Find the Integers

Discussion in 'Algebra' started by nycmathguy, Jan 26, 2022.

  1. nycmathguy

    nycmathguy

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    I want to know if the set up is correct.

    There are three consecutive odd integers. Three times the largest is seven times the smallest. Find the integers.

    Set Up:

    Three consecutive odd integers:

    2n + 1, 2n + 3, 2n + 5.

    Let 2n + 5 = largest integer.

    Let 2n + 1 = smallest integer.

    3(2n + 5) = 7(2n + 1)

    Yes?
     
    nycmathguy, Jan 26, 2022
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  2. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Jan 26, 2022
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  3. nycmathguy

    nycmathguy

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    Ok. I wasn't too sure about this one. Not sure? After months playing with Precalculus, not sure?
     
    nycmathguy, Jan 26, 2022
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  4. nycmathguy

    Country Boy

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    You could have written the "three consecutive odd integers" as x, x+ 2, and x+ 4. 3(x+ 4)= 7x.
    3x+ 12= 7x
    Subtracting 3x from both sides:
    4x= 12
    Dividing both sides by 4:
    x= 3.

    The three consecutive odd integers are 3, 5, and 7.

    Check: "three times the largest" is 3(7)= 21 and "seven times the smallest" is 7(3)= 21.
     
    Country Boy, Jan 26, 2022
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  5. nycmathguy

    nycmathguy

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    So, am I right or wrong? Let me guess: WRONG! SHAME ON ME!!
     
    nycmathguy, Jan 26, 2022
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  6. nycmathguy

    MathLover1

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    the general form of an odd integer can be expressed as n is n=2k+1, where k is an integer, and a pair of any consecutive odd integers are also 2 units apart
    2k+3 is the second
    2k+5 is the third odd integer
     
    MathLover1, Jan 26, 2022
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  7. nycmathguy

    MathLover1

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    you are right
     
    MathLover1, Jan 26, 2022
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  8. nycmathguy

    nycmathguy

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    Thank you. This Country Boy guy can't stop criticizing my effort to present correct calculations.
    Remember: Precalculus on my days off only.
     
    nycmathguy, Jan 27, 2022
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  9. nycmathguy

    Country Boy

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    I pointed out a different way of doing the problem. That is NOT "criticism"!

    If you write the three consecutive odd integers as n, n+ 2, and n+ 4 then "three times the largest is seven times the smallest is "3(n+ 4)= 7n". 3n+ 12= 7n. 4n= 12. n= 3.

    Your formulation gave "3(2n + 5) = 7(2n + 1)". 6n+ 15= 14n+ 7. 15= 8n+ 7. 8n= 8. n= 1. But you had written the first number as 2n+ 1 so the first number is 2+ 1= 3, as before.

    I would also encourage you (not intended as criticism!) to check the answer:
    The three consecutive integers are 3, 5, and 7. "3 times the largest" is 3 times 7= 21. "7 times the smallest" is 7 times 3= 21.
     
    Country Boy, Jan 28, 2022
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  10. nycmathguy

    nycmathguy

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    A different way is ok by me.
     
    nycmathguy, Jan 29, 2022
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