Number properties question

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yousufa: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Fri Aug 24, 2012 9:09 am

by yousufa » Mon Jul 22, 2013 5:18 pm

Hey guys, having a lot of trouble with quant section of the practice exams. Heres one I couldn't figure out:

If n is a positive integer and the product of all the intergers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

10

11

12

13

14

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Source: Beat The GMAT — Problem Solving | Mon Jul 22, 2013 5:18 pm

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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 » Mon Jul 22, 2013 5:33 pm

yousufa wrote:
If n is a positive integer and the product of all the intergers 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

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yousufa: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Fri Aug 24, 2012 9:09 am

by yousufa » Mon Jul 22, 2013 6:53 pm

Thanks a lot for the help Brent.

Just wondering, how do you know which prime factors' product results in the desired number? Basically, how did you know 2,3,3,5,11 were the factors for 990? It would probably take a while to arrive at that combination from just trial and error.

Thanks

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Mon Jul 22, 2013 7:20 pm

Hi yousufa,

Finding the prime factors of a number is a pretty straight-forward idea: Find ANY number that goes into the larger number and then divide it out, break it down into primes, then continue repeating that process until you have a list of primes.

In the case of 990, you might spot that 2, 3, 5, 10, 99, etc. are divisible to start.

Let's say you divide out a 10:

Now you have 990 = 10 x 99

Break the 10 down into 2 x 5

Break the 99 down into 9 x 11

Break the 9 into 3 x 3

So, 990 = 2 x 5 x 3 x 3 x 11

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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hemant_rajput: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 22, 2012 7:13 amFollowed by: 13 members; Thanked: 46 times; GMAT Score:700

by hemant_rajput » Tue Jul 23, 2013 5:39 am

the question say product of all integer from 1 to n is the factor of 990 that means n! is the product of 990.
expanding 990 - 9 * 10 * 11. we know that 11 is the prime number and n! is of the form 1*2*3*...*n,
so we need one occurrence of 11 and thus the least possible value of n is 11 because it include other factor 9 and 10.

I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.

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yousufa: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Fri Aug 24, 2012 9:09 am

by yousufa » Tue Jul 23, 2013 7:35 am

[email protected] wrote:Hi yousufa,

Finding the prime factors of a number is a pretty straight-forward idea: Find ANY number that goes into the larger number and then divide it out, break it down into primes, then continue repeating that process until you have a list of primes.

In the case of 990, you might spot that 2, 3, 5, 10, 99, etc. are divisible to start.

Let's say you divide out a 10:

Now you have 990 = 10 x 99

Break the 10 down into 2 x 5

Break the 99 down into 9 x 11

Break the 9 into 3 x 3

So, 990 = 2 x 5 x 3 x 3 x 11

GMAT assassins aren't born, they're made,
Rich

Ah! Thanks a lot Rich. I completely forgot about basic factoring. Back to studying!

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