Posts: 80
Own Kudos [?]: 45 [0]
Given Kudos: 14
Concentration: General Management, Entrepreneurship
GMAT 1: 650 Q42 V37
GRE 1: Q630 V680
GPA: 3.16
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   06 Nov 2010, 10:53
Took me 0:37 to do 1440x2 and divide that by the smallest choice to see if it fit. 2880/8=360, so A.

(I know my methods are unorthodox, but what can you do.)
User avatar
L
IanStewart
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9240 [1]
Given Kudos: 91
 Q51  V47
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   07 Nov 2010, 08:37
1
Kudos
Expert Reply
rockzom wrote:
Took me 0:37 to do 1440x2 and divide that by the smallest choice to see if it fit. 2880/8=360, so A.

(I know my methods are unorthodox, but what can you do.)


'Unorthodox' is one way of putting it. :)

I'm not sure I understand why you've done what you did, but we're not looking here for the smallest answer choice which divides some multiple of 1440; it's coincidence that you get the right answer if you try that. If '4' or '5' had instead been the first answer choice, you'd find that 2880/4 or 2880/5 is an integer, but those aren't correct answers to the question. There's a reason Bunuel and Karishma carried out the steps they did.
User avatar
SOURH7WK
Manager
Manager
Joined: 15 Jun 2010
Posts: 242
Own Kudos [?]: 1174 [2]
Given Kudos: 50
Concentration: Marketing
Schools:IE'14, ISB'14, Kellogg'15
 Q47  V26 GMAT 2: 540  Q45  V19 GMAT 3: 580  Q48  V23
GPA: 3.2
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Send PM
Reviews Badge
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   20 Oct 2012, 23:08
2
Kudos
kapsycumm 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


K= multiple of 1440 ie product of some integer P X 1440. &
K= n! = 1.2.3.4......n

Now lets factorize 1440 = 12 * 12 * 10
Now my objective is to find this multiple 12*12*10 with min possible numbers in n!. Lets go on now......
n= 1*2*(3*4)*5*6*7*8(4 X 2)
Hence 8! is sufficient to get a multiple of 12*12*10

Hence Answer A.
User avatar
P
JeffTargetTestPrep
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3043
Own Kudos [?]: 6271 [4]
Given Kudos: 1646
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   15 May 2017, 17:30
3
Kudos
1
Bookmarks
Expert Reply
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


Let’s break 1440 into prime factors:

1440 = 144 x 10 = 12 x 12 x 10 = 2^5 x 3^2 x 5^1

Thus, k/(2^5 x 3^2 x 5^1) = integer.

We also know that k is the product of all integers from 1 to n inclusive, or in other words, k = n!.

Let’s check our answer choices:

A. If n = 8, then k = 8! and 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 = 7 x 5 x 3^2 x 2^7 does contain five 2s, two 3s and one 5.

Answer: A
avatar
B
Tajmir
Intern
Intern
Joined: 06 May 2020
Posts: 7
Own Kudos [?]: 3 [0]
Given Kudos: 7
Send PM
n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   21 Jul 2020, 05:14
Hope, this illustration will work.

Posted from my mobile device
Attachments

20200721_181124.jpg
20200721_181124.jpg [ 2.97 MiB | Viewed 27113 times ]

avatar
B
KalyanChamarthi
GMAT Tutor
Joined: 04 Feb 2020
Posts: 32
Own Kudos [?]: 17 [1]
Given Kudos: 14
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   21 Jul 2020, 05:27
1
Kudos
Expert Reply
Factorise 1440 : = 10 X 144
2x5 x 12 x 12
2X5x 3 x 4 x 6 x 2
1 x 2x3x 5 x 6 x 8 ( to get consequtive number 1 to n = you need 4 and 7
Now K is the Multiple of 1440 : There fore K can be = 1440 x 28
and also 1 x 2 x 3x4x5x6x7x8 = 1440 X 28
So n should be latest : 8
User avatar
L
Fdambro294
VP
VP
Joined: 10 Jul 2019
Posts: 1392
Own Kudos [?]: 542 [0]
Given Kudos: 1656
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   22 Aug 2020, 21:38
Tajmir wrote:
Hope, this illustration will work.

Posted from my mobile device




that has to be the neatest penmanship I have ever seen... lol
User avatar
L
MathRevolution
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10161
Own Kudos [?]: 16592 [0]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   23 Aug 2020, 00:08
Expert Reply
n is a positive integer, and k is the product of all integers from 1 to n inclusive which means k = n!

K is a multiple of 1440 => k = 1440 * p

1440 = 2*2*2*2*2*3*3*5

=> 1440 = \(2^5\) * \(3^2\) * 5

n! = \(2^5\) * \(3^2\) * 5 * p

We need at least five (2's), two(3's) and one(5)

=> 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2 = \(2^5\) * \(3^2\) * 5 * \(2^2\) * 7 = 1440 * \(2^2\) * 7

Answer A
User avatar
P
MHIKER
Director
Director
Joined: 14 Jul 2010
Status:No dream is too large, no dreamer is too small
Posts: 972
Own Kudos [?]: 4927 [0]
Given Kudos: 690
Concentration: Accounting
Send PM
n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   23 Dec 2020, 08:48
Top Contributor
shrive555 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


The question tells k! is divided by \(1440\)

So k have to have all the factors that \(1440\) will have.

\(The \ factors \ of \ 1440=2^5, 3^2\ & \ 5\)

From the answer choices \(8!\) contains \(= 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2=2^7, 3^2, 5, \ & \ 7\)

The answer is \(A.\)
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32633
Own Kudos [?]: 821 [0]
Given Kudos: 0
Send PM
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink] New post   03 Mar 2024, 07:37
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.
GMAT Club Bot
gmatclubot
Re: n is a positive integer, and k is the product of all integers from 1 [#permalink]
03 Mar 2024, 07:37
Moderators:
Bunuel
Math Expert
92883 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions