BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

least possible value of N

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Thu Nov 01, 2012 7:52 pm
Thanked: 7 times

by FLUID » Wed Nov 07, 2012 8:29 pm
B166418 wrote:If N is posstive integer and product of all integers from 1 to N is multiple of 990 ,what is least possible value of N

sorry no options available
if you factorize 990 it should be 9 * 10 * 11

Least value of N should be 11
Thanks,

"Take Risks in Your Life.
If you Win, you can Lead!
If you Loose, you can Guide!".
Top
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Sat Mar 26, 2011 4:59 am
Thanked: 4 times

by B166418 » Wed Nov 07, 2012 9:14 pm
Again u r spot on...
but just describe little bit how we can conclude 11 as answer if wefactorize 990 it should be 9 * 10 * 11
Top
Senior | Next Rank: 100 Posts
Posts: 52
Joined: Mon Aug 13, 2012 11:53 pm
Location: Mumbai
Thanked: 1 times

by jkaustubh » Wed Nov 07, 2012 10:01 pm
990 = 11*3*3*2*5


the highest prime number here.

Hence to achieve 990 as the product on N numbers, we need minimum one 11.

This implies that the least possible value shall be 11 for the product to be a multiple of 990
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 07, 2012 10:51 pm
B166418 wrote:If N is posstive integer and product of all integers from 1 to N is multiple of 990 ,what is least possible value of N
A) 10
B) 11
C) 12
D) 13
E) 14
Here are the original answer choices to the question.

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
Top
Post new topic Post Reply
• Page 1 of 1