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Re: [MGRE 5lb. Book] Divisibility and Primes
Originally Posted by
weilun85
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
First, if n is an integer and n^{3} is divisible by 24, then we can say that n^{3} = 24k for some integer k.
We can also rewrite this as: (n)(n)(n) = 24k for some integer k
Now let's rewrite 24 as a product of primes to get: (n)(n)(n) = (2)(2)(2)(3)k
This tells us that there is at least one 2 "hiding" in the prime factorization of n
It also tells us that there is at least one 3 "hiding" in the prime factorization of n.
So, if we're certain that there is at least one 2 and one 3 "hiding" in the prime factorization of n, we can be certain that n is divisible by 2 and by 3.
In other words, we can be certain that n is divisible by 6.
Of course it COULD be divisible by a number greater than 6, but the question asks us to determine the largest number that must be a factor of n.
Answer: SPOILER: C
Cheers,
Brent