Given a number with digits of the form

  X_N ...X_4 X_3 X_2 X_1 X_0

If the sum of these digits is equal to a number that is divisible by 3 then the number is divisible by 3.  Proof follows

Starting with    X_1 X_0  with  X_0 which must be 0.3,6,9 then  

  X_1 + X_0 is said to be divisible by 3 thus  X_1 + X_0 = 3n_1   Where n_1 is an integer

  10X_1 + X_0     should be divisible by 3.

  Rearranging the terms of 

      X_1 + X_0 = 3n_1 

Yields

     X_1 = 3n_1 - X_0 

Substituting in 

    10X_1 + X_0  

Yields

    10( 3n_1 - X_0 ) + X_0  

Simplifying yields

    30n_1 - 9X_0    

And therefore it is a sufficient condition that if X_1 + X_0 be divisible by 3 then    X_1 X_0 is divisible by 3 and by extension   X_N ...X_4 X_3 X_2 X_1 X_0 is divisible by 3 if   X_N+ ...+X_4+ X_3+ X_2+ X_1+ X_0  is divisible by 3.

As you might guess the same applies to all powers of 3 with 9 being the most useful next integer.

 

Categories: Math

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