NandishSS wrote:If m is a positive integer and m² is divisible by 48, then the largest positive integer that must divide m is?
(A) 3
(B) 6
(C) 8
(D) 12
(E) 16
ASIDE-----------------------------------
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)
--------------------------------------
m² is divisible by 48
48 =
(2)(2)(2)(2)(3)
This tells us that m² =
(2)(2)(2)(2)(3)(?)(?)(?)(?)(?)
NOTE: the various (?)'s includes other possible prime numbers in the prime factorization of m²
So, we know how the prime factorization of m² looks.
What does this tell us about the prime factorization of m?
First, since there are FOUR 2's in the prime factorization of m², we know that the prime factorization of m will include at least TWO 2's, since
(2)(2) x
(2)(2) =
(2)(2)(2)(2)
Likewise, since there is ONE 3 in the prime factorization of m², we know that the prime factorization of m will include at least ONE 3.
So, we can be certain that m =
(2)(2)(3)(?)(?)(?)(?)(?)
Since
(2)(2)(3) =
12, we can be certain that
12 is a divisor of m.
Answer:
D
Brent Hanneson - Creator of GMATPrepNow.com
