Oct 14 08:00 PM PDT  11:00 PM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R2.**Limited for the first 99 registrants. Register today! Oct 15 12:00 PM PDT  01:00 PM PDT Join this live GMAT class with GMAT Ninja to learn to conquer your fears of long, kooky GMAT questions. Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Dec 2009
Posts: 9

If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
Updated on: 07 Jul 2019, 23:23
Question Stats:
81% (01:15) correct 19% (01:23) wrong based on 2285 sessions
HideShow timer Statistics
If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n? A. 10 B. 11 C. 12 D. 13 E. 14 A. 8 B. 9 C. 10 D. 11 E. 12
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by nailgmattoefl on 28 Dec 2009, 11:47.
Last edited by Bunuel on 07 Jul 2019, 23:23, edited 4 times in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 58320

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
28 Dec 2009, 12:22
If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n? A. 10 B. 11 C. 12 D. 13 E. 14 We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger).. Answer: B. Hope it's clear.
_________________




Intern
Status: Getting ready for the internship summer
Joined: 07 Jun 2009
Posts: 33
Location: Rochester, NY
Schools: Simon
WE 1: JPM  Treasury

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
18 Sep 2010, 15:18
This question tests the concept of prime factorization. In order to figure out if 990 can be a divisor of the unknown product you need to determine all of the prime factors that combine to successfully divide 990 without leaving a remainder.
The prime factors of 990 are 2, 3, 3, 5, 11
Since 11 is a prime factor of 990, you need to have the number present somewhere in the unknown product. Answer A is the product of integers from 1 to 10 and therefore does not meet the necessary criteria. Answer B includes 11 and therefore is the answer.




Manager
Joined: 26 May 2005
Posts: 160

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
27 Feb 2010, 21:50
divanshuj wrote: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value for n?
a) 10 b) 11 c) 12 d) 13 e) 14 990 = 2 * 5 * 3 * 3* 11 so product from 1 to n should have all the above numbers. n = 11 will have all the above numbers as factors of 1 * 2* ..... * 11 B



Manager
Joined: 16 Apr 2010
Posts: 181

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
02 Aug 2010, 00:00
Hi,
The question says that 990 is a factor of n!. 990 = 9*110 = 9*11*10 For 990 to be a factor of n!, n should at least be 11 so that it includes the common factor 11 that is found in 990.
regards, Jack



Manager
Joined: 06 Oct 2009
Posts: 50

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
02 Aug 2010, 07:18
990= 2x3x3x5x11 So 11 has to be one of the prime factor So 11!
_________________
+1 kudos me if this is of any help...



Retired Moderator
Joined: 02 Sep 2010
Posts: 728
Location: London

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
18 Sep 2010, 15:59
The least possible value of such a n will be at least the highest prime factor of 990, which is 11 The only other thing to check is if the rest of the factors of 990, can be obtained by multiplication of numbers less than 11. Otherwise n may be greater than 11 But that is easy to see since 990=9*10*11. So taking all the numbers below 11 will suffice.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 58320

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
13 Feb 2011, 18:53



Manager
Joined: 19 Jul 2011
Posts: 89
Concentration: Finance, Economics
GPA: 3.9

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
24 Aug 2011, 08:13
You need to primarily find the primes of 990: 2,3,3,5,11 The integer must contain all these primes By knowing 11 is an prime, we know the answer can be B,C,D or E as these all contain 11. 11=1,2,3,4,5,6,7,8,9,10,11 11 contains the 2,3,5 and 11 explicitly and the second 3 comes from the 6 (2*3), therefore the answer is 11. This is because we know a number that contains all the primes of 990 will be a multiple of 990. Hope this helps
_________________
Show Thanks to fellow members with Kudos its shows your appreciation and its free



Manager
Joined: 19 Jul 2011
Posts: 89
Concentration: Finance, Economics
GPA: 3.9

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
24 Aug 2011, 09:29
The answer would be the number which contains all the factors of the figure. The series of 1 to 11 contained all the factors of 990. 2,3,3,5 and 11. Because 990 is made up from these 5 numbers such that 2*3*3*5*11 = 990 If you times the other numbers (1,4,6,7,8,9,10) in the series you will get a multiple of 990 from this. I picked 11 because it asked for the least possible value of n. Picking 12,13,14 would result in a multiple of 990 but the number is more than the least possible value of 11. The best way to interpret problems is through practice, you will come to understand questions better over time. Always remember GMAT questions are simple questions made to look complicated.
_________________
Show Thanks to fellow members with Kudos its shows your appreciation and its free



Intern
Joined: 09 Feb 2011
Posts: 8

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
24 Aug 2011, 09:50
just entered this in google and you're right! (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11) / 990 = 40 320
meanwhile (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / 990 = 3 665.45455 wouldn't work
It makes sense because the highest prime factor of 990 must also be a factor of any multiple of 990.
I guess I need to do more practice problems involving multiples and number properties.



Manager
Joined: 20 Nov 2010
Posts: 122

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
03 Sep 2011, 11:11
Since we need to find the minimum value of n, So the greatest prime factor of 990 will be the answer. 990 = 2 * 3^2 * 5 * 11. Hence the minimum value of n such that n! ( 1*2* ... *n) is divisible by 990 is 11.
_________________



Senior Manager
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 423
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
25 Jan 2013, 02:29
Bunuel wrote: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10 B. 11 C. 12 D. 13 E. 14
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. to cement my understanding, maximum value of n can be anything, right?
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Math Expert
Joined: 02 Sep 2009
Posts: 58320

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
25 Jan 2013, 04:51
Sachin9 wrote: Bunuel wrote: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10 B. 11 C. 12 D. 13 E. 14
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. to cement my understanding, maximum value of n can be anything, right? The maximum value of n is not limited at all. For all n more than or equal to 11, n! will be a multiple of 990.
_________________



Manager
Joined: 21 Jun 2011
Posts: 56
Location: United States
Concentration: Accounting, Finance
WE: Accounting (Accounting)

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
15 Jul 2013, 00:29
Bunuel wrote: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10 B. 11 C. 12 D. 13 E. 14
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. Hi Bunuel, I am not sure why have excluded 10, which seems to be one of the factors (5*2).



Math Expert
Joined: 02 Sep 2009
Posts: 58320

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
15 Jul 2013, 00:33
davidfrank wrote: Bunuel wrote: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10 B. 11 C. 12 D. 13 E. 14
We are told that \(n!=990*k=2*5*3^2*11*k\) > \(n!=2*5*3^2*11*k\) which means that \(n!\) must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of \(n\) is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger)..
Answer: B.
Hope it's clear. Hi Bunuel, I am not sure why have excluded 10, which seems to be one of the factors (5*2). n! must be a multiple of 990=90* 11. Now, if n=10, then it wont have 11 as its factor, so it cannot be a multiple of a number which HAS 11 as its factor. Hope it's clear.
_________________



SVP
Joined: 14 Apr 2009
Posts: 2273
Location: New York, NY

Re: If n is a positive integer and the product of all the
[#permalink]
Show Tags
26 Feb 2014, 11:11
Here's a more elaborated thinking response to what you all have summarized above  but for those who need the extra steps, try this.
990* multiple = product of 1 to n 990 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n 11 * 3 * 3 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n
We need to find the smallest value for n. We know in the factor tree, any number can be broken down to its lowest level in the factor tree. And that factor tree will comprise of all prime factors.
The factors on the left side are: 11 and 3 and 'multiple'  the multiple can be anything.
We know that we want 990 to go into the right side (we don't care how many times).
The 990 consists of 11 and 3^2. So essentially it's prime factors are 11 and 3. These lowest level prime factors must go into the right side  so the right side at the very least must contain the lowest prime factors from the left side.
So, we see 1 * 2 * 3 * 4 * 5 * ... * n
contains a 3.
But it does not contain an '11'. 11 is the answer.
Hope that helps.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9699
Location: Pune, India

Re: If n is a positive integer and the product of all the
[#permalink]
Show Tags
26 Feb 2014, 19:30
flood wrote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
a) 10 b) 11 c) 12 d) 13 e) 14
I got this one right but I'm not too clear on the logic behind it. Can you multiply two numbers (other than 11) and get an 11? No because 11 is prime. If 11 is a part of the product of 3 numbers, say a*b*c, it must have been one of the numbers. Given that the product is 990x (where x is some integer). When we start breaking down 990 into factors, we start with 9*11*10. We see that 9 and 10 will be further broken down into smaller factors but 11 will not be. It is prime. So one of the numbers which multiplied to give 990x must have been 11. Since consecutive integers 1 to n were multiplied, n is either 11 or greater than 11 to include 11 in the numbers multiplied. Least possible value of n is 11.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2974
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
20 Oct 2015, 04:31
Bunuel wrote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?
(A) 8 (B) 9 (C) 10 (D) 11 (E) 12
Kudos for a correct solution. product of all the integers from 1 to n = n! 990 = 9*10*11 i.e. n! must be divisible by 9, 10 and 11 all for n! to be divisible by 990 i.e. minimum value of n must be 11 for n! to be divisible by 11 and 10 and 9 both P.S. 11!= 11*10*9*8*7*6*5*4*3*2*1 n = 11 Answer: option D
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



SVP
Status: It's near  I can see.
Joined: 13 Apr 2013
Posts: 1685
Location: India
Concentration: International Business, Operations
GPA: 3.01
WE: Engineering (Real Estate)

Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
Show Tags
20 Oct 2015, 05:28
Bunuel wrote: If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?
(A) 8 (B) 9 (C) 10 (D) 11 (E) 12
Kudos for a correct solution. My Solution:
Prime factorization of 990 gives us = 2 x 3 x 3 x 5 x 11
Therefore, if n! is divisible by 990 it must contain all above factors.
Option A , B and C don't contain 11 as factor. So least value of n must be 11 as 11! contains all possible factors.
Answer is D
_________________
"Do not watch clock; Do what it does. KEEP GOING."




Re: If n is a positive integer and the product of all integers from 1 to n
[#permalink]
20 Oct 2015, 05:28



Go to page
1 2
Next
[ 27 posts ]



