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Prime Factors

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lachlanc Rising GMAT Star Default Avatar
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Prime Factors Post Sat Nov 08, 2008 12:50 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

    1) m has more than 9 positive factors
    2) m is a multiple of p^3

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    logitech GMAT Titan
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    Post Sat Nov 08, 2008 12:56 pm
    lachlanc wrote:
    If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

    1) m has more than 9 positive factors
    2) m is a multiple of p^3
    IMO it is H

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    cramya GMAT Titan Default Avatar
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    Post Sat Nov 08, 2008 1:28 pm
    Quote:
    If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

    1) m has more than 9 positive factors
    2) m is a multiple of p^3
    It would be B)


    Q: Is m a multiple of p^2 t

    Given: p and t are the only prime factors of m

    Stmt I

    FYI:
    We can obtain m's factors by adding 1 to the exponents of p and t and multiplying them

    Eg: No of factors of 6 are four(1,2,3 and 6)


    Express 6 as the product of its prime factors

    6= 2 ^ 1 * 3 ^1 (dropping the bases ; adding 1 to the 2's and 3's exponents and multiplying together)

    (1+1) (1+1) = 4


    m has more 9 positive factors. This could be p ^ 1 t ^ 9 Still m would have more than 9 positive factors. p^2 t is not a multiple of m

    or it could be p^ 2 t ^ 3 Then m is a multiple of p^2 t

    We dont know for sure INSUFF

    Stmt II

    m is a multiple of p^3 so m contains p ^ 3 in it somewhere(if not m cannot be a multiple of p^3). It is also given t is a prime factor therefore m will be a multiple of p^2 t

    SUFF

    B)

    Let me know if u still have questions

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    Post Sat Nov 08, 2008 1:44 pm
    Cramya, please help me to understand the statement II. I am still not clear about where the t is in P^3

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    cramya GMAT Titan Default Avatar
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    Post Sat Nov 08, 2008 1:48 pm
    t is not in p^ 3 ; t is in the given info (t is also a prime factor of m)

    Bidisha800 GMAT Destroyer! Default Avatar
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    Post Sat Nov 08, 2008 9:45 pm
    m is multiple of P^3

    P^2*t= P^3.P^(2t-3)=m*P^(2t-3)

    Therefore m must be a multiple of P^2*t

    (B)

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    Post Sat Nov 08, 2008 9:49 pm
    Bidisha800 wrote:
    m is multiple of P^3

    P^2*t= P^3.P^(2t-3)=m*P^(2t-3)

    Therefore m must be a multiple of P^2*t

    (B)
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    nitin86 Really wants to Beat The GMAT! Default Avatar
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    Post Thu Nov 13, 2008 9:10 am
    cramya wrote:
    Quote:
    If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

    1) m has more than 9 positive factors
    2) m is a multiple of p^3
    It would be B)


    Q: Is m a multiple of p^2 t

    Given: p and t are the only prime factors of m

    Stmt I

    FYI:
    We can obtain m's factors by adding 1 to the exponents of p and t and multiplying them

    Eg: No of factors of 6 are four(1,2,3 and 6)


    Express 6 as the product of its prime factors

    6= 2 ^ 1 * 3 ^1 (dropping the bases ; adding 1 to the 2's and 3's exponents and multiplying together)

    (1+1) (1+1) = 4


    m has more 9 positive factors. This could be p ^ 1 t ^ 9 Still m would have more than 9 positive factors. p^2 t is not a multiple of m

    or it could be p^ 2 t ^ 3 Then m is a multiple of p^2 t

    We dont know for sure INSUFF

    Stmt II

    m is a multiple of p^3 so m contains p ^ 3 in it somewhere(if not m cannot be a multiple of p^3). It is also given t is a prime factor therefore m will be a multiple of p^2 t

    SUFF

    B)

    Let me know if u still have questions
    @cramya,

    I think answer should be D.
    from the logic that you have provided for calculating the factors of any number.

    So, the logic was, if a number M has only 2 prime factors , we can write it as M = (P)^a * (T)^b, where P and T are prime factor and a and b are powers of these prime factor respectively.

    Now, Total number of positive factors of M can be calculated as
    (a+1) * (b+1),
    [This is by the same logic, as used by @cramya to calculate the factors of 6]

    Now, if according to STMT 1, M has 9 positive factors, then that means
    Total number of factors of M = 9 = (a+1) * (b+1)

    Now, (a+1) * (b+1) = 9, means that various values that (a+1) and (b+1) can take to make 9 are

    1) 9 * 1
    2) 1 * 9
    3) 3 * 3

    but if any of (a+1) or (b+1) takes value as 1, then that mean that mean either a or b has to be equal to ZERO.

    But according to the question, a and b can't be zero, as both P and T are prime factors of M.

    Hence, (a+1) and (b+1) both are equal to 3, or each of a and b is equal to 2

    Hence M = (P)^2 * (T)^2

    Hope this helps

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    Post Thu Nov 13, 2008 9:18 am
    the question says....MORE THAN 9 FACTORS

    your solution is for exactly 9 factors

    (a+1)(b+1) > 9

    a=1, b=4........one of the possibilities
    a=2, b=3........another

    I is insufficient

    stop@800 GMAT Destroyer!
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    Post Thu Nov 13, 2008 9:34 am
    Incorrect: Please ignore
    I assumed Option A to be
    "It has 9 factors".

    ============

    Avoid so much calculations.

    Lets look at what the question says:

    m is multiple of p and t
    we do not to what powers but each exists at least once.

    We need to find is m multiple of p^2*t
    that means do we have an extra p or not


    A
    9 factors
    9 can only be derived by 3*3 [using product of integers]
    hence m has to be of the form p^2 t^2
    hence extra p is in m
    so Suff

    B
    multiple of p^3
    we just needed an extra p
    but here we got two Smile
    Sufficient

    Ans IMO is D


    The catch here is:
    People will commit mistake by not dividing 9 into factors.


    cramya,
    p ^ 1 t ^ 9 will have
    (1+1)(9+1) = 20 factors

    remeber the a,b rule

    p1^a p2^b will have
    (a+1)(b+1) factors



    Last edited by stop@800 on Thu Nov 13, 2008 10:40 am; edited 1 time in total

    lachlanc Rising GMAT Star Default Avatar
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    Post Thu Nov 13, 2008 10:31 am
    OA is B. You the man, cramya.

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