Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 288

n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
04 Nov 2010, 18:15
Question Stats:
67% (01:41) correct 33% (02:53) wrong based on 929 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
I'm the Dumbest of All !!



Math Expert
Joined: 02 Sep 2009
Posts: 46284

Re: property of Integers [#permalink]
Show Tags
04 Nov 2010, 18:28



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8102
Location: Pune, India

Re: property of Integers [#permalink]
Show Tags
04 Nov 2010, 18:35
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
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 288

Re: property of Integers [#permalink]
Show Tags
04 Nov 2010, 19:52
Got it Thanks B !! Thanks Karisham
_________________
I'm the Dumbest of All !!



Manager
Status: Planning to retake.
Affiliations: Alpha Psi Omega
Joined: 25 Oct 2010
Posts: 86
Concentration: General Management, Entrepreneurship
GRE 1: 1310 Q630 V680
GPA: 3.16

Re: property of Integers [#permalink]
Show Tags
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.)
_________________
Did I help you? Please give me kudos.
Each moment of time ought to be put to proper use, either in business, in improving the mind, in the innocent and necessary relaxations and entertainments of life, or in the care of the moral and religious part of our nature.
William Andrus Alcott



Manager
Joined: 19 Aug 2010
Posts: 69

Re: property of Integers [#permalink]
Show Tags
06 Nov 2010, 11:36
rockzom, but if the first answer is not the correct one, you'll have to divide with all the 5 answer choices. Then it will take you more than 37 seconds...



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1345

Re: property of Integers [#permalink]
Show Tags
07 Nov 2010, 08:37
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.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Intern
Joined: 03 Nov 2010
Posts: 3

Re: property of Integers [#permalink]
Show Tags
07 Nov 2010, 20:28
i have problem in integer..please can anyone help me



Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 288

Re: property of Integers [#permalink]
Show Tags
08 Nov 2010, 08:21
annie89 wrote: i have problem in integer..please can anyone help me Read Number Theory topic in Math Book of Gmat Club. That's a wonderful book. you can find the link in ' Bunel's ' signature. well let me make it more easy for you. here is link http://gmatclub.com/forum/gmatmathbook87417.htmlBest of luck
_________________
I'm the Dumbest of All !!



Senior Manager
Joined: 15 Jun 2010
Posts: 336
Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

Re: the smallest possible value of n [#permalink]
Show Tags
20 Oct 2012, 23:08
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.
_________________
Regards SD  Press Kudos if you like my post. Debrief 610540580710(Long Journey): http://gmatclub.com/forum/from600540580710finallyachievedin4thattempt142456.html



VP
Joined: 02 Jul 2012
Posts: 1194
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: the smallest possible value of n [#permalink]
Show Tags
21 Oct 2012, 00:29
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
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Thanks To The Almighty  My GMAT Debrief
GMAT Reading Comprehension: 7 Most Common Passage Types



Intern
Joined: 15 Jul 2014
Posts: 2

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
22 Jul 2014, 21:47
Hi Bunuel, I have a question here. I use the same method as you > finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck... Thanks for your help in advance. Bunuel wrote: 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).



Math Expert
Joined: 02 Sep 2009
Posts: 46284

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
23 Jul 2014, 02:20
nueyada wrote: Hi Bunuel, I have a question here. I use the same method as you > finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck... Thanks for your help in advance. Bunuel wrote: 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). No, that's not correct. You got the correct answer by fluke.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
28 Dec 2016, 07:38



Intern
Status: One more try
Joined: 01 Feb 2015
Posts: 49
Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
28 Dec 2016, 08:40
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 N=8!=720*56 K multiple of 1440 Hence as per options a) is ans
_________________
Believe you can and you are halfway thereTheodore Roosevelt



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3511
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
28 Dec 2016, 11:49
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 k = 1 * 2 * 3 * 4..............n Or, k = n! Further we know k = m*1440 Or, k = m*( 2*6! ) Now, we have  n! = m*( 2*6! ) Or, n! = m *2*6*5*4*3*2*1 Thus, The nearest factorial of n! will be 8 Hence, correct answer will be 8!
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club  Rules for Posting in QA forum  Writing Mathematical Formulas Rules for Posting in VA forum  Request Expert's Reply ( VA Forum Only )



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
09 Jan 2017, 22:12



Manager
Status: Just redeemed Kudos for GMAT Club Test !!
Joined: 14 Sep 2013
Posts: 93
Location: Bangladesh
GPA: 3.56
WE: Analyst (Commercial Banking)

n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
16 Feb 2017, 10:10
k is a multiple of 1440. That means K must have at least all the factors of 1440. For this reason, 6 or 7 can't be the value of n. n must comprise the extra factor '2'.
_________________
______________ KUDOS please, if you like the post or if it helps "Giving kudos" is a decent way to say "Thanks"
Master with structure  Numerical comparison [source: economist.com] https://gmatclub.com/forum/masterwithstructurenumericalcomparison233657.html#p1801987



Intern
Joined: 22 Aug 2016
Posts: 1

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
21 Feb 2017, 01:27
Hi
Thanks for the answers  very useful.
I had a question please: if n! is equal to k is equal to 1440x then doesn't that imply that 1440/n is equal to the product of all preceding numbers?
So if n could be 8 then 1440 divided by 8 should be the product of all number 1 through to 8.
But that isn't the case. Am I missing something? Any advice would be greatly appreciated!
Thank you, Rukshan



NonHuman User
Joined: 09 Sep 2013
Posts: 7027

Re: n is a positive integer, and k is the product of all integer [#permalink]
Show Tags
02 Mar 2018, 06:40
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 Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: n is a positive integer, and k is the product of all integer
[#permalink]
02 Mar 2018, 06:40






