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A lot of integer property questions can be solved using prime factorization.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
Hi Brent,Brent@GMATPrepNow wrote:A lot of integer property questions can be solved using prime factorization.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
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
Since 11 is a factor of 990, we need 11 to be included in the product. So, n cannot equal 10.prada wrote:Hi Brent,Brent@GMATPrepNow wrote:A lot of integer property questions can be solved using prime factorization.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
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
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
Thanks Rich[email protected] 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
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