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8 November, 12:19

Find all sets of three consecutive even integers whose sum is greater than 102 and less than 116

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  1. 8 November, 12:29
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    So three consecutive integers 2k, 2 (k+1),2 (k+2)

    there sum will be

    2k+2 (k+1) + 2 (k+2)

    2k+2k+2+2k+4

    6k+6

    so setting the inequality

    102<6k+6<116

    subtracting 6 from all sides

    96<6k<110

    16
    since k is an integer

    the only integers between 16 and 18.3

    are 17 and 18

    so

    when k=17

    2 (17),2 (17+1),2 (17+2)

    34,36,38

    when k=18

    2 (18),2 (18+1),2 (18+2)

    36,38,40

    so to check are soultion we substitute

    34+36+38=108 which is between 102 and 116

    36+38+40=114 which is between 102 and 116

    so our soultion is correct

    the final answer will be

    {{36,38,40},{38,40,42}}
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