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multiple of 990?

This topic has 3 expert replies and 3 member replies
josh80 Senior | Next Rank: 100 Posts Default Avatar
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multiple of 990?

Post Sun Dec 15, 2013 3:18 pm
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?

10
11
12
13
14

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GMAT/MBA Expert

Post Sun Dec 15, 2013 3:54 pm
josh80 wrote:
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 = 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

_________________
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theCodeToGMAT Legendary Member
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Post Sun Dec 15, 2013 10:37 pm
n! = 11 x 5 x 2 x 3 x 3 x _

From RHS, we need one 11, 5, 2 & 3x3

So, n! must comprise of prime number "11"

For that we need minimum n=11 .. as if n = 10.. then "11" will not be there..

So, {B}

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prada Senior | Next Rank: 100 Posts Default Avatar
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Post Thu May 19, 2016 3:30 pm
Brent@GMATPrepNow wrote:
josh80 wrote:
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 = 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
Hi Brent,

Could we not say the answer is (a) or 10 since in the PF (2)(3)(3)(5)(11) 2x5=10? I did the same methodology as you but I concluded that it would be (a) since with the pf we can come up with 10

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Post Thu May 19, 2016 3:40 pm
prada wrote:
Brent@GMATPrepNow wrote:
josh80 wrote:
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 = 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
Hi Brent,

Could we not say the answer is (a) or 10 since in the PF (2)(3)(3)(5)(11) 2x5=10? I did the same methodology as you but I concluded that it would be (a) since with the pf we can come up with 10
Since 11 is a factor of 990, we need 11 to be included in the product. So, n cannot equal 10.

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
Come see all of our free resources

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GMAT/MBA Expert

Post Thu May 19, 2016 3:40 pm
Hi prada,

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. 10! does NOT have the "11" that we need.

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

Final Answer: B

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

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Contact Rich at Rich.C@empowergmat.com

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prada Senior | Next Rank: 100 Posts Default Avatar
Joined
08 Dec 2010
Posted:
64 messages
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1
Post Thu May 19, 2016 3:46 pm
Rich.C@EMPOWERgmat.com wrote:
Hi prada,

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. 10! does NOT have the "11" that we need.

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

Final Answer: B

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

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