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21 November, 02:09

The some of three consecutive integers is 33 more than the least of the integers. Find the integers

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  1. 21 November, 02:35
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    Let the smallest number = x

    The next higher number would be x + 1

    The 3rd number would be x + 2

    The sum of the 3 numbers: x + x+1 + x+2

    33 more than the smallest number is written as x+33

    Now you have:

    x + x+1 + x+2 = x+33

    Simplify the left side:

    3x + 3 = x + 33

    Subtract 3 from each side:

    3x = x + 30

    Subtract 1x from each side:

    2x = 30

    Divide both sides by 2:

    X = 30/2

    x = 15

    The three numbers are 15, 16 and 17
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