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anonymous

  • one year ago

Find three consecutive odd integers whose sum is 72 more than the smallest integer.

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  1. welshfella
    • one year ago
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    let smallest integer be x then x + x + 2 + x + 4 = x + 72

  2. anonymous
    • one year ago
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    ^^ and add -2 and -4 to both sides. That results in 3x=66+x Subtract x from each side, and you get 2x=66 Therefore, x=33!

  3. zzr0ck3r
    • one year ago
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    Same thing we did last time, but use the smallest number on the right hand side of the equality.

  4. anonymous
    • one year ago
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    Okay so the smallest consecutive odd integer is X. but I'm still confused on the 2nd consecutive odd integer and the largest integer. Would the 2nd be x+2?

  5. anonymous
    • one year ago
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    Yes. and the third integer would be (x+4).

  6. welshfella
    • one year ago
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    the next odd number is 2 after the first Remember thers is an even number in between

  7. anonymous
    • one year ago
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    I would be careful, @welshfella . Your expressions for the three integers does not require them to be odd. The solution will be odd only by luck. Generically, three consecutive odd integers are best represented by 2x+1, 2x+3, and 2x+5, where x is an integer. This GUARANTEES that the integers are odd.

  8. anonymous
    • one year ago
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    Thank you all so much! I understand it clearly now

  9. welshfella
    • one year ago
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    yes @ospreytriple - I took the chance. I'm lazy that's my problem!

  10. anonymous
    • one year ago
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    Me too! :)

  11. anonymous
    • one year ago
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    lol.

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