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Divisibility

This topic has 2 expert replies and 3 member replies
GmatKiss Legendary Member Default Avatar
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Divisibility

Post Sat May 05, 2012 11:41 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If n = 20! + 17,then n is divisible by which of the following?

      I.15

      II.17

      III.19

    A.None
    B.I only
    C.II only
    D.I and II
    E.II and III

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    aneesh.kg Master | Next Rank: 500 Posts
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    Post Sat May 05, 2012 2:51 pm
    n = 20! + 17

    20! is a multiple of 15, 17 as well as 19.
    So lets write 20! as 15X, 17Y and 19Z, where X, Y and Z are integers.

    n = 15X + 17
    n = 15X + 15 + 2
    n = 15(X + 1) + 2
    (Comparing with Dividend = Divisor*Quotient + Remainder)
    i.e., n leaves a remainder of 2 when divided by 15.
    I ruled out.

    n = 17Y + 17
    n = 17(Y + 1)
    (Comparing with Dividend = Divisor*Quotient + Remainder)
    i.e., n leaves a remainder of 0 when divided by 17, or it is completely divisible by 17.
    II is correct.

    n = 19Y + 17
    (Comparing with Dividend = Divisor*Quotient + Remainder)
    i.e., n leaves a remainder of 17 when divided by 19.
    III ruled out.

    II only.
    (C) is the answer.

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    amanpreet Junior | Next Rank: 30 Posts
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    Post Sat Apr 20, 2013 8:56 am
    20! 17
    -- + --
    15 15


    20! 17
    --- + -- this one is possible so Ans is c
    17 17


    20! 17
    -- + --
    19 19

    karan.7045 Junior | Next Rank: 30 Posts Default Avatar
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    Post Sun Apr 21, 2013 8:31 am
    Since 20! is divisble by 15,17 and 19 but as it is added by 17 so it is nly divisble by 17 so C is the answer.

    Post Sun Apr 21, 2013 6:21 pm
    GmatKiss wrote:
    If n = 20! + 17,then n is divisible by which of the following?

      I.15

      II.17

      III.19

    A.None
    B.I only
    C.II only
    D.I and II
    E.II and III
    Answer choice I: is 20! + 17 divisible by 15?
    20! + 17 = (20)(19)(18)(17)(16)(15)(other stuff) + 15 + 2
    = (15)(some number + 1) + 2
    So, if we divide (15)(some number + 1) + 2 by 15, the remainder will be 2
    So, 20! + 17 is NOT divisible by 15


    Answer choice II: is 20! + 17 divisible by 17?
    20! + 17 = (20)(19)(18)(17)(other stuff) + 17
    = (17)(some number + 1)
    If we divide (17)(some number + 1) by 17, the remainder will be 0
    So, 20! + 17 IS divisible by 17


    Answer choice III: is 20! + 17 divisible by 19?
    20! + 17 = (20)(19)(other stuff) + 17
    = (19)(some number) + 17
    If we divide (19)(some number) + 17 by 19, the remainder will be 17
    So, 20! + 17 is NOT divisible by 19

    Answer = C

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    Post Wed Jun 24, 2015 3:08 am
    GmatKiss wrote:
    If n = 20! + 17,then n is divisible by which of the following?

      I.15

      II.17

      III.19

    A.None
    B.I only
    C.II only
    D.I and II
    E.II and III
    Solution:

    We are given that n = 20! + 17 and need to know whether n is divisible by 15, 17, and/or 19. To determine this, we rewrite the given expression for n using each answer choice.

    Thus, we have:

    Does (20! + 17)/15 = integer?

    Does (20! + 17)/17 = integer?

    Does (20! + 17)/19 = integer?

    We now use the distributive property of division over addition to determine which of these expressions is/are equal to an integer.

    The distributive property of division over addition tells us that (a + c)/b = a/b + c/b. We apply this rule as follows:

    I.

    Does (20! + 17)/15 = integer?

    Does 20!/15 + 17/15 = integer?

    Although 20! is divisible by 15, 17 is NOT, and thus (20! + 17)/15 IS NOT an integer.

    We can eliminate answer choices B and D.

    II.

    Does (20! + 17)/17 = integer?

    Does 20!/17 + 17/17 = integer?

    Both 20! and 17 are divisible by 17, and thus (20! + 17)/17 IS an integer.

    We can eliminate answer choice A.

    III.

    Does (20! + 17)/19 = integer?

    Does 20!/19 + 17/19 = integer?

    Although 20! is divisible by 19, 17 is NOT, so (20! + 17)/19 IS NOT an integer.

    We can eliminate answer choice E.

    Thus, II is the only correct statement.

    Answer:C

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