Rule for Divisibility by 9A 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. ExamplesA.) 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.ProofThe 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 a First, we can state that x = a
_{0} + a_{1}×10
+ a_{2}×10^{2} + a_{3}×10^{3}... + a_{n}×10^{n}Next if we let s be the sum of its digits then s = a _{0} + a_{1} + a_{2} + a_{3} + ... + a_{n .}So x - s = (a _{0} - a_{0}) + (a_{1 }×_{ }10
- a_{1}) + (a_{2}×10^{2} - a_{2}) + ... +
(a_{n}×10^{n} - a_{n}) = a _{1}(10 - 1) + a_{2}(10^{2} -
1) + ... + a_{n}(10^{n} - 1).If we
let b x - s = a It follows that all numbers b Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa. |

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