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Thread: real hard GMAT question - 800score!

  1. #1
    Eager! cinghal1's Avatar
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    Question real hard GMAT question - 800score!

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    Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?

    a. 10%
    b. 25%
    c. 50%
    d. 90%
    e. 100%

    PLEASE EXPLAIN YOUR ANSWER!

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    Trying to make mom and pop proud EEMan's Avatar
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    I'm not sure about the answer, but IMO, if 678,463 was a multiple of 6^K, 678,463 would be divisible by 6. But since 678,463 is not divisible by 6, so definitely the probability that 678,463 is not a multiple of 6^K is 100%.

    Are you sure that Z=6^K? or K^6?

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    Eager! cinghal1's Avatar
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    EEMan - super explainition.
    I guess you are right.
    I just got so intimated by the text of the Q - couldnt think its so simple!!

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    Given: 6^k

    It means at the bare minimum 2 should be a factor (2 is factor 6 in 6^k). As the last digit in the given number 678463 is not divisible by 2 the result is 100%

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    Hey man.

    Since 678,463 is not divisible by 6. That means the only case where it could be a multiple of z is when k (in 6^k) is 0, as that would make your z value equal to 1, and we know that anything is divisible by 1.
    For k to be equal to 0, our integer value which can range from 0-10 can only be 0 for 678,463 to be a mutiple of.

    Since there are 5 numbers in the set, there can be 5 events where our z value is 0.
    The total number of possible events is: (the size of our set)x(the number of available integer values) since there are 5 numbers in our set and 11 available integer values(0,1,2,3,4,5,6,7,8,9,10) the total number of possible events is : 11*5=55.

    (probability of our number being a multiple of z) = (the number of events where our z value is 0)/(the total number of events) = 5/55 = .091

    probablility of our number NOT being a multiple z) = 1 - (probability of our number being a multiple of z) = 1 - (.091) = 0.9

    the answer is therefore 90%.

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    Actually the integers range from 0-9 giving us a probability of exactly 90%. But i agree with your method.

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    Re: real hard GMAT question - 800score!

    The credit for answering this question goes to sadz4u. But, here is a small correction. Actually, we have 10 first non-negative integers which are the numbers from 0 to 9. Then, if we consider the probability of our required number NOT being a multiple of Z then there are 9 such integers out of 10.

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