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Math Help - 3 Consecutive Odd Integers

  1. #1
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    3 Consecutive Odd Integers

    What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by magentarita View Post
    What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?
    Let the largest of the odd integers be 2x+1, then the others are 2(x-1)+1 and 2(x-2)+1 (where x is an integer).

    So we are told that: [2(x-2)+1][2x+1]=3[2(x-1)+1]+6

    This is a quadratic in x, so solve it and choose the root (or roots) that makes sense.

    CB
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    Hello, magentarita!

    Another approach . . .


    What is the largest of three consecutive odd integers if the product
    of the first and third integers is 6 more than three times the second integer?
    Consecutive odd integers "go up by two's".

    Let x = the smallest odd integer.
    Then (x+2)\text{ and }(x+4) are the next two odd integers.


    . . \underbrace{\text{Product of first and third}} _{x(x+4)}\; \underbrace{\text{is}}_{=} \; \underbrace{\text{6 more than}}_{6\;\; +} \; \underbrace{\text{3 times second}}_{3(x+2)}


    And we have: . x(x+4) \;=\;3(x+2)+6 \quad\Rightarrow\quad x^2 + x - 12 \:=\:0


    Solve the quadratic equation (and take the odd root).

    And remember, they asked for the largest of the three.

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  4. #4
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    ok.......

    Quote Originally Posted by CaptainBlack View Post
    Let the largest of the odd integers be 2x+1, then the others are 2(x-1)+1 and 2(x-2)+1 (where x is an integer).

    So we are told that: [2(x-2)+1][2x+1]=3[2(x-1)+1]+6

    This is a quadratic in x, so solve it and choose the root (or roots) that makes sense.

    CB
    Great work!
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  5. #5
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    ok.........

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Another approach . . .

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

    Let x = the smallest odd integer.
    Then (x+2)\text{ and }(x+4) are the next two odd integers.


    . . \underbrace{\text{Product of first and third}} _{x(x+4)}\; \underbrace{\text{is}}_{=} \; \underbrace{\text{6 more than}}_{6\;\; +} \; \underbrace{\text{3 times second}}_{3(x+2)}


    And we have: . x(x+4) \;=\;3(x+2)+6 \quad\Rightarrow\quad x^2 + x - 12 \:=\:0


    Solve the quadratic equation (and take the odd root).

    And remember, they asked for the largest of the three.
    I save your replies for future reference.
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