SOLUTION: if the sum of the digits of a 3 digit number is divisible by 9, prove that the 3 digit number is also divisible by 9

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Question 995573: if the sum of the digits of a 3 digit number is divisible by 9, prove that the 3 digit number is also divisible by 9
Answer by ikleyn(52781) About Me  (Show Source):
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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  Divisibility by 9 rule  in this site.