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

Question

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 Thanks in advance

Brent@GMATPrepNow · Answer

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

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

Rich.C@EMPOWERgmat.com · Answer

Hi lucas211,

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

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

Contact Rich at [email protected]

Matt@VeritasPrep · Answer

If we've got n! = 9 * 10 * 11, we need n! to contain 9, 10, and 11 somewhere in its prime factorization.

Since 11 can only appear in multiples of 11, we need a multiple of 11. The smallest positive one is 11 itself, so n ≥ 11, and the least possible n is 11 itself.

shwetagoyal121 · Answer

Brent@GMATPrepNow 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 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 The answer would be 11. Look at this question from the prospective options angle as well. The first probable answer is 10 ; (because that is the smallest value) Product of all the integers till n= 990 x (Some other number say Y) Case I: Now if n=10, whether the product is a multiple of 990 will depend on whether Y has 11 as a factor or not. case II: now if n=11 (next smallest value) the product has all the factors to make it a multiple of 990 irrespective of the value of Y. Hence 11 is the right answer.

SampathKp · Answer

lucas211 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

Thanks in advance

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]

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

Scott@TargetTestPrep · Answer

lucas211 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

Thanks in advance

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]

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

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