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Divisibility test

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krishna239455 Really wants to Beat The GMAT! Default Avatar
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Divisibility test Post Tue Mar 27, 2012 8:37 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Problem is stated below:
    x is a positive number
    (9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is Y divisible by 5?
    A)5 is a factor of x
    B)x is a integer

    I have doubt whether this can be a data suffeciency problem because,
    I can simplify the above expression as: 9^x(1+9^1+9^2+9^3+9^4+9^5)=y
    and i know that even power of 9 gives digit with 1 in units place and odd poer gives digit with 9 in units place. With this logic if i add all the terms in the bracket i should get digit with 0 in units place. This digit will always be divisible by 5. No need of statements given below. Kindly correct me if i am wrong!!!!

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    Mike@Magoosh GMAT Instructor
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    Post Tue Mar 27, 2012 3:23 pm
    Hi, there. I'm happy to give my two cents with this. Smile

    Prompt:
    x is a positive number
    (9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is y divisible by 5?


    I totally agree with your approach. . . .

    y = (9^x)*(1+9^1+9^2+9^3+9^4+9^5)

    The parenthesis indeed is a multiple of 10, so must be divisible by 5. If x = 0 or a positive integer, than y will definitely be divisible by 5.

    IF x is a positive rational number, y will not be a whole number. If x is a positive irrational number, for example a logarithm, then y could be either a non-integer or even an integer not divisible by 5 --- (for example, if x = log(17/5)/log(9) --- but that's miles and miles beyond what you need to know for the GMAT.)

    I am deeply suspicious of this question, because the "catch" in it seems to depend on an understanding of logarithms, which again is way way beyond the realm of GMAT math. I think someone was trying too hard to write a challenging DS question, and lost all track of what math is actually tested on the GMAT.

    So, in short, I agree with you that this is a flawed DS question.

    Here's a much more GMAT-like DS question involving laws of exponents.
    http://gmat.magoosh.com/questions/1003
    When you submit your answer to that question, the following page will have a complete video explanation. At Magoosh, each of our over 800+ questions has its own video explanation. If you are looking for high quality GMAT prep material at a surprisingly affordable price, now is particular good time to check out Magoosh, because we are having a sale that ends later this week.

    Does everything I've said make sense? Let me know if you have any further questions.

    Mike Smile

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    krishna239455 Really wants to Beat The GMAT! Default Avatar
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    Post Tue Mar 27, 2012 10:44 pm
    Thanks Mike

    rahulvsd Really wants to Beat The GMAT! Default Avatar
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    Post Thu Mar 29, 2012 8:44 am
    Hi Mike,

    Could you please explain how Statement 2 in the Magoosh example (from the link you have provided) is insufficient ?

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    Mike@Magoosh GMAT Instructor
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    Post Thu Mar 29, 2012 11:45 am
    rahulvsd wrote:
    Hi Mike,

    Could you please explain how Statement 2 in the Magoosh example (from the link you have provided) is insufficient ?
    Prompt: If x and y are integers and xy ≠ 0, what is the value of [x^(-2y)]/[y^(2x)]?

    Well, that expression simplifies a little to

    [x^(-2y)]/[y^(2x)] = 1/[x^(2y)*y^(2x)] = 1/[x^y*y^x]^2

    Statement #2: xy = y/x

    Well, with this expression, first multiply by x to clear the fraction, and we get:

    (x^2)*y = y

    Since we know y does not equal zero, we can divide by y, which results in:
    x^2 = 1
    x = +/-1
    (Always remember you need to include the +/- when you take the squareroot of a variable.)

    Thus, we know x = +/-1 and we have absolutely no idea what y is --- y could be absolutely an integer other than zero. That's why statement #2 is insufficient.

    Does this make sense? Let me know if you have any further questions.

    Mike Smile

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    amit.trivedi@ymail.com GMAT Destroyer!
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    Post Fri Mar 30, 2012 1:47 am
    Well there are 2 cases in here...

    Case (i): when x is a positive integer divisible by 5, in that case the value of x can only be 0, or multiples of 5.

    In that case see the equation and overall there are 6 terms i.e even number of terms.

    we know that number 9 has a cyclicity of 9 and 1. only 2 numbers

    if x = 0, then the terms last digit would be - 1,9,1,9,1,9. the total comes out to be 10 in short the last digit has a 0 which is divisible by 5.

    same for 5 and 10, the total comes out to be 10 and the last digit has a 0 in the end.

    In case i, the answer comes out to be A, as we do not want the statement 2.


    Case (ii): The value of x can be a negative multiple of 5, in that case the overall Y will be in decimal terms...

    IN case of decimal terms it is very difficult Eg: 9^-1 which is (1/9)

    and also 9^-2, the answer is (1/81) so when you will add the 6 numbers you will still get that the overall numerator has 0 in the last place.

    the numerator has 0 in its last place, the denominator is 9 and when you multiply the overall fraction by 5 the answer is a decimal fraction left by the denominator 9 but the numerator is divisible 5.

    In this case the answer comes out to be E, then as the resulting number is having 9 as the numerator...


    Hope my explanation is helpful to you guyzzzzz


    Thank YOU..

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    sanju09 GMAT Instructor
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    Post Fri Mar 30, 2012 4:15 am
    krishna239455 wrote:
    Problem is stated below:
    x is a positive number
    (9^x)+(9^(x+1))+(9^(x+2))+(9^(x+3))+(9^(x+4))+(9^(x+5))=y, Is Y divisible by 5?
    A)5 is a factor of x
    B)x is a integer

    I have doubt whether this can be a data suffeciency problem because,
    I can simplify the above expression as: 9^x(1+9^1+9^2+9^3+9^4+9^5)=y
    and i know that even power of 9 gives digit with 1 in units place and odd poer gives digit with 9 in units place. With this logic if i add all the terms in the bracket i should get digit with 0 in units place. This digit will always be divisible by 5. No need of statements given below. Kindly correct me if i am wrong!!!!
    Received a PM from amit.trivedi@ymail.com to comment:


    We can take

    y = 9^x (1 + 9 + 9^2 + 9^3 + 9^4 + 9^5) = 9^x {(9^6) - 1}/8.

    The part {(9^6) - 1}/8 is a positive integer divisible by 5 because

    (9^6) - 1 = {(9^3) - 1} {(9^3) + 1} = {8 (81 + 9 + 1)} {10 (81 - 9 + 1)}.

    Now, only one thing can hamper y to be divisible by 5, and that is if x is other than a whole number. Let’s browse the statements:

    I. If 5 is a factor of x, then y is divisible by 5 if x is a whole number and not divisible by 5 if x is a negative integer. Insufficient

    II. If x is an integer, then y is divisible by 5 if x is non negative and not divisible by 5 if x is negative. Insufficient

    Combining the two statements, we know that x is an integer divisible by 5; and still, y is divisible by 5 if x is non negative and not divisible by 5 if x is negative. Insufficient

    Take E

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    amit.trivedi@ymail.com GMAT Destroyer!
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    Post Fri Mar 30, 2012 10:15 am
    Thank You Sanju 09!!! this means that my calculations were right...

    I took some time in the trial and error, but finally got it...

    Thanks once again...

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