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Math Help - Consecutive integers (SAT)

  1. #1
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    Consecutive integers (SAT)

    Hi,

    If j, k, and n are consecutive integers such that 0 < j < k < n and the units digit of the product jn is 9, what is the units digit of k?

    A. 0
    B. 1
    C. 2
    D. 3
    E. 4

    I guessed E once but my final answer is C.
    The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    I guessed E once but my final answer is C.
    The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.
    j,k,n are bigger than zero yes, but might not be 1,2,3 as you have suggested. Instead make them m, m+1 and m+2.
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  3. #3
    Super Member Quacky's Avatar
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    Edit: Nevermind, was too slow at typing
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    Quote Originally Posted by pickslides View Post
    j,k,n are bigger than zero yes, but might not be 1,2,3 as you have suggested. Instead make them m, m+1 and m+2.
    I just realized that only j, k, and n are consecutive.
    jn = 9

    j = 9

    n = 11

    9 \times 11 = 99

    Number intervening 9 and 11 is 10, therefore units digit of k is 0.
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  5. #5
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    Also these pairs work

    19 21
    29 31
    39 41
    49 51
    59 61
    ... ...
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  6. #6
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    This is a problem where you just want to start listing and look for the pattern

    1 2 3
    2 3 4
    3 4 5
    4 5 6
    ......

    In less than 30 seconds you should discover that j=9, k=10, n=11 are the first set of numbers that "work."
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  7. #7
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    Quote Originally Posted by Hellbent View Post
    Hi,

    If j, k, and n are consecutive integers such that 0 < j < k < n and the units digit of the product jn is 9, what is the units digit of k?

    A. 0
    B. 1
    C. 2
    D. 3
    E. 4

    I guessed E once but my final answer is C.
    The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.
    Also, since the integers are m, m+1, and m+2, you can use modular arthimetic.

    m(m+2)\equiv 9 \ \mbox{(mod 10)}

    Therefore, m = 9+10t \ \ t\in\mathbb{Z} \ \ \ t\geq 0.

    j=9+10t, \ k=(9+10t)+1, \ n=(9+10t)+2
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