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Math Help - Integer Problem

  1. #1
    Mad
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    Integer Problem

    Five times the smallest of three consecutive odd integers is ten more than twice the largest. Find the integers.


    EDIT: Mind you, I just need help with setting it up. I can solve it, I'm just having a tough time writing it down properly.
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  2. #2
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    Hello, Mad!

    Five times the smallest of three consecutive odd integers
    is ten more than twice the largest. Find the integers.

    Consecutive odd integers "go up by 2's".


    Let . \begin{array}{ccc}x &=& \text{1st number} \\ x+2 &=& \text{2nd number} \\ x+4 &=& \text{3rd number}\end{array}


    \underbrace{\text{Five times smallest}}\;\;\underbrace{\text{is}}\;\;\underbrac  e{\text{twice largest}}\;\underbrace{\text{plus 10}}

    . . . . 5 \times x \qquad\qquad =\qquad2(x+4)\qquad + 10


    And there is your equation: . 5x \;=\;2(x+4) + 10

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  3. #3
    Mad
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    Thanks, I see how it's done now.


    EDIT: Wait...shouldn't there be an x+2 in there somewhere?
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  4. #4
    Moo
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    Quote Originally Posted by Mad View Post
    Thanks, I see how it's done now.


    EDIT: Wait...shouldn't there be an x+2 in there somewhere?
    Hello,

    Nope, because they're only asking you things about the smallest and the largest of the three integers :-)
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  5. #5
    Mad
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    Ah, I see. Thanks.
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  6. #6
    Senior Member Twig's Avatar
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    hmm

    Uh, doesnt 5x = 2x + 18 give x = 6

    werent the integers supposed to be odd?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mad View Post
    Five times the smallest of three consecutive odd integers is ten more than twice the largest. Find the integers.


    EDIT: Mind you, I just need help with setting it up. I can solve it, I'm just having a tough time writing it down properly.
    Quote Originally Posted by Twig View Post
    Uh, doesnt 5x = 2x + 18 give x = 6

    werent the integers supposed to be odd?
    You are correct. (Good spot! )

    So let's redo this with the smallest number being 2x + 1 and the largest being 2x + 5:
    5(2x + 1) = 2(2x + 5) + 10

    10x + 5 = 4x + 10 + 10

    6x = 15

    As there is no integer solution for x the problem does not have a solution. (In fact you can figure this out just by doing an "even-odd" argument: 5 times an odd number is odd; 2 times an odd number plus 10 is even. So the two sides can't be equal.)

    -Dan
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  8. #8
    Senior Member Twig's Avatar
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    hey

    hi dan

    yeah, I just felt that there was something off about this

    Yeah smart thinking with the odd times odd..

    Odd*odd = odd

    Odd*even = even
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Twig View Post
    hi dan

    yeah, I just felt that there was something off about this

    Yeah smart thinking with the odd times odd..

    Odd*odd = odd

    Odd*even = even
    You thought there was something "odd" about it?

    (Ahem. Sorry about that.)

    -Dan
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  10. #10
    Senior Member Twig's Avatar
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    lol

    haha yeah! =)

    Def. something odd about it
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