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Thread: OG 10th Q48

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    OG 10th Q48

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    If n is a positive integer, then n(n+1)(n+2) is

    a) even only when n is even
    b) even only when n is odd
    c) odd whenever n is odd
    d) divisible by 3 only when n is odd
    e) divisible by 4 whenever n is even

    Please, explain me how to solve such problems. Thanks.

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    n(n+1)(n+2) is a fancy way of saying the product of three consecutive integers

    if n is odd, n+1 is even, and n+2 is odd
    n(n+1)(n+2) = odd x even x odd = even x odd = even

    if n is even, n+1 is odd, and n+2 is even
    n(n+1)(n+2) = even x odd x even = even

    (if you don't understand this, you should review number properties of odd's and evens)

    from above, n is even regardless if n is even or odd; thus A and B are false we can eliminate them.

    when n is odd, n(n+1)(n+2) is even so C is false.

    Now we have choices D and E left.

    choice D:
    any three consecutive positive integers will have one of the numbers divisible by 3.

    proof:
    when a positive integer is divided by 3, it has remainders of 0 (i.e. divisible by 3), 1, or 2
    if n has a remainder of 0, it is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3
    if n has a remainder of 1, n+1 has a remainder of 2, and n+2 has a remainder of 0. n+2 is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3
    if n has a remainder of 2, n+1 has a remainder of 0, and n(n+1)(n+2) is divisible by 3

    Eliminate D.

    We could pick by process of elimination but for the sake of completion:

    If n is even, n can be expressed as 2k, for k an integer
    if n = 2k
    then n + 2 is also even and n + 2= 2m for m an integer

    n(n+1)(n+2) = 2k(n)(2m) = 4knm
    here knm is an integer, so 4knm and consequently n(n+1)(n+2) is divisible by 4.

    enjoy!

    p.s. for more OG alternate explanations, click here
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    another way you can do this problem is to try and find counterexamples:

    n(n+1)(n+2)

    a) even only when n is even
    if n=3 3 x 4 x 5 = 60 n is odd and the product is even FALSE.

    b) even only when n is odd
    if n=2 2 x 3 x 4 is clearly even FALSE

    c) odd whenever n is odd
    from a, when n is odd the product is even FALSE

    d) divisible by 3 only when n is odd
    if n=6, 6 x 7 x 8 is divisible by 6 and so also must be divisible by 3 FALSE

    e) divisible by 4 whenever n is even
    Process of elimination, only choice left.

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    Thank you very much.

    Gmathintsdotcom, i have some more question if you dont mind. I dont want to open a new thread. So I think i could post them right here. Could I use your further help?

    OG 10th Q80
    Jack is now 14 years older than Bill. If in 10 years Jack will be twice as old as Bill, how old will Jack be in 5 years?

    a 9
    b 19
    c 21
    d 23
    e 33

    Q92

    Which of the following CANNOT be the greatest common divisor of two positive integers x and y?

    a 1
    b x
    c y
    d x-y
    e x+y

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    sure.

    OG 10th Q80
    Jack is now 14 years older than Bill. If in 10 years Jack will be twice as old as Bill, how old will Jack be in 5 years?

    a 9
    b 19
    c 21
    d 23
    e 33

    j = jack's age now
    b = bill's age now
    j = 14 + b (i.e. jack is 14 years older than bill)

    (j+10)=2(b+10)
    j+10 = 2b + 20
    j = 2b + 10

    substitute
    14 + b = 2b + 10
    b = 4

    bill is now 4
    jack is now 18

    in 5 years, jack will be 23. The answer is D.



    Q92 (This is question 98 in the OG Quant Review 2nd Edition - Green Book)

    Which of the following CANNOT be the greatest common divisor of two positive integers x and y?

    a 1
    b x
    c y
    d x-y
    e x+y

    a) Can 1 we be the greatest common divisor of two positive integers x and y?
    Yes. Suppose x and y are prime. (e.g. the gcd of 2 and 3 is 1)

    b) yes, suppose x = y or x = 2 and y = 4

    c) yes, suppose x = y or y = 2 and x = 4

    d) yes, suppose x = 4 and y = 2, the GCD is 2 and x-y = 2

    All of these can be the GCD so we could stop here.

    e) A number cannot be divisible by a larger number. Could 5 be divisible by 17. No. 5 cannot be divisible by any number larger than 5.

    because x and y are positive, x + y > x and x + y > y

    y / (x+y) is <1 and will not be an integer and hence not divisible

    x / (x+y) is <1 and will not be an integer and hence not divisible

    The answer is E.
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    gmathintsdotcom thank you!

    I really need some tutor like you

    Still have a lot of questions. If you dont mind AGAIN I would post my questions right here. Would be very happy and pleasant to learn something from you.

    P.S. By the way, I also do need some advanced book for math theory. All books that I bought are full bull *****. Barrons, McGraw, even OG. All I got in that books are basics which dont help at all. Could you suggest me something really effective to read?

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    If x=a/3+b/3^2+c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

    A. 1/27
    B. 1/9
    C. 4/27
    D. 2/9
    E. 4/9

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    sure.

    If x=a/3+b/3^2+c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

    A. 1/27
    B. 1/9
    C. 4/27
    D. 2/9
    E. 4/9

    Rewrite the equation
    x = a/3 + b/9 + c/27

    if it helps, further we could rewrite it as
    x = (1/3)a + (1/9)b + (1/27)c
    x = (9/27)a + (3/27)b + (1/27)c

    a, b, c can be either 0 or 1

    starting with choice A:
    1/27 possible? Yes, let c=1 and a=0, b=0

    choice B
    1/9 or 3/27 possible? Yes, let b=1 and a=0, c=0

    choice C
    4/27 possible? Yes, let b=1 and c=1 and a=0


    choice D
    2/9 or 6/27 possible? No possible way.

    choice E
    4/9 or 12/27 possible? Yes, let a=1 and b=1 and c=0

    Choice D is not possible.
    to get 6/27, clearly a cannot be 1.
    if b=1 and c=0, we end up with 3/27
    if b=1 and c=1, we end up with 4/27
    if b=0 and c=1, we end up with 1/27

    As far as books for math theory, I would say Manhattan GMAT books are pretty good and all you will need (for quant at least). I have written about my gmat experience here

    While I have not used them personally, Veritas Prep and EZ Solutions books do get good reviews on Amazon so they may be worth checking out.

    best.
    Visit my gmat website: - www.gmathints.com

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    Quote Originally Posted by fr743 View Post
    If n is a positive integer, then n(n+1)(n+2) is

    a) even only when n is even
    b) even only when n is odd
    c) odd whenever n is odd
    d) divisible by 3 only when n is odd
    e) divisible by 4 whenever n is even

    Please, explain me how to solve such problems. Thanks.
    Plug in some values for n

    Let n be 1 then n(n+1)(n+2) is = 1*2*3 =>6

    Let n be 2 then n(n+1)(n+2) is = 2*3*4 =>24

    Let n be 3 then n(n+1)(n+2) is = 3*4*5 => 60

    Let n be 4 then n(n+1)(n+2) is = 4*5*6 => 120

    Let n be 5 then n(n+1)(n+2) is = 5*6*7 => 210

    Now go thru the options.

    a) even only when n is even

    We can find that it is not true , irrespective of the value of n ( be it odd or even) the product is always even


    b) even only when n is odd

    We can find that it is not true , irrespective of the value of n ( be it odd or even) the product is always even

    c) odd whenever n is odd

    False see the values of n = 1 , 3 , 5 ; U will find the value is always even

    d) divisible by 3 only when n is odd

    All the numbers are divisible by 3 , irrespective of the values of n ( Be it odd or even)

    e) divisible by 4 whenever n is even

    True , all of them are divisible by 4 except n =5 , so we can claim all even values of n are divisible by 5

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    Quote Originally Posted by fr743 View Post
    gmathintsdotcom thank you!

    I really need some tutor like you

    Still have a lot of questions. If you dont mind AGAIN I would post my questions right here. Would be very happy and pleasant to learn something from you.

    P.S. By the way, I also do need some advanced book for math theory. All books that I bought are full bull *****. Barrons, McGraw, even OG. All I got in that books are basics which dont help at all. Could you suggest me something really effective to read?
    Kaplan is usually a very good source for tough questions. I also heard that their CD that comes with several books is pretty helpful!

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