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Math Help - solve

  1. #1
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    solve

    Hi !

    please forgive my english because I am from KUrdistan /Iraq.

    Thank you

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  2. #2
    Senior Member vincisonfire's Avatar
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    In a more simple way, what you wrote means
    (i) the sum of the digits of a number is divisible by 3 if and only if the number is divisible by 3.
    (ii) the sum of the digits of a number is divisible by 9 if and only if the number is divisible by 9.
    (i) Let a number divisible by 3 be  N = a_na_{n-1}...a_0 then  N = 0 \text{ mod } 3
    The number in front of the mod is the rest after dividing by the modulo number.
    We can write any number as a_n10^n+a_{n-1}10^{n-1}+ ... + a_0 .
    Remark that 10 = 1 \text{ mod } 3 .
    Therefore  10^n = 1 \text{ mod } 3 and  a10^n  = a\cdot 1=a \text{ mod } 3 .
    It follows a_n10^n+a_{n-1}10^{n-1}+ ... + a_0 = a_n+a_{n-1}+ ... + a_0 = 0 \text{ mod } 3
    And you have your answer that is 3|N \Longleftrightarrow 3| \sum a_i
    (ii) Since  10 = 1 \text{ mod } 9 you can draw the same conclusions.
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