Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Least possible value

Expert replies
by saidov.mikhail » Thu Oct 03, 2013 1:27 am
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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
Join the discussion
Source: — Problem Solving |

by theCodeToGMAT » Thu Oct 03, 2013 1:51 am
n! = 990 x _

990 = 9 x 10 x 11 = 3^2 x 2 x 5 x 11

Now, "n!" must contain "11" so the least possible answer is 11!

if you take "n" as 10 then "11" is missing..

and if you take "n" as 12, then even though it also contains 11 but it's not the least possible value.

Answer [spoiler]{B}[/spoiler]
R A H U L
Join the discussion

by mevicks » Thu Oct 03, 2013 1:52 am
saidov.mikhail 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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
The fancy way saying n! is "product of all the integers from 1 to n"

Given: n! is divisible by 990 (Factorize: 11*3*3*2*5)
So the division of n!/990 should be an integer

The least value of n should cancel out all the factors of 990
Here we see that if we take n to be 11
11! will contain all the factors of 990

11! = 11*10*9(this will cancel out two 3s)*8*7*6*5*4*3*2*1

Hence answer is [spoiler]b) 11[/spoiler]

Regards,
Vivek
Join the discussion

by GMATGuruNY » Thu Oct 03, 2013 3:12 am
saidov.mikhail 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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
990 = 9*10*11.

The product of all the integers from 1 to n, inclusive = n!.
We can plug in the answers for n.
Since the question stem asks for the LEAST POSSIBLE VALUE of n that is divisible by 9*10*11, start with the SMALLEST answer choice.

Answer choice A: n=10
10*9*8*7*6*5*4*3*2*1)/(9*10*11).
Only the values in blue cancel out.
11 cannot divide into 10!.
Eliminate A.

Answer choice B: n=11
(11*10*9*8*7*6*5*4*3*2*1)/(9*10*11).
Success!
All of the values in blue can divide into 11!.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by Brent@GMATPrepNow » Thu Oct 03, 2013 5:46 am
saidov.mikhail 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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Similarly, we can say:
If N is is a multiple of k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
So, the answer is B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion