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7 March, 13:29

I am a 5 digit number between 52,000 and 53,000. The sum of my digits is 21. My ones digit is a multiple of 3. My tens digit is equal to my hundreds digit.

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  1. 7 March, 13:51
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    I am 52446.

    Step-by-step explanation:

    Let the digits be a, b, c, d, and e.

    Then the number is abcde.

    We have five conditions:

    (1) a = 5

    (2) b = 2

    (3) a + b + c + d + e = 21

    (4) e = 3, 6, or 9

    (5) c = d

    From (3), 5 + 2 + c + d + e = 21

    7 + c + d + e = 21

    c + d + e = 14

    From (5), 2c + e = 14

    c = ½ (14 - e)

    If e = 3, c = ½ (14 - 3) = 11/2. Not possible.

    If e = 9, c = ½ (14 - 9) = 5/2. Not possible.

    If e = 6, c = ½ (14 - 6) = 8/2 = 4.

    The solution is

    a = 5, b = 2, c = 4, d = 4, e = 6 ⟶ 52446

    The number is 52446.
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