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Cloud moderators

19/12/2017 at 17:42
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How many positive three-digit integers have one digit equal to the average of the other two digits?




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    Dao Trong Luan Coordinator 19/12/2017 at 18:36

    average = 1 --> sum 1st,3rd add to 2 --> 1 + 1, 2 + 0 --> 2 numbers
    average = 2 --> sum add to 4 --> 1 + 3, 2 + 2, 3 + 1, 4 + 0 --> 4 numbers
    average = 3 --> sum add to 6 --> 1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1, 6 + 0 --> 6
    average = 4 --> sum add to 8 --> 1 + 7, 2 + 6, 3 + 5, 4 + 4, 5 + 3, 6 + 2,............................ 7 + 1, 8 + 0 --> 8 numbers
    average = 5 --> sum add to 10 --> 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4,............................ 7 + 3, 8 + 2, 9 + 1 --> 9
    ave = 6 --> sum add to 12 --> 3 + 9, 4 + 8, 5 + 7, 6 + 6, 7 + 5, 8 + 4,
    ............................ 9 + 3 --> 7
    ave = 7 --> sum add to 14 --> 5 + 9, 6 + 8, 7 + 7, 8 + 6, 9 + 5 --> 5

    ave = 8 --> sum add to 16 --> 7 + 9, 8 + 8, 9 + 7 --> 3 numbers
    ave = 9 --> sum add to 18 --> 9 + 9 --> 1 number
    2 + 4 + 6 + 8 + 9 + 7 + 5 + 3 + 1
    appears to be the sum of the digits from 1 to 9
    sum digits = n(n+1)/2 = 9(10)/2 = 90/2 = 45 numbers
    111, 210
    123, 222, 321, 420
    135, 234, 333, 432, 531, 630
    147, 246, 345, 444, 543, 642, 741, 840
    159, 258, 357, 456, 555, 654, 753, 852, 951
    369, 468, 567, 666, 765, 864, 963
    579, 678, 777, 876, 975
    789, 888, 987
    999

    Answer: 45 numbers
     

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