Prove that a number k Z is divisible by 9 if and only if th

Prove that a number k ? Z is divisible by 9 if and only if the sum of its digits is divisible by 9.

Solution

: Suppose we have n such that the sum of the digits of n is divisible by 9. Let a be the hundred’s digit of n, b the ten’s digit, and c the one’s digit. Then n = 100a+10b+c. Now suppose that the sum of the digits of n is divisible by 9. Then, a+b+c = 9k, k Z. Adding 99a+9b to both sides of the equation, we get 100a+10b+c = n = 9k+99a+9b = 9(k+11a+b). So n is divisible by 9. In this case the converse of the theorem is also true: If n is divisible by 9, the sum of its digits is divisible by 9, too. In other words, the sum of the digits of n is divisible by 9 if and only if 1 n is divisible by 9. In general, to prove P Q, you have to do two proofs: You must show that P = Q and then, separately, you must also show that Q = P.

We already proved above that if the sum of the digits of n is divisible by 9 then n is divisible by 9. So we only need to prove the converse. We use the same notation for the digits of n as we used in the previous proof: n is divisible by 9 = n = 9l,l Z = 100a+10b+c = 9l = 99a+9b+(a+b+c) = 9l = a+b+c = 9l 99a9b = a+b+c = 9(l 11ab) = a+b+c = 9k, k = l 11ab Z = a+b+c is divisible by 9. Note that, in this simple example, the proof of Q = P is essentially the same as the proof of Q = P “run backwards.” In such a case, you may be able to get away with proving both of the implications at the same time (using the symbol at every step of the proof). However, be very careful if you do this: for the proof to be legitimate, the steps have to make just as much sense backwards as forwards! (Go back and read the last proof again, starting with the last line and ending with the first, and convince yourself that it also works backwards.) To avoid potential pitfalls, it is recommended that you always prove a statement of the form P Q using two separate proofs. This will in any case be necessary in more interesting examples, where the proofs of P = Q and of Q = P might look very different