Date: Tue, 16 Feb 1999 19:20:16 -0600 (CST) Name: Razzi
Who is asking: Student
Question: I've been having a hard time trying to solve the following problem and I was wondering if you could help me. For any positive integer a let S(a) be the sum of its digits. Prove that a is divisible by 9 if and only if there exist a positive integer b such that S(a)=S(b)=S(a+b).
Hi
909900 <-- b Now add a and b ignoring the carries to get c. Continuing with the example
909900 <-- b _______ 2208840 <-- c - The digit 9 does not appear in c.
- If B has a 9 in some position then the digit of c in that position is one less than the digit of a in that position.
- In every other position the digits of a and c are the same.
For the converse suppose that S(a)=S(b)=S(a+b). We are going to use modular arithmetic but you could also use the procedure of casting out nines. The crutial fact here is that S(a+b) is congruent to S(a)+S(b) modulo 9. Thus, working modulo 9, S(a) is congruent to S(a)+S(b) which is S(a)+S(a)=2S(a). Thus S(a) is congruent to 2S(a) and hence S(a) is congruent to 0 modulo 9. Therefore a is divisible by 9.
Cheers, |

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