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Thread: Consecutive integers?

  1. #1

    Thread Starter
    Dazed Member
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    Consecutive integers?

    Ive been working out of a high school math book that my dad grabbed for me(since he is a custodian in a high school) and im up to the chapter involving equations with consecutive integers.
    It dosen't seem that bad but i am not sure if the result of this equation is correct. it should be but i just want to double check.

    The ages in years of three brothers are consecutive. The sum of their ages is 39 decreased by the age of the youngest. What are their ages?

    I come up with the following....

    a + (a + 1) + (a + 2) = 39 - a

    3a + 3 = 39 - a

    3a + a + 3 = 39 - a + a

    4a + 3 = 39

    4a + 3 - 3 = 39 - 3

    4a/4 = 36/4

    a = 9

    So their ages must be 9,10,11

    Does this look correct?

  2. #2
    Fanatic Member
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    Yes, it appears correct.
    Digital-X-Treme
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    [VBCODE]Debug.Print Round(((1097) - ((55 ^ 5 + 311 ^ 3 - 11 ^ 3) _
    / (68 ^ 5))) ^ (1 / 7), 13)[/VBCODE]

  3. #3

  4. #4
    sql_lall
    Guest

    Talking A little trick

    A handy trick when using consecutive integers is to have "a" as one of the middle terms.
    i.e. let there ages be a-1, a, a+1
    This way, if you have to use their product, it is
    =a*[(a+1)*(a-1)] -diff. of squares
    =a(a^2-1)
    =a^3 - a

    This is much easier to work with than:
    =a*(a+1)*(a+2)
    = a*(a^2+3a+2)
    = a^3+3a^2+2a

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