Rule for Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. For large numbers this rule can be applied again to the result. In addition, the final iteration will result in a 9.
A.) 2,880: 2 + 8 + 8 + 0 = 18, 1 + 8 = 9, so 9| 2,880.
B.) 3,564,213: 3+5+6+4+2+1+3=24, 2+4=6, so 9 does NOT divide 3,564,213.
The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.
For any integer x written as an· · · a3a2 a1a0 we will prove that if 9|(a0 + a1+ a2+ a3 ... + an), then 9|x and vice versa.
First, we can state that
x = a0 + a1×10 + a2×102 + a3×103... + an×10n
Next if we let s be the sum of its digits then
s = a0 + a1 + a2 + a3 + ... + an .
x - s = (a0 - a0) + (a1 × 10 - a1) + (a2×102 - a2) + ... + (an×10n - an)
= a1(10 - 1) + a2(102 - 1) + ... + an(10n - 1).
If we let bk = 10k - 1, then bk = 9...9 (9 occurs k times) and bk =9(1…1) and we can rewrite the previous equation as
x - s = a1(b1)+ a2(b2)+ ... + an (bn)
It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk (which is x-s) is also divisible by 9.
Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa.
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