sandy wrote:
If n is an integer and \(n^3\) is divisible by 24, what is the largest number that must be a factor of n?
(A) 1
(B) 2
(C) 6
(D) 8
(E) 12
-----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 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)
-----ONTO THE QUESTION!---------------------
If n³ is divisible by 24, then we can say that n³= 24k for some positive integer k
Since 24 = (2)(2)(2)(3), we can write:
n³= (2)(2)(2)(3)k for some positive integer k
So, what does this tell us about n?
Since there's an 8 hiding in the prime factorization of n³, we can conclude that 8 is a divisor of n³, which means
2 must be a divisor of nIt also tells us that
3 must be a divisor of nThese are the only two guaranteed conclusions we can draw about the divisors of n
If 2 and 3 must be divisors of n, then we can be certain that and is divisible by 6
Answer: C