be one.

Does anyone use casting out nines anymore?

]]>and

is theIf

is the smallest integer such that , , then , , and , .**Thus the sum of digits is **

Example:

This is equal to

(8 is the no. of digits ofJulianthemath that is the correct way to start, there are three patterns to be found depending on the initial three digit value you subtract.

]]>3+3+3=9

6+6+7=19

9+19=28

Don't know if correct, jackme12. Reply me if correct or not. ]]>

If c and c' represent the two complementary numbers then casting out nines on these two will

always produce tens complements. For example COD(255)=3 and COD(745)=7;

COD(111)=3 and COD(889)=7; COD(100)=1 and COD(900)=9 (9 is equivalent to 0 when casting

out nines.); COD(257)=5 and COD(743)=5. ]]>

Let be a three-digit number.

Then

.If

, we write .Then the sum of digits is

.If

, we have .If

, the sum of digits is .If

, then and the sum of digits is .So the sums are

(i) 28 if

(ii) 19 if and

(iii) 10 if and ]]>

For example 1000 - 255 = 745

2+5+5 = 12

7+4+5 = 16

12+16 = 28

Bonus points for multiple patterns,

Bonus bonus points for anyone who can find a direct proof.