If n is a positive integer and the product of all integers

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rsarashi: Master | Next Rank: 500 Posts; Posts: 186; Joined: Sat Dec 24, 2016 12:38 amFollowed by: 3 members; Thanked: 5 times

by rsarashi » Tue Mar 07, 2017 9:17 am

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

OAB

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Mar 07, 2017 9:31 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

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.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Mar 07, 2017 9:39 am

If n is a positive integer and 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 = (2)(2)(2)(3)
70 is divisible by 5 <--> 70 = (2)(5)(7)
330 is divisible by 6 <--> 330 = (2)(3)(5)(11)
56 is divisible by 8 <--> 56 = (2)(2)(2)(7)

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

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by [email protected] » Tue Mar 07, 2017 10:36 am

Hi rsarashi,

For a number to be a multiple of 990, that number must have the exact same prime factors (including duplicates) as 990 (but may have "extra" prime factors as well).

This prompt adds the extra stipulation that N has to be as SMALL as possible, so after factoring 990, we need to find the smallest product that "holds" the 2, 5, 11, and two 3s that make up 990.

1(2)(3)(4)....(11) is the smallest product that does that, so N = 11.

Final Answer: B

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Rich

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Tue Mar 07, 2017 10:45 pm

rsarashi 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

OAB

Hi rsarashi,

When we see numbers 22, 33, 44, 55, 66, 77, 99, what does click our mind?

It's that they all are the multiples of the prime number 11. Thus, 990 must also be a multiple of '11.'

Thus, all the options except option A are correct; however, option B: 11 is the least possible value of n.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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