Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
Kudos for a correct solution.
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 NConsider these examples:
24 is divisible by
3 because 24 = (2)(2)(2)
(3)Likewise, 70 is divisible by
5 because 70 = (2)
(5)(7)
And 112 is divisible by
8 because 112 = (2)
(2)(2)(2)(7)
And 630 is divisible by
15 because 630 = (2)(3)
(3)(5)(7)
For more on this concept, see the 2nd half of this free video: http://www.gmatprepnow.com/module/gmat- ... /video/825So, 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
I’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is…
Students often get questions wrong NOT because they lack the skills to solve the question, but because they don’t understand what the GMAT is testing Learn more