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

Redeem

PS- multipe 990

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 275
Joined: Wed Jul 02, 2008 4:19 am
Thanked: 4 times

PS- multipe 990

by Xbond » Fri Sep 11, 2009 10:50 am
Hi there,

Could you give me your ELEGANT, QUICK, and EFFICIENT method to resolve (with these concepts and with finding the solution).


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
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 77
Joined: Sun Jun 21, 2009 10:25 am
Location: Germany
Thanked: 7 times

by Nermal » Fri Sep 11, 2009 12:16 pm
IMO B (11)

in order to be multiple of 990, n! (this is the product of numbers from 1 to n inclusive) has to have at least the same prime factors as 990.

990=2*3*3*5*11

11 is the greatest prime factor of 990 and 11! includes all the other factors 2,3^2,5.
Top
Master | Next Rank: 500 Posts
Posts: 275
Joined: Wed Jul 02, 2008 4:19 am
Thanked: 4 times

by Xbond » Mon Sep 14, 2009 12:50 am
many thks

OA is B
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 » Thu Oct 10, 2019 5:46 am
Xbond 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
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

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Sun Oct 13, 2019 5:05 pm
Xbond wrote:Hi there,

Could you give me your ELEGANT, QUICK, and EFFICIENT method to resolve (with these concepts and with finding the solution).


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
We are given that n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990. Thus:

n! / 990 = integer

To determine the minimum value of n, we need to break 990 into its prime factors.

990 = 10 x 99 = 5 x 2 x 3^2 x 11

Thus:

n! / (5 x 2 x 3^2 x 11) = integer

Since n! must be divisible by 2, 5, 9, and 11, the minimum value of n must be 11. Recall that 11! is divisible by 11 and by any positive integer less than 11. In other words, 11! / 990 = integer.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1