Gmat_mission wrote: ↑Fri Jan 08, 2021 4:30 am
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
Answer:
B
Source: Official Guide
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
Consider 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: https://www.gmatprepnow.com/module/gmat- ... /video/825
So, 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
