ASIDE-----------------------------------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
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)
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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
