Question 995573
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Let us consider an arbitrary three-digit number  "abc". 


This three-digit number  "abc"  is   100*a + 10*b + c.   (It is how our decimal system works). 


Now, regroup it:   100*a + 10*b + c = (99*a+a) + (9*b+b) + c = (99*a + 9*b) + (a + b + c). 


The additive  (99*a + 9*b)  is divisible by  9.  Therefore,  the divisibility by  9  of our original number  "abc"  depends on and is determined solely by the last additive  (a + b + c),  which is the sum of the digits of the number  "abc". 



This  "divisibility by 9"  rule is true not only for  3-digit numbers:  it is true for all integer numbers.


For the proof see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-9-rule.lesson>Divisibility by 9 rule</A>&nbsp; in this site.