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Thread: [MGRE 5lb. Book] Divisibility and Primes

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    Smile [MGRE 5lb. Book] Divisibility and Primes

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    Hi all,

    I encountered the following problem in pg 534 of the Manhatten GRE 5lb. book and is unable to understand the given explanation. Appreciate if anyone could shed some light on this.

    If n is an integer and n3 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

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    Re: [MGRE 5lb. Book] Divisibility and Primes

    Quote Originally Posted by weilun85 View Post
    If n is an integer and n3 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 n3 is divisible by 24, then we can say that n3 = 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

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