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GMATPrep - Practice Test #1 - Problem Solving - Need Help

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dominate11 Newbie | Next Rank: 10 Posts Default Avatar
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GMATPrep - Practice Test #1 - Problem Solving - Need Help

Post Wed Feb 28, 2007 7:19 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Can someone please help explain these questions.

    1. 2^(4-1)^2 / 2^(3-2) = ?

    3. For all positive integers m, <m> = 3m when m is odd and <m> = .5m when m is even. Which of the following is equivalent to <9> X <6> ?

    11. In n is a postive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n? 1) 10 2) 11 3) 12 4) 13 5) 14

    I appreciate any guidance...

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    Neo2000 Legendary Member
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    Post Wed Feb 28, 2007 9:03 am
    For 1
    2 ^(3)^2 = 2^9/2 = 2^8 = 256

    dominate11 wrote:
    3. For all positive integers m, <m> = 3m when m is odd and <m> = .5m when m is even. Which of the following is equivalent to <9> X <6> ?
    <9> implies that m is odd => <9> = 3(9) = 27
    Similarly for <6> = (.5)6 = 3

    So finally you get 27 X 3 = 81



    Last edited by Neo2000 on Wed Mar 07, 2007 8:00 am; edited 1 time in total

    gabriel Legendary Member
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    Post Fri Mar 02, 2007 8:24 am
    dominate11 wrote:
    Can someone please help explain these questions.

    1. 2^(4-1)^2 / 2^(3-2) = ?

    3. For all positive integers m, <m> = 3m when m is odd and <m> = .5m when m is even. Which of the following is equivalent to <9> X <6> ?

    11. In n is a postive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n? 1) 10 2) 11 3) 12 4) 13 5) 14

    I appreciate any guidance...
    ok so the first has been answered by neo.. so let me get to the other two

    3.) since <m>= 3m when m is odd ... therefore < 9> = 3*9 since 9 is odd... also <m> = 0.5*m when m is even ... therefore <6>=.5*6 since 6 is even... therefore the answer to <9> X <6> = 27*3=81.... btw i am assuming that X means the multiplication..

    11.) okay 990= 9*11*10... therefore from 1 to n we need to have 9,10 and 11 and the least value for such a n is 11.... because the sequence of numbers from 1 to 11 includes all the three no ...

    2 more questions in the same league that should help u understand the concept better... find the least n such that the product of numbers from 1 to n is a multiple of 350..... also find the least n such that the product from 1 to n is a multiple of 455... try solving them and let me know if u dont get it ...

    anuroopa Junior | Next Rank: 30 Posts Default Avatar
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    Post Wed Mar 07, 2007 4:01 am
    Hi
    I am not sure if Neo's answer is right - according to the exponent rule- (x^y)^z = x^zy

    So,
    so it should be 2^6 - 2^1= 2^5

    Neo, could you elaborate on ho u got ur answer

    Neo2000 Legendary Member
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    Post Wed Mar 07, 2007 7:33 am
    anuroopa wrote:
    Hi
    I am not sure if Neo's answer is right - according to the exponent rule- (x^y)^z = x^zy
    While this is correct, what the question is asking for is x^(y)^z which is how i got the answer.

    Hope this helps. If not, do let me know Smile

    gabriel Legendary Member
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    Post Wed Mar 07, 2007 7:58 am
    anuroopa wrote:
    Hi
    I am not sure if Neo's answer is right - according to the exponent rule- (x^y)^z = x^zy

    So,
    so it should be 2^6 - 2^1= 2^5

    Neo, could you elaborate on ho u got ur answer
    Nope..... neo has got it right... the only thing wrong with his answer is that 2^8 = 256... not 64

    Neo2000 Legendary Member
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    Post Wed Mar 07, 2007 8:01 am
    gabriel wrote:
    [
    Nope..... neo has got it right... the only thing wrong with his answer is that 2^8 = 256... not 64
    whatcha talking about?? i did write 2^8 = 256



    <wicked grin>

    anuroopa Junior | Next Rank: 30 Posts Default Avatar
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    Post Wed Mar 07, 2007 8:46 am
    thnks neo - I understood the q wrong

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